If K is a field and p is an irreducible polynomial over K, there is some splitting field F/K that contains all the roots of p. In fact the splitting field can be constructed by adjoining the roots of p in succession.
The roots of p(x) are called conjugates, and these conjugates are indistinguishable. There is no way to tell one from another. Like special relativity, there is no correct frame of reference. As a corollary, all the roots of p(x) have a common multiplicity. If there are 7 instances of u in p(x) there are 7 instances of v in p(x), and so on.
The irreducible polynomial p(x) over K is separable if its roots, in the splitting field F/K, are all distinct. In other words, the common multiplicity is 1. Each root appears once, and there are n different roots, all conjugate to each other.
The algebraic element u in F/K is separable if the irreducible polynomial associated with u is separable.
By this definition, u in K is always separable, having the irreducible polynomial xu.
The extension F/K is separable if every u in F is separable. Thus every separable extension is algebraic.
The field K is perfect if every algebraic extension of K is separable. We will show that Q and Z/p are perfect, which is why most fields are separable.
An inseparable polynomial is irreducible, but not separable. The common multiplicity is greater than 1. An inseparable element u in F/K is algebraic, but not separable. An inseparable extension is algebraic, but not separable.
You could study algebraic extensions for a long time and never run into an inseparable extension, But you'll find some here. Let K be a field with characteristic 0 and let p be an irreducible polynomial over K. Let F/K be the splitting field for p(x).
Suppose the root u appears more than once in p(x). In other words, the complete factorization of p(x) in the field F includes (xu)m, where the multiplicity m is at least 2.
Let q be the formal derivative of p, then take the gcd of q and p to find any repeated factors. Since xu appears in both q and p, the gcd is nontrivial, and p is not irreducible after all. This procedure always works when the characteristic of K is 0, because the derivative q(x) does not simply disappear. In contrast, the derivative of x7+3 drops to 0 mod 7.
In summary, a field with characteristic 0 is perfect. All its algebraic extensions are separable. K could be the rationals, or various algebraic extensions of the rationals, or the reals, or the complex numbers, or perhaps fields like Q(x,y,z), the rationals adjoin three indeterminants. If K is a finite field, and F/K is algebraic, then F/K is separable.
Let n = F1, whence F is cyclotomic, adjoining the nth roots of 1. In fact F is the roots of xn1, (along with 0). Since the formal derivative is nxn1, having no factors in common with xn1, all these roots are distinct. For any u in F, u and its conjugates are distinct, belonging to some irreducible factor of xn1. This makes u separable, and F separable. Every finite extension of K is separable, the algebraic closure of K is separable, and K is a perfect field. An algebraic extension F over K is galois iff it is separable and normal. This was proved earlier, but let's revisit the proof here.
If F/K is galois, build a polynomial with one copy of all the conjugates of u that live in F. Each automorphism c of F permutes these roots, and fixes any expression that is symmetric in these roots, such as their sum or product. This includes every coefficient of p. since the coefficients of p are fixed by all the automorphisms of F, they all live in K. Thus p is the irreducible polynomial of u, having precisely one copy of each conjugate of u. That makes F normal and separable.
Conversely, separable implies at least one other conjugate v distinct from u, and normal implies v is part of F. Some homomorphism moves u to v, and this extends to an automorphism of F (because F is normal), hence the automorphisms move every u not in K, and F is galois.
Every normal extension of a perfect field is galois. In this context, normal and galois are synonymous.
But there's more. Recall that F/K is normal if it is generated by all the roots of a set of polynomials. If each generator implies all its conjugates, then F becomes normal and every u in F implies all its conjugates. the generators drive everything else. This idea is powerful, and a bit surprising. You can review the proof here.
The same is true for separability. If E is generated by separable elements then E is separable across the board. To get us started, increase E up to F, so that F is generated by all the conjugates of all the generators. The conjugate of a separable element is separable, so there is no trouble here. Prove F is separable, and E, a subfield of F, is separable.
F has the advantage of being normal. F is the splitting field for a set of polynomials, and all those polynomials just happen to be separable. We're going to prove F is galois, and by the above, F becomes separable. The goal, then, is to prove F is galois.
As a warmup, let the conjugates of one separable polynomial generate F over K. In other words, F is the splitting field for an irreducible polynomial p(x).
A field isomorphism c, fixing K, maps u to some other conjugate v in p(x). This can be parlayed into an automorphism on F, compatible with c. The proof, in chapter 7, is rather technical, and will not be reproduced here. It asserts the existence of a compatible automorphism, but this time we're going to augment that proof with some bookkeeping, To count the number of automorphisms on F. This begins by counting the number of isomorphisms c mapping u to v. There are n choices for c, n conjugates of u, all present in F because F is normal  where n is the degree of p(x). At the same time, the dimension of K(u)/K is also n. So far the number of isomorphisms agrees with the dimension of the extension K(u) over K.
At this point I'm going to pause, so that you can review the earlier proof. Note how c is extended, if necessary, through one irreducible polynomial after another, and as we'll see, the dimension, at each step, always agrees with the number of choices for extending c. We're really proving F is galois by brute force, counting all the isomorphisms and finding it is equal to the dimension of F/K. I'll take one more step, and then you can take it from there.
As you recall, the isomorphism c maps u to v in n different ways, and K(u) maps onto K(v) accordingly. If v is not part of K(u) then v is adjoined to K(u) via another irreducible polynomial, q(x), a factor of p(x) over K(u). Let m be the degree of q(x). Thus K(u)(v) has dimension m over K(u). At the same time, c is extended by mapping v to w, where w is any root of c(q)(x), and this can be done in m different ways. Once again the number of choices for the isomorphism c agrees with the dimension of the extension K(u)(v) over K(u).
If K(u,v) = K(v,w) then c becomes an automorphism, otherwise on we go. At each step the field extends through a new irreducible polynomial, and the degree of this polynomial, call it d, becomes the dimension of the next extension, as well as the number of choices for extending the isomorphiism c. The dimension over K and the isomorphism count are both multiplied by d. This ends when c becomes an automorphism on L, a subfield of F. At this point the number of automorphisms on L agrees with the dimension of L, and L is galois over K.
If L = F we are done; but if not then find another root of p, call it u, in the sense that we are starting over. Find the irreducible polynomial q(x) containing u, where q is a factor of p(x) over L, and map u to v where v is any root of c(q)(x). What happens? the dimension of L(u) over L equals the degree of q(x) equals the number of choices for extending c. If L(u) = L(v) then c becomes an automorphism on L(u), otherwise extend the field to L(u)(v) and map v to one of several roots, commensurate with the degree of the polynomial with root v and coefficients in L(u). Continue this process until once again c becomes an automorphism on a larger subfield of F.
Eventually c becomes an automorphism on F. The number of possible automorphisms agrees with the dimension of F, and F is galois.
If F is a splitting field for two polynomials, adjoin the roots of the first polynomial to build a normal, galois subfield somewhere between K and F, then adjoin the roots of the second polynomial, or perhaps the roots of an irreducible factor of the second polynomial if that polynomial factors over F. As the second polynomial rolls in, the number of automorphisms increases as the dimension increases. This continues through any finite set of polynomials.
Finally let E be generated by an infinite set of separable elements. Suppose E contains an inseparable element u. Since u is built using a finite set of generators, u belongs to a finite extension within E. focus on these generators and their conjugates, building a finite subfield F within E. F is now finite, and normal, and galois via the previous bookkeeping proof. Thus F is separable over K. yet F still contains u, which is inseparable. This is a contradiction, hence any field generated by separable elements is separable. The separable elements in a field extension F/K form a separable subfield S. This is a corollary of the previous theorem. If u and v are separable then K(u,v) is a separable extension, u+v is separable, u*v is separable, and u/v is separable. The separable elements form a field within F, the separable subfield of F. If F is finite over K, F is a simple extension K(u) iff F has a finite number of intermediate fields. Here u is the "primitive element".
This theorem is true if K is a finite field, because F/K is galois, with a finite number of subfields, and F is also generated by a root of an irreducible factor of xm1, where m is F1. So assume K is infinite.
Let F/K have a finite number of intermediate fields. Choose u in F so that the dimension of K(u) over K is maximal. Suppose this remains a proper subfield of F, and select any v from FK(u). Consider all intermediate fields K(u+av) where a is taken from K. Since K is infinite, there must be a distinct pair a and b such that K(u+av) = K(u+bv). It follows that (u+av)(u+bv) = (ab)v is in K(u+av). Divide by ab to get v, then multiply by a, and subtract from u+av to get u. Thus K(u+av) contains K(u), and also v, which contradicts the selection of u. Therefore there is some u with K(u) = F, and F is a simple extension.
Conversely, assume F = K(u). Let u satisfy the monic irreducible polynomial p(x). If E is an intermediate extension, factor p(x) in E, and find the irreducible factor q(x) such that q(u) = 0. Adjoin the coefficients of q to K to obtain a subfield of E. Call this subfield L, and suppose L is properly contained in E. Of course q remains irreducible over L, and q(u) is still 0. The dimension of L(u) over L is the degree of q. Yet, the dimension of E(u) over E is also the degree of q, and L(u) and E(u) are both F. Therefore E and L have the same dimension over K, and since E contains L, they are equal. The subfield E is generated by the coefficients of the irreducible polynomial q(x).
The polynomial p(x) factors uniquely in the ring F[x]. These prime factors clump together in finitely many ways to build intermediate polynomials q(x). Therefore there are finitely many intermediate fields.
As a corollary, all separable finite extensions are simple. Let F/K be such an extension and let E/K be its normal closure. Now E is normal, and finite over K, and E contains F. E/K is galois, with a finite galois group G. Intermediate extensions correspond to the subgroups of G, and there are finitely many of these. Therefore there are finitely many extensions between F and K, and F/K is a simple extension K(u) for some primitive element u. Let u be algebraic over the base field K, such that u is a root of the irreducible polynomial q(x). (I'm going to use p for the prime characteristic, so I'll use q for the irreducible polynomial.) If u is the only root of q(x), then u, and q, are purely inseparable. In other words, the common multiplicity is the same as the degree of q(x).
Technically, an element u in K is purely inseparable, with polynomial xu, but we usually require a multiplicity > 1, so that q(x) is nontrivial, and u does not belong to K. In other words, a purely inseparable element is inseparable as per the earlier definition, i.e. not separable.
The field extension F/K is purely inseparable if every u in FK is purely inseparable. Let's try to characterize such an extension.
Let u be purely inseparable over K. We already know K cannot have characteristic 0, nor can K be finite, since the algebraic extensions over these fields are all separable. Thus K includes a transcendental extension of Z/p.
If q(u) = 0, then q is (xu)e for some exponent e. (Remember that  really means + when p = 2.) Here e is the common multiplicity of the single root u, and the degree of q. Write e as mpr, where p does not divide m. Raise xu to the p power first, remembering that (a+b)p = ap+bp. Do this r times for pr. Hence q can be rewritten as x to the pr  u to the pr, all raised to the m.
Think of q as (ab)m and apply the binomial theorem. Consider the term mbam1. (Remember that a is a power of x and b is a power of u.) Since q(x) has coefficients in K, mb lies in K, and b lies in K. So ab represents a polynomial that is already in K. If q is irreducible, then m = 1, and q is xu raised to some power of p. Thus u to the pr lies in K. Here is an illustration with p = 5. The powers of u form a basis for F/K. 

Conversely, let r be the least exponent such that u to the pr lies in K. Build q as above. In other words, q(x) is xu raised to the pr. If q is not irreducible then (xu)e is a proper factor of q, with coefficients in K. We showed above that e has to be a power of p. However, r is minimal with u to the pr in K; therefore q is irreducible.
In summary, u is purely inseparable if it is the pr root of v for some v in K, and F is a purely inseparable extension if every u in F has some r such that u to the pr lies in K.
Yes this really happens: let u5 = t, over the transcendental field (Z/5)(t).
It is interesting to watch this fail when K is finite. The frobenius homomorphism xp is an automorphism that can be reversed. The pth root of anything in K is already in K. There are no purely inseparable elements over a finite field. However, when K is transcendental the map xp is not onto, not an automorphism, and purely inseparable extensions are possible.
Another characterization is based on formal derivatives. Let u be an inseparable element over the base field K. Let q be the irreducible polynomial with root u. Thus u has multiplicity > 1 in q. Let r be the derivative of q. Compute gcd(q,r) to find all the repeated factors. We know there are some, since u is inseparable. However, a gcd would divide q, whence q could not be irreducible. This apparent contradiction is resolved only when r drops to 0 in the field K. The derivative of q is 0, so all the exponents on the terms of q(x) are divisible by p. Divide these exponents by p and replace u with up, so that the new u is a root of the new q, whose degree has been divided by p. Repeat this process as long as u remains inseparable. In conclusion, you can raise an inseparable element to the pr power, for some r, and get a separable element.
With this in mind, a purely inseparable extension is an algebraic extension with all its separable elements in K. One direction is obvious, so assume the separable elements are all in K. Every u in FK is inseparable, and by the above lemma, you can raise u to a power of p and find a separable element, which lies in K. Hence every u satisfies u to the pr in K for some r, and F is purely inseparable.
Examples are good, so let's build an inseparable element that is not purely inseparable. Once again let K be the rational functions in t with coefficients mod 5. Let q(x) = x10  2tx5 + t22.
Let u be the fifth root of t, and let c be the square root of 2 in the finite field of order 25. Raise u+c to the fifth power and get t+4c, or tc. Raise uc to the fifth and get t+c. Multiply these together and get t22. Therefore the unique factorization of q is (xuc)5 * (xu+c)5.
If q is not irreducible then some, but not all, of these factors clump together to give a product whose coefficients lie in K. Put four factors together and get u4, along with lesser powers of u; this cannot lie in K. Similarly three or two factors cannot clump together. The only option is 5 and 5. At the same time, the product of 5 factors leaves ±c as a constant, which is not in K. Therefore q is irreducible.
There are two distinct roots, u+c and uc, with a common multiplicity of 5. Raise either of these roots to the fifth power and get tc and t+c, which are separable over K, but not part of K. thus u+c and uc are inseparable elements, but not purely inseparable, and q is an inseparable polynomial, but not purely inseparable.
Finally, use the characterization of an inseparable element to prove the separable subfield S of F/K is closed. Let u be an element that is separable over S, and suppose u is not separable over K. Raise u to a power of p and find a separable element in S. This makes u purely inseparable over S, which is a contradiction. The separable elements over S are already in S, and S is closed. When raising u±v, u*v, or u/v to the p power, raise u and v to the p power first, then apply the arithmetic operator. This is the frobenius homomorphism again. Therefore, if u and v are purely inseparable, there is a sufficiently large r that shows u+v is purely inseparable. The same for u*v and u/v, hence the field generated by purely inseparable elements is purely inseparable. Also, the set of purely inseparable elements in the extension F/K forms a field, a purely inseparable extension of K inside F.
If F is purely inseparable over K, and u is purely inseparable over F, then u is the root of something in F, which is the root of something in K, hence u is purely inseparable over K. The composition of purely inseparable extensions is purely inseparable.
If K is countable, such as (Z/p)(x,y,z), (polynomials in x y and z with coefficients mod p), adjoin the pth roots of elements in K, one by one, and repeat, to create a complete purely inseparable extension of K. If K is uncountable, the complete purely inseparable extension exists, and is a subset of the closure of K. As such, it is unique up to isomorphism.
Given F/K algebraic, you have a choice;
build the separable subfield S, or the purely inseparable subfield P.
These both exist in F, and are disjoint (other than K).
Bring in S first, if you like,
whence all the separable elements of F lie in S, and as shown above, F/S is purely inseparable.
Every algebraic extension is the composition of a separable extension followed by a purely inseparable extension.
If you prefer, bring in P first, and then suppose u, outside of P, is inseparable over P. Raise u to some power to find a separable element in P, but all the separable elements live in K. Thus u is purely inseparable after all, and is part of P. Everything over P is separable. Every algebraic extension is the composition of a purely inseparable extension followed by a separable extension. 

By definition, the dimension of S/K is the separable degree, and the dimension of P/K is the inseparable degree. It would probably be better to call these separable and inseparable dimensions, but that is not the established terminology. Here is a finite field extension with infinitely many subfields inside. We are placing infinitely many shelves into a finite cabinet, which seems impossible, but it's not.
By the primitive element theorem, the extension has to be inseparable. Let K = (Z/p)(x,y) for any prime p, say 5. All fifth powers have exponents divisible by 5, so x has no fifth root in K. Let u be the fifth root of x, and let v be the fifth root of y. Let F = K(u,v). Since F is generated by purely inseparable elements it is a purely inseparable extension over K.
Consider the fifth powers of all the elements in K(u). K(u) really looks like polynomials in u and y with coefficients mod 5. You can take the fifth power by multiplying each exponent by 5. That is the frobenius homomorphism. So u → x, u2 → x2, and so on. At the same time, y → y5, and all exponents on y are divisible by 5. This means K(u) does not include v, since v5 = y, and y is not one of the fifth powers. The roots u and v are independent.
v has dimension 5 over K, but does v still have dimension 5 over K(u)? It's dimension is greater than 1, not being in K(u), and at most 5, since it is the fifth root of y. And v is purely inseparable over K(u), so its dimension is some power of 5. That gives v a dimension of 5 over K(u), and F/K has dimension 25.
Suppose K(w) = K(u,v), so that w, purely inseparable, satisfies w25 ∈ K. Yet everything in K(u,v), when raise to the fifth power, lies in K, so w5 lies in K, and w does not have order 25 after all. The extension F/K is not simple. There are then infinitely many intermediate extensions, as illustrated below.
Every intermediate field is purely inseparable with dimension 5. The fields K(u) and K(v) are two examples. For any integer j, consider the extension K(u+vxj). This is the fifth root of x+yx5j. All such extensions are purely inseparable with dimension 5. Suppose K(u+vxi) also contains u+vxj. It then contains the difference (xixj)v, and hence v, and hence u. The extension is F, and cannot have dimension 5, which is a contradiction. All such extensions are distinct, and F/K has an infinite number of intermediate fields. The field of complex numbers, denoted C, is algebraically closed. Every polynomial with complex coefficients has a complex root, thus roots can be extracted one by one until the polynomial splits. This is the fundamental theorem of algebra.
If you have studied complex analysis, then you have probably seen the proof based on Liouville's theorem. Here is another, based on separable fields and galois theory. Of course there is a little bit of continuity involved; you can't prove a theorem about a complete metric space without a touch of δ ε somewhere under the covers.
The intermediate value theorem provides a positive square root for every positive real number, and at least one root to any odd degree polynomial in the reals as x moves from ∞ to +∞. Therefore every irreducible polynomial in the reals has even degree.
The existence of real square roots implies a complex square root for every point in the complex plane. This follows from Demoivre's Formula. Thus there is no extension of C with dimension 2.
To show C is closed it is sufficient to show there are no finite dimensional extensions of C. If E is such an extension it is also a finite separable extension over R. Separable, because the characteristic of R is 0. Take the normal closure to get a galois extension F, which is still finite.
Since C/R has dimension 2, the galois group G, for F over R, has even order. The first sylow theorem provides a subgroup H whose order is a power of 2, while the index of H is odd. Let L be the extension of R that is fixed by H. This takes place inside a galois extension, so the dimension of L/R is the index of H in G, which is odd. Since L/R is finite and separable, apply the primitive element theorem and write L = R(u). The irreducible polynomial associated with u has odd degree, yet an odd degree polynomial has a root in R. The polynomial must have degree 1, and L = R. Thus G = H, and G is a power of 2.
If G = Z/2 then F = C. For larger G, F properly contains C, and F/C is galois.
Let G be the galois group of F/C, and let H be a subgroup of G with index 2. The field fixed by H is a quadratic extension of C, which is impossible. Therefore C is closed.
It is not a surprise then that C is the only proper field extension of R. Let F be a proper extension of R, generated by u. Let p(x) have root u. We know that p is irreducible in R, yet it splits in C. Let v be a root of p(x) taken from the extension C/R. Build a field isomorphism from R(u) onto R(v), both satisfying p(x). Since v is outside of R, R(v) = C. Thus we have an isomorphism from R(u) onto all of C. Now R(u) is closed, and F is algebraic over R(u), hence F = R(u), which is isomorphic to C. The only extension of the reals is the complex numbers.
As a corollary, the only irreducible polynomials over the reals are quadratic. Since C is separable, the two roots are distinct. You know them better as conjugates in the complex plane.
x2 + 4 = (x+2i) * (x2i)
If L is a finite galois extension of K, and M is an arbitrary extension,
L+M over M is finite and galois,
with galois group isomorphic to the group of L over L∩M.
Here L+M is the compositum, the smallest field containing L and M,
and L∩M is their intersection, which I will call E.
This theorem is trivially true if M contains L, so assume M does not contain L.
Since L is finite and galois, apply the primitive element theorem and let L = K(u). Being normal, L contains all the conjugates of u. L is also galois over E, with L = E(u). L has all the conjugates of u, relative to its irreducible polynomial over K, and it certainly has all the conjugates of u relative to its irreducible polynomial over E, which is a factor of its polynomial over K. Start with M and adjoin u. The result contains M and L, and is L+M. L+M contains all the conjugates of u, which are distinct, hence L+M is normal, and separable, and galois over M. Note that M/K need not be a finite extension; it doesn't even have to be algebraic. Build a map from the M automorphisms of L+M onto the E automorphisms of L. How do we do this? Given an automorphism of L+M that fixes M, restrict this automorphism to L. L is normal, so the resulting homomorphism on L has to keep L within L, hence an automorphism on L. Of course it automatically fixes E. This map is a group homomorphism. In other words, two automorphisms on L+M can be composed, then restricted to L, and the result is the composition of the two restricted automorphisms on L. 

If an automorphism fixes both M and L, it fixes L+M. The kernel of the group homomorphism is trivial. The galois group of L+M over M embeds in the galois group of L over E.
Surjective is the only tricky part. Start with an automorphism of L that maps u to v, where u and v are conjugates of the same irreducible polynomial p(x), having coefficients in E. None of the roots of p(x) lie in E, or in M. Suppose p factors over M. Let g(x) be one of these factors. Write g as the product of binomials xuj, for various conjugates of u. The coefficients of g lie in L. They also lie in M, hence they lie in E. Therefore p factors over E. Since p is irreducible, this is a contradiction. Thus p remains irreducible over M.
Now u and v are conjugates of the same irreducible polynomial over M. This means M(u) and M(v) are isomorphic field extensions of M. Mapping u to v induces a field isomorphism from M(u) onto M(v), which extends to an automorphism on L+M (L+M being normal over M), which then restricts to a field isomorphism from E(u) onto E(v), which is the desired automorphism of L. The map is onto, and the galois groups are isomorphic.
Since the order of the galois group equals the dimension of the galois extension, the dimension of L+M over M equals the dimension of L over E. This can also be seen by the degree of p(x), which is the same over E and over M.
Two subgroups within a larger group are disjoint if they have only the identity element in common.
This is a bit odd, since disjoint sets have nothing in common  but you'll get use to it.
Two ring extensions of R in a larger ring S are disjoint if they have only R in common. To field extensions L/K and M/K are disjoint if they have only K in common. This is standard terminology, but here comes a new adverb. If L/K is a finite field extension and M/K is an arbitrary field extension, L and M are linearly disjoint if (L+M)/M has the same dimension as L/K. Let u be an element in LK and adjoin u to M. The irreducible polynomial that establishes the dimension of K(u)/K may factor over M. In other words, M(u)/M could have a smaller dimension than K(u)/K. Its dimension can't be any larger. Continue adjoining the generators of L/K, and the dimension of L+M over M is bounded by the dimension of L over K. If L and M are not disjoint, select the first u from M intersect L. Now K(u) is a proper field extension of K, but M(u) is trivial. From this point on, the dimension of the extension of M is less than the dimension of the corresponding extension of K. L and M cannot be linearly disjoint unless they are disjoint. 

Assume the dimensions of L/K and M/K are relatively prime. The dimension of L+M over K is divisible by both these dimensions, and is at least their product. However, the dimension of L+M over M is at most the dimension of L over K, hence the dimension of L+M over K is at most the product of the two dimensions. This forces dimensional equality, and L and M are linearly disjoint.
Assume L is finite galois, and apply a base change, as described in the previous section. The dimension of L+M over M equals the dimension of L over L∩M. If L is finite galois, the extensions are disjoint iff they are linearly disjoint.
Continue the above, where L and M are linearly disjoint, and L is finite and galois. Consider a subfield of L. For convenience, write it as a simple extension K(u). A subgroup of the galois group of L over K fixes K(u), and a subgroup of the galois group of L+M over M fixes M(u). These galois groups are isomorphic. Review the isomorphism described in the previous section, and note that the subgroups fixing M(u) and K(u) correspond. They are the same size, hence M(u)/M has the same dimension as K(u)/K. Every subfield of L is linearly disjoint with M.
Now let's turn this around. If L/K and M/K are field extensions, and L is a finite separable extension, and M is disjoint from the normal closure of L, then L and M are linearly disjoint. This is true for the normal closure of L, hence it is true for the subfield L.
If we're careful, we can remove the word separable from the above. Proceed by induction on the inseparable degree. The theorem holds when L is separable, having inseparable degree 1. Let N be the normal closure of L, disjoint from M. Let u be a purely inseparable element in L, with order ps. Let q be the irreducible polynomial xu to the ps. Suppose q has a proper factor g(x) over M. The reasoning is the same as that shown in the previous section; the coefficients of g lie in L, and in M, hence in K. Thus M(u)/M has the same dimension ps.
Suppose N contains z, not in K(u), and M(u) also contains z. Write z as a linear combination of powers of u with coefficients from M.
z = m0 + m1u + m2u2 + …
Since N contains u and z it contains m0. N and M intersect in K, so replace m0 with k0. Subtract k0 from z, divide by u, and now N contains m1. Replace m1 with k1, and march down all the coefficients, until z actually lives in K(u). Thus N and M(u) have only K(u) as their intersection. By induction, L and M(u) are linearly disjoint over K(u). Fold in the dimension of K(u) over K, which is the same as M(u) over M, and L and M are linearly disjoint. As long as L is finite, with normal closure disjoint from M, we're good to go.
In a bit of role reversal, let M be finite and let L be arbitrary, with the normal closure of L disjoint from M. A purely inseparable element of M can be tacked on top of L, just as a purely inseparable element of L was tacked on top of M in the previous paragraph. The reasoning is the same, thus it is sufficient to prove these extensions are linearly disjoint when M is separable.
Let M be the simple extension K(u). Let u have irreducible polynomial q(x), and try to factor q over L(u). Suppose q(x) = g(x) * h(x). Now g is a product of linear terms, using some of the conjugates of u. These conjugates of u are algebraic over K, and the same holds for the coefficients of g, and of h. The coefficients form a finite set of algebraic elements over K, contained in L. Let these coefficients generate the intermediate finite extension L′ inside L. Now q factors into g*h over the field L′. However, L′ is finite, and its normal closure is disjoint from M, hence the extensions are linearly disjoint. Adjoining u to K or to L′ yields the same dimension, hence q(x) does not factor over L′, nor does it factor over L.
In summary, L and M are linearly disjoint if M is disjoint from the normal closure of L, and at least one of the extensions is finite. Let L and M be disjoint, finite, and galois. By base change, the galois group of L+M over M is that of L over K. These automorphisms fix M, and run independently of, and commute with, the automorphisms of M. Together they form a group whose order equals the dimension of L+M over K. Thus the galois group of L+M over K is the direct product of the galois groups of L and of M.
Conversely, if the galois group of F is a direct product, take the two fixed fields L and M, fixed by the normal summands of G. These galois extensions are disjoint, since their intersection is fixed by all of G. Thus they are linearly disjoint. The field L+M lives inside F, and has the proper dimension, hence it is equal to F.
This extends to a finite direct product of groups. If G = H1*H2*H3, then F is the join of the subfield fixed by H1, and the subfield fixed by H2*H3. Once again these two subfields are disjoint and linearly disjoint. The first subfield has galois group H2*H3, and can be separated into two subfields under H1, one fixed by H2 and one fixed by H3. Thus the three pairwise disjoint subfields that combine to form F are fixed by H1*H2, H1*H3, and H2*H3. Given an algebraic extension F/K, you can adjoin P first, the purely inseparable subfield of F, or you can start with S, the separable subfield of F. The inseparable degree of F is the dimension P/K, but is this well defined? What if you adjoin S first, then adjoin the rest of F, which is purely inseparable over S. Is the inseparable degree the same?
P and S are disjoint, and P is its own normal closure. If P or S is finite, then the two extensions are linearly disjoint. Let E = P+S. The dimension of P is the same whether P stands alone or P is put on top of S. You can adjoin the generators of P, or S, in either order. We only need show E = F.
Suppose there is a z in F that is not in E. If z is separable over E, then adjoin S first, then P. Everything above S is purely inseparable over S. (This was described in an earlier section.) Since z is purely inseparable over S it is purely inseparable over E. The order of z may be a lesser power of p, as its polynomial factors over E, but still z is purely inseparable. This is a contradiction.
On the other hand, suppose z is inseparable over E. Adjoin P first, then S. Everything above P is separable over P, thus z satisfies some irreducible polynomial q(x) over P, with distinct roots. Perhaps q factors further over E, but still the irreducible piece of q that contains z has distinct roots, and z is separable over E. This is a contradiction.
Since z cannot float between E and F, E = F. Every algebraic extension is the compositum of its separable subfield and purely inseparable subfield, which are linearly disjoint. Combine the generators of the two subfields to build F. Furthermore, the inseparable degree is the dimension of the purely inseparable extension, whether the purely inseparable generators are at the bottom or the top.
In fact, you can adjoin separable and purely inseparable layers in alternation, and the product of the individual inseparable degrees is the inseparable degree of F. Swap the first two layers, if need be, so that P is on the bottom. Think of P as the new base field. Separable elements follow from there, and then perhaps another layer of purely inseparable elements. Let this be a field extension of P. Swap the separable and inseparable layers, so that the purely inseparable elements are on the bottom, then merge these with P to make P larger. Repeat this process until all the inseparables are on the bottom and all the separables are on top. Dimensions are preserved along the way, thus the inseparable degree is strictly a function of F over K.
In a tower of extensions, the inseparable degrees multiply. Consider F/E/K. Put the purely inseparable subfield at the top of E/K, and at the bottom of F/E. Now F/K is a tower of 4 extensions: the separable subfield of E, the inseparable subfield of E, the inseparable subfield of F, and the separable subfield of F. Combine the second and third extensions, which are both pure inseparable, and their dimensions multiply. Swap this with the separable subfield below if you like, to put all the purely inseparable elements back on the bottom. Thus the inseparable degree of F/K is the inseparable degree of E/K times the inseparable degree of F/E. The same result holds for the separable degree. 

This section presents another definition of norm and trace in terms of field extensions, which is consistent with the two definitions sited above. Thus we will have three equivalent definitions for trace and norm, at least for certain matrices.
I will allow the extension to be inseparable, which some textbooks do not address. That's why this section is here, under separable fields, instead of somewhere else.
If you haven't had enough of norm and trace, there are even more general definitions involving projective modules, ideals or fractional ideals in a ring, and so on, but let's not go there today.
Let F be a finite extension of K. The norm of an element u is the product of the images of u under all K isomorphisms from F into the algebraic closure of K. Actually it is enough to map F into the normal closure of F, since u isn't going to go anywhere outside the normal closure of F. This is sometimes more convenient, and doesn't call upon that pesky axiom of choice. 

Notice that the definition requires a finite extension and an element u in that extension. Change either u or the extension and the norm will change.
Let's look at a simple example. Consider x32 = 0. Let u be the real cube root of 2 and adjoin u to the rationals. This is a field extension of dimension 3. That's certainly finite, so on we go.
There are no "other" cube roots of u in F, so it looks like there is nothing to do; but norm requires us to map u into its normal closure. Adjoin the cube roots of 1, and the complex cube roots of 2 appear. Map u to any of these three cube roots of 2. This uniquely determines the map from F into its normal closure. Thus there are three isomorphisms of F, and three images of u. Multiply these three images of u together to get 2.
The trace is the sum of the images of u. The cube roots of 2 are placed symmetrically about a circle in the complex plane, hence their sum is 0. The trace is 0.
If the extension F/K is inseparable, the norm must be raised to the inseparable degree, and the trace is multiplied by the inseparable degree. This is a bit of a fudge factor, but it makes things work out properly.
If F/K is inseparable, the inseparable degree is a power of p, where p is the characteristic of K. When this is multiplied by the trace, the result is 0. The trace is always 0 in an inseparable extension.
Given two elements u and v in an extension F/K, Show that norm(uv) = norm(u)*norm(v), and trace(u+v) = trace(u)+trace(v). This because each isomorphism of F into its normal closure respects addition and multiplication. And if F/K is inseparable, the inseparable degree just goes along for the ride.
If F/K is galois then F is already normal and separable, and the isomorphisms of F are all automorphisms of F. The norm of u, written u, is the product of the images of u, which are all conjugates of u living in F. This is how norm was introduced in chapter 1. Recall the gaussian integers. There are two automorphisms, the trivial automorphism that leaves things where they are, and complex conjugation that maps i to i. The norm of u is then u times u, the product of the two images of u. The eisenstein integers have the same norm, while Cyclo 8 has a 4part norm, corresponding to the 4 automorphisms of the eighth roots of 1 over Q. These examples are all galois, but F/K need not be galois, in which case we must map u to other conjugates of u that might lie outside of F.
To get things started, let u lie in K. By definition, every K isomorphism fixes K, and u. The trace is a multiple of u and the norm is a power of u.
Let L be the purely inseparable extension of K inside F. The isomorphisms cannot move an inseparable element, hence L is fixed. The K isomorphisms are the same as the L isomorphisms, so count the L isomorphisms instead.
Select an element v in F, but not in L, and build an isomorphism by mapping v onto one of its conjugates in the normal closure of F. This determines the isomorphism on all of L(v). Next adjoin w, and map w to one of its conjugates. This creates an isomorphism on L(v)(w). Continue this process until the isomorphism maps all of F onto something equivalent to F in the normal closure. The isomorphism is determined by the image of v, then the image of w, and so on up the line. At each step a conjugate is selected, a root of an irreducible polynomial. The number of roots always equals the degree of the irreducible polynomial, which is the dimension of the intermediate extension. This is the bookkeeping proof described in the Separable Subfield section above. By combinatorics, the number of isomorphisms is the product of the number of conjugates at each step, or the product of the dimensions, or the dimension of F/L.
The norm of u is the product of the images of u (which are all u) under all possible isomorphisms of F. The dimension of F/L counts the number of isomorphisms, hence u is raised to this power. Finally, bring in the inseparable degree. This is the dimension of L/K. Put this all together and u = un, where n is the dimension of F/K.
Similarly, trace(u) = n×u, where n is the dimension of F/K.
Now place u above L, whence u is separable. The isomorphisms of F map u to one of its conjugates, then v, then w, as before. The norm of u is the product of the conjugates of u, raised to the dimension of F/L(u), then raised to the dimension of L/K. This is the product of the conjugates of u, raised to the n/d, where n is the dimension of F/K and d is the number of distinct conjugates of u. 

Notice that this formula works even if u is in L. There is but one conjugate of u, and it is raised to the n.
The trace of u is the sum of the conjugates of u, times n/d, where n is the dimension of F/K and d is the number of distinct conjugates of u. This works whether u is in L or above L.
Once u is brought in, the rest of F simply raises the norm to a higher power. If E is a further extension of F, u in E is U in F raised to the dimension of E/F. Similarly, the trace of u is multiplied by the dimension of E/F.
Let N be the normal closure of F. Thus N/L is galois. The product of the conjugates of u does not change when the conjugates of u are permuted. This is exactly what happens under each automorphism of N. Thus the automorphisms of N fix the product of the conjugates of u, and this product lies in L. Raise this product to a higher power and the result still lies in L. Finally, raise this again to the inseparable degree, and U lies in K.
A similar proof shows the trace of u lies in K. As mentioned earlier, the trace is 0 if F is partly inseparable.
Let u be an element in E/F/K. Since the norm of u, relative to E/F, lies in F, it is reasonable to talk about the norm of the norm, from E to F and then from F down to K. In a tower of extensions, the norm of the norm is the norm, and the trace of the trace is the trace. I'll illustrate with the norm; the proof for trace is essentially the same.
Let the inseparable degree of F/K be i1, and let the inseparable degree of E/F be i2. As shown in the previous section, the inseparable degree of E/K is i1×i2, which I will call i3. The norm of u in E/K is now the product of all the images of u, under all the isomorphisms of E/K, raised to the i3.
Every isomorphism of E into its normal closure implies, by restriction to F, an isomorphism of F into its normal closure. The set of E isomorphisms forms a group G, and the set of E isomorphisms that fix F forms a subgroup H of G.
Let M be a map from F onto a field isomorphic to F, inside the normal closure of F. Premultiply M by any isomorphism in H and find another isomorphism that is identical to m when restricted to F. In other words, the E isomorphisms that restrict to M(F) are a left coset of H in G. There is such a coset, because m can be extended up to an isomorphism on E.
These groups are finite, and the cosets of H all have the same size. For any map M, the isomorphisms in the class M, i.e. the isomorphisms that look like M when restricted to F, correspond 1 for 1 with the elements of H.
Let u be any element in E and take its norm relative to E/K. We can multiply the images of u in any order we like. Consider the isomorphisms in H first. When multiplied together, these images give the norm of u with respect to E/F. (You may need to raise this product to i2 if E/F is inseparable.) Let v be the norm of u in E/F.
Next consider the isomorphisms in the coset determined by some map M. The product of the images of u, under this coset, is the image of v under M. This holds for every map M, i.e. every isomorphism of F. When these are multiplied together, the result is the product of the images of v, for every isomorphism of F, which is the norm of v, or the norm of the norm of u. (Again you might need to raise everything to the i1, but that just makes the combined exponent i3 as it should.) Therefore the norm of the norm is the norm.
How does all this connect to matrices? The connection is the irreducible polynomial of u. Let F/K be a simple separable extension with primitive element u and irreducible polynomial p(x). In this case the norm and trace can be calculated directly. The polynomial is the product of xr, where r ranges over all the conjugates of u, i.e. all the roots of p(x). The lead coefficient is 1, and the next coefficient is minus the sum of the roots. Hence the opposite of the second coefficient gives the trace. Similarly, the constant coefficient, multiplied by (1)n, gives the norm. Here n is the degree of p, or the dimension of F/K.
If u is purely inseparable, then the single root u appears n times. There is but one isomorphism, mapping u to u. Raise the single image u to the nth power, the inseparable degree, and once again the norm is the constant coefficient times (1)n.
That takes care of u separable and inseparable. The formula holds for any simple extension K(u) whose irreducible polynomial is known.
Let F = K(u), whence F is a K vector space of dimension n, whose basis is given by the powers of u. Multiplication by u implements a linear transformation on the K vector space. It maps 1 → u, u → u2, u2 → u3, and so on up to un1 → un, which becomes a linear combination of the lower powers of u according to p(x).
Build a matrix that implements multiplication by u. If M is the matrix, x*M (where x is a linear combination of powers of u) becomes ux. The first row of M is the image of 1, which is u. Thus the first row of M is [0,1,0,0,0,…]. The second row is the image of u, or [0,0,1,0,0,…]. This pattern continues, with ones just above the main diagonal. The last row is un, expressed as lower powers of u according to p(x).
What is the determinant of M? Apply det[2](M), the permutation definition, and only one permutation produces a nonzero product. Multiply the ones above the main diagonal by the lower left entry. Call this lower left entry c, hence the determinant is c times the parity of the permutation. The parity is even if n is odd, odd if n is even.
If u has irreducible polynomial p(x), then p has constant term c, as shown by the lower left entry of M. The norm of u, with irreducible polynomial p, is the constant coefficient of p times (1)n. Thus u = (1)n+1c, which agrees with the matrix definition.
Now return to the matrix M and consider its trace. This is the sum of the diagonal elements, which is the lower right entry, which is minus the second coefficient in the irreducible polynomial p, which is the trace according to p(x). Once again the two definitions agree.
Switch to another basis b for F as a K vector space. Let Q be the matrix that represents each bi as a linear combination of powers of u. Multiplication by u is now QM/Q. Yet this has the same trace and norm as M. For any basis b of F over K, the matrix that implements multiplication by u has a norm that coincides with the product of the conjugates of u, and a trace that coincides with the sum of the conjugates of u.
Once the norm of u is computed with respect to K(u), u can be computed relative to a larger field E. Raise the norm to the jth power, where j is the dimension of E over K(u). This was described earlier.
How does this look when u acts on all of E? Does the norm still agree with the determinant of the larger matrix? Again, any basis will do  so select a convenient one. Start with K(u), and let the powers of u build a basis for K(u)/K. Then adjoin other elements, building a basis v1 through vj for E/K(u). The cross product of these basis elements, the powers of u times v1 through vj, builds a basis for E/K. Now consider multiplication by u. This acts on K(u), and modifies the coefficients on v1 through vj. By unique representation, this must be the action of u on E. In other words, u does not move v1 to v2, or anything inconvenient like that. The matrix that implements multiplication by u is block diagonal, with j blocks for v1 through vj. The determinant of each block is the norm of u in K(u)/K, as described above. There are j blocks, so the determinant of the entire matrix is the norm raised to the j, which is the norm of u in E/K. Similarly, the trace is multiplied by j.
For any element u in a finite extension, the algebraic definitions of trace and norm agree with the matrix definitions, where the matrix implements multiplication by u.
In this example, the extension is K(u,v), where u satisfies x52x42x27x9, and v is anything quadratic.
0  1  0  0  0  0  0  0  0  0 
0  0  1  0  0  0  0  0  0  0 
0  0  0  1  0  0  0  0  0  0 
0  0  0  0  1  0  0  0  0  0 
9  7  2  0  2  0  0  0  0  0 
0  0  0  0  0  0  1  0  0  0 
0  0  0  0  0  0  0  1  0  0 
0  0  0  0  0  0  0  0  1  0 
0  0  0  0  0  0  0  0  0  1 
0  0  0  0  0  9  7  2  0  2 
Let R be a ufd and let K be its fraction field. Typically R is the integers and K is the rationals. Adjoin y to R, where y satisfies some irreducible monic polynomial p(x) over R. By Gauss' lemma, p is irreducible over K. Thus K[y] becomes a field, in fact the fraction field of R[y].
If R[y] happens to be another ufd you can do this all over again. Let R[y,z] build a new ring, with fraction field K[y,z]. Assume this has been done finitely many times to produce a ring S with fraction field F. Typically it is done once, adjoin y, and S = R[y].
The norm is a multiplicative homomorphism from F into K; restrict this map to S and find a useful homomorphism from S into R.
All this notation may seem rather abstract, but you've seen these concepts before. Adjoin i to the integers to get the gaussian integers. These two rings live within the rationals and the complex rationals respectively. The norm maps the complex plane into the real line, and this same nor maps the gaussian integers into the integers. This homomorphism helped us characterize the gaussian primes over the integer primes. Recall that 2+i is prime upstairs, because its norm, 5, is prime downstairs. This generalizes to other rings and fields.
First show that the image of S lies in R. Write a member of S as a linear combination of basis elements with coefficients in R. After all, S is R adjoin one or more elements, so everything in S looks like a vector with coefficients in R. The norm of u is then the norm of the matrix that implements multiplication by u. This matrix has entries in R, and its determinant lies in R. Thus the norm of u lands somewhere in R.
If u is a unit in S then write uv = 1. Take norms, and u*v = 1 in R. Thus the norm of u is a unit in R. Units upstairs map to units downstairs.
If x and y are associates in S, with y = some unit times x, then x and y are associates in R.
Sometimes this can be reversed. Assume all the conjugates of y lie in S. Since y generates F, this makes F/K a normal extension. The isomorphisms from F into its normal closure are really automorphisms on F. Anything in S, based on R and y, winds up in S. Now let the norm of u be a unit in R. If u*v = 1 in R, then u times all its conjugates times v yields 1 in S, making u a unit in S. Thus u is a unit iff its norm is a unit. This happens, for example, when y is the square root of something. S includes y, so we're ok. It also happens for every cyclotomic extension, since every conjugate of y is a power of y. This explains, in part, why quadratic and cyclotomic extensions are well characterized.
Since factorization into nonunits upstairs implies factorization into nonunits downstairs, irreducible downstairs implies irreducible upstairs. This recapitulates the earlier example of 2+i prime, because its norm 5 is prime. But be careful, don't call the irreducible element upstairs prime unless you know S is a ufd. We adjoined the square root of 5 to Z and found that 3 was irreducible but not prime.