As you might guess, a cyclotomic number field is a number field that is also a cyclotomic extension. Adjoin y, the nth root of 1 to Q and to Z. Remember that y is a root of ζn(x), (a zeta polynomial), which has degree φ(n). Gauss proved this polynomial is irreducible over Q, and over Z. The conjugates of y are the powers of y, where the exponent is coprime to n.
Since gauss' lemma applies, a polynomial p is irreducible over Z iff it is irreducible over Q. Let R be a ring of fractions between Z and Q. If p splits over Z it splits over R, and it splits over Q, and it splits over Z. Thus p is irreducible over Z iff it is irreducible over R. Apply this to ζ, and ζ remains irreducible under any localization of Z.
A global field could be a number field based on Q, or it could be an extension of K, where K = (Z/p)(t). Adjoin y to K, and the extension is the same as adjoining y to Z/p, and then adjoining the indeterminant t. Note that t has not suddenly become algebraic over (Z/p)(y), else it would be algebraic over Z/p. (A tower of algebraic extensions is algebraic.) Therefore the cyclotomic global field is the extension of Z/p by y, giving another finite field, then t is adjoined.
Adjoin the cube root of 1 to Z and get a quadratic extension that is a ufd. The only units are the 6th roots of 1, lying on the unit circle. Since this particular cyclotomic extension is a quadratic extension, adjoining the square root of 3, the splitting problem has been solved. We know, for each prime p, whether p (as an element or as a principal ideal) splits upstairs. 2 remains prime, 3 is the square of a prime ideal in the complex extension, and p (odd) splits into distinct prime ideals, having conjugate generators, iff 3 is a square mod p, iff p is a square mod 3 (quadratic reciprocity), iff p = 1 mod 3. By coincidence the p mod 3 test also handles 2, which is not 1 mod 3, and does not split. The disposition of p depends only on p mod 3.
Adjoin the fourth roots of 1 and find the gaussian integers, another quadratic extension (square root of 1) that is also a ufd. Once again the splitting problem has been solved. 2 is the square of the ideal generated by 1+i, and p (odd) splits into distinct prime ideals having conjugate generators iff 1 is a square mod p, iff p = 1 mod 4. The disposition of p depends only on p mod 4.
For n = 5, or n beyond 6, φ(n) is greater than 2. The cyclotomic extension is not a quadratic extension any more, and we won't be able to take advantage of that machinery.
Assume n is an odd number > 3, and let y be the nth root of 1. Note that 1+y = (1y2)/(1y) is a ratio unit. The same is true of 1+yi = (1y2i)/(1yi), as long as i is less than n. This because gcd(i,n) = gcd(2i,n).
These units, 1 + various powers of y, lie on a circle of radius 1 centered at 1. The first unit, 1+y, is farthest to the right, and very close to 2 for large n.
Let u be such a unit, and suppose u lies on the original unit circle as well. Intersect the circles and find the 6th root of 1. If n is a multiple of 3, 1+yn/3 is a torsion unit, the 6th root of 1. All the other units about 1 are torsion free.
Don't assume these units are independent of one another; they are not. By Dirichlet's unit theorem, there are only phi(n)/21 fundamental units. I'll illustrate with n = 5, whence there is 1 fundamental unit. The units mod the powers of y form a cyclic group isomorphic to Z. Thus 1+y and 1+y2 must be related. Multiply 1+y times 1+y2 and get y4. Once we adjust for the powers of y, 1+y and 1+y2 are inverses. As for 1+y4, it's the same as 1+y times y4. (This holds for any prime n; the units closest to 2 are equivalent via yn1.) Finally, 1+y times 1+y3 = y2.
Let n be an odd prime, and let y be the nth root of 1. Use the powers of y, from 1 to yn2, as a basis. Then write the matrix that implements multiplication by 1y. In other words, a row vector of coefficients, times this matrix, is the same as multiplication by 1y. This matrix has 1 down the main diagonal, 1 on the superdiagonal, and 1 on the bottom row, except for the lower right entry, which = 2. In general, a+by yields a matrix with a down the main diagonal, b on the superdiagonal, b on the bottom row, and ab in the lower right corner. (a and b are drawn from the base ring.) Here is the matrix when n = 5.
a  b  0  0 
0  a  b  0 
0  0  a  b 
b  b  b  ab 
Find the determinant of such a matrix using induction on n. Remember that the matrix is n1 by n1. The formula is (an ± bn) / (a+b). This is clear when n = 2 and the 1×1 matrix is ab, or a2  b2 over a+b. For a larger matrix, remove the first row and column, and the cofactor becomes a times the determinant of the n2 by n2 matrix, based on n1. Expand the quotient into an2  an3b + an4b2  … bn2, and multiply by a. The cofactor in the lower left becomes ±bn1. Add these two cofactors and get (an ± bn) / (a+b). Since n is odd, we can simply write this as (an + bn) / (a+b).
The determinant is the norm of a+by, which is the product of the conjugates of a+by.
Realize that the quotient form is merely a notational convenience. When a = 1 and b = 1, it looks like we're dividing by 0. The true formula is the expanded homogeneous polynomial of degree n1. Evaluate this at 1y, a = 1 and b = 1, and get n. The product of the conjugates is ±n. Since factors can be grouped into pairs by complex conjugation, angles sum to 0, and the product is indeed n.
If you don't like the matrix proof, remember that the product over the conjugates of xy is the ζ polynomial, which is the sum of xi as i runs from n1 down to 0. We are interested in the product of the conjugates of a+by. This is b times (a/b)y. Substitute a/b in for x, and multiply by (b)n1, and get the earlier formula.
Let M be the ideal generated by 1y. Since the index equals the norm, the quotient ring has size n. Its characteristic divides n, and since n is prime, the characteristic equals n. In other words, the quotient ring is Z/n, which is a field. M is a maximal ideal, and its generator, 1y, is irreducible.
Since 1y generates n, M contains n, and M lies over the maximal ideal generated by n.
Every maximal ideal in a commutative ring is prime, hence 1y is prime.
Each conjugate of 1y has the same norm, i.e. the product of the conjugates, and it generates its own maximal ideal. The product of these n1 maximal ideals gives the principal ideal {n} within Z[y].
Divide 1y by any of its conjugates and find a ratio unit. All the conjugates are associates, and all of the n1 maximal ideals coincide. Thus {n} = Mn1. n is totally ramified.
A relatively simple matrix implements division by 1y. Specifically, Mi,j = nj for j ≥ i, and j for j < i, with a uniform denominator of n. Multiply this by the matrix we built earlier to get the identity matrix. Thus x is divisible by 1y iff x*M (writing x as a row vector on the left) has integer components.

* 

= 

However, there is an easier way to tell if x is divisible by 1y. Write x using the powers of y as a basis, and let s be the sum of the n1 coefficients. Using synthetic division starting with the constant term, compute x/(1y). After n1 steps we have a remainder of s times yn1. This is an associate of s. 1y divides x iff it divides s. If 1y divides s then take norms, and n divides sn1. Since n is prime, n divides s. Conversely, if n divides s then 1y divides s. Therefore x belongs to the maximal ideal generated by 1y iff its coefficients sum to a multiple of n.
this leads to a natural ring homomorphism. Let M be the maximal ideal generated by 1y. As shown above, x is in M iff the sum of the coefficients of x, denoted s(x), reduced mod n, is equal to 0. In fact, s implements a natural ring homomorphism from Z[y] onto the integers mod n, with M as kernel. Addition under s is straight forward. Multiply x by any integer k, and s(x) is multiplied by k. Multiply x by y and the coefficients shift. The last coefficient, on yn2, is negated and applied to 1 through yn2. Apply s, and this coefficient is multiplied by (n1), which is 1 mod n. Thus s(xy) = s(x) = s(x)×1 = s(x)×s(y). Apply the distributive law and s respects multiplication. Furthermore, s maps the integers to themselves. It is therefore a ring epimorphism from the cyclotomic extension onto Z/n. The kernel is the maximal ideal M. Hence Z/n is naturally isomorphic to the cosets of M, with the integers from 0 to n1 acting as canonical coset representatives.
Consider the norm of x, which is the product of the conjugates of x. Conjugation is a ring automorphism that maps M onto itself, i.e. the only prime ideal lying over n. It therefore induces an automorphism on the quotient ring Z/n. Yet conjugation fixes the integers, which are the canonical representatives for the cosets of M. Thus conjugation becomes the identity map on Z/n. In other words, conjugation does not change the coset of M. It preserves the value of s(x) mod n.
Since each conjugate of x has the same value s(x), the product of all n1 conjugates is s(x) raised to the n1. This is 0 if x lies in M, and 1 otherwise, by Fermat's little theorem. Therefore x is 0 mod n for x ∈ M, and 1 mod n otherwise.
Localize about n, and the ring downstairs becomes a dvr, whose one and only maximal ideal is generated by n. Upstairs we find an integral extension with primes over primes. Of course 0 lies over 0. The only prime over n is the localization of M, generated by 1y. Every prime ideal is principal, and that makes Z[y], localized about n, a pid. With just one prime ideal, it is a dvr, lying over the dvr downstairs. If the valuation of n is 1, the valuation of 1y is 1/(n1).
Let n be an odd prime, and let y be the nth root of 1. Z[y] is a simple extension, so express the discriminant as the square of the product of the differences between the conjugates of y. The conjugates of y are the powers of y from 1 to n1.
Start with yy2 times yy3 times yy4 etc, and factor out y. This gives yn2 times 1y times 1y2 … 1yn2. Then do the same for y2yi, factoring out y2. This gives y2 to the n2, times 1y through 1yn1, sans 1yn2. Do this for each power of y.
The factors of y become y times y2 times y3 … times yn1, all raised to the n2. The sum of the exponents, from 1 to n1, is (n2n)/2. This is divisible by n, and yn = 1, hence this expression drops to 1 and goes away.
The factor 1y appears in every expression, except the last. Similarly, 1yn1 appears in every expression except the first. Each 1yi appears n2 times. Regroup, and the result is the product of the conjugates of 1y, raised to the n2 power. In the last section we showed the norm of 1y is n. Therefore the discriminant is nn2.
The cyclotomic extension Z[y] is integral over Z, but is it integrally closed in its fraction field Q(y)? It is, but the proof is not trivial. It relies on the properties of discriminants under localization, which is why we needed to find the discriminant of Z[y].
Being integrally closed is a local property. Prove integral closure under each localization, and that will do the trick.
Localize about n, and the result is a dvr. This was discussed in the previous section. A dvr is integrally closed, so we can move on to a prime p other than n.
Localize about p, and the discriminant does not change. This means the discriminant is nn2, localized at p. Since n is a unit under this localization, the discriminant is a unit. Therefore the extension is integrally closed. That completes the proof.
It follows that Z[y] is dedekind, and each localization about a prime ideal p in Z becomes a pid over a dvr.
Let n be an integer > 2, and let y be the nth root of 1. Adjoining y to Q builds a field extension of dimension φ(n). This because y satisfies the irreducible polynomial ζn, which has degree φ(n).
An automorphism moves y to one of its conjugates, yi, for each i coprime to n. That's φ(n) field automorphisms in an extension of dimension φ(n), hence a galois extension.
The fixed field of the galois group G is Q, and when the same automorphisms are applied to the ring Z[y], the fixed ring is Z. If some linear combination of powers of y is fixed then it is equal to a rational number, and that implies a lesser minimum polynomial of y over Q.
For any prime ideal p in Z, G acts transitively on the prime ideals lying over p. Thus the uniform degree equation applies. The same holds under localization. In fact I will typically localize about p, since the splitting problem does not change under localization.
Let n be an odd prime and let p be another prime that is 1 mod n. Let f be an nth root of 1 mod p. This is feasible since p1 is divisible by n. Thus fn = 1 mod p. If fn = 1 mod p2, expand (f+p)n by the binomial theorem. The second term, npfn1 is not divisible by p2. Hence we can replace f with f+p, and fn is 1 mod p, but not 1 mod p2.
As a polynomial, the roots of xn1 are 1 and all the conjugates of y. Expand fn1 into f1 times fy times fy2 times … fyn1. This product is kp, where k does not divide p. Localize, and let the elements generate ideals. Since f is a nontrivial nth root of 1 mod p, it is not 1 mod p, and f1 is a unit in the localized ring. Pull this out of the equation, leaving P the product of n1 ideals. If these are proper ideals then by the degree equation, these ideals are prime, with residue degree 1.
Let J be the principal ideal generated by fy. The norm of fy is kp, an associate of p. Since fy is not a unit, fy is not a unit either. J is a proper ideal lying over p, and is prime, as described above.
Suppose two prime ideals coincide. Without loss of generality, let fy and fy2 generate the same ideal. Pull back to the original cyclotomic extension, where the corresponding prime ideal is generated by fy and p. This also generates fy2, hence it generates yy2. Multiply by a power of y and get 1y, which lies over n. Thus n is also generated, and since n and p span 1, we have the entire ring. This is a contradiction, hence the n1 prime ideals lying over p are distinct. The extension is unramified at p.
Now let p be greater than 1 mod n. This is actually a generalization of the above, and should perhaps be called p ≠ n.
There are three factors that combine to solve the splitting problem: the ramification degree, the residue degree, and the number of primes lying over p. Their product is the dimension of the extension, or n1, Let's start with the residue degree.
There is no particular reason to localize about p, since the two residue fields, before and after localization, are the same.
Let Q be a prime lying over p, and mod out by p and Q, giving the two residue fields. The field downstairs has order p. It is of course the integers mod p. The field upstairs, call it K, has order pl. K is the cosets of Q in S, and is generated by the image of y adjoin Z/p. If two different powers of y map to the same thing in K, their difference lies in Q. This puts 1 minus a power of y in Q and n is in Q, which is a contradiction. Thus y generates all the nth roots of 1 in K. The order of k, minus 1, is divisible by n, and since y generates all of K, K is the smallest field with this property. Therefore l is the least integer such that pl = 1 mod n. l is the order of p mod n. When p was already 1 mod n, as above, l = 1. The base field already has the n roots of 1. At the other extreme, p could be primitive mod n, and you might have to raise p to the n1 to get 1 mod n. l, the residue degree, is some factor of n1, the order of p mod n.
Since the discriminant is a power of n, it becomes a unit after localization. Referring to a theorem in a future chapter, p is unramified. In other words, the ramification degree is 1. Thus there are (n1)/l distinct primes over p, with residue degree l. The splitting problem has been solved when n is prime.
If n is odd and p is a prime that does not divide n, residue degree is calculated in exactly the same way. The image of y in the finite field K has to produce the n roots of 1. Thus l, the residue degree, is the order of p mod n. More on this below.
This section explores some cyclotomic extensions that are ufds. n = 4 yields the gaussian integers, and n = 6 yields the eisenstein integers; These are familiar ufds.
When n = 5 the discriminant is 125, and that produces a minkowski bound of 1.7. There are no prime ideals of index 1.7, hence this extension is a ufd.
When n = 7 the discriminant is 75, and the minkowski bound is 4.13. Restrict attention to primes lying over 2 and 3. These primes have order 3 and 6 mod 7, respectively. The residue degree is at least 3. The index of the overlying prime ideal is at least 8, and that's larger than 4. Therefore the extension is a ufd.
When n = 11 The minkowski bound is 58.9, so consider primes p, with order l mod 11, such that pl is below 59. (Remember that the ramified prime lying over 11 is principal, so we don't have to worry about that one.) 2, for instance, has order 10, and that's inert and principal. 3 has order 5, but 35 = 243, larger than 58. The only candidate is p = 23, with residual degree 1. Let 5+y and 23 generate Q, lying over P. (The norm of 5+y is (511+1)/6, which is divisible by 23 but not 232.) If there is a gcd, it divides 23  4×(5+y), or 34y. Add y*(5+y) to this to get 3+y+y2. Then subtract 5+y to get 2+y2. Add 3+y+y2 back in to get 1+y+2y2. Add y2*(2+y2) to get 1+y+y4. This has a norm of 23 (calculation), hence the ideal is principal, and the extension is a ufd.
When n = 8, the extension has dimension 4. The four conjugates of y lie on the 45 degree lines. Take the square of the product of their differences to find a discriminant of 256. The extension is integrally closed at all odd primes.
The minkowski bound is 2.43, so we only need look at primes of index 2, lying over 2. Verify that the norm of 1y is 2. Also, 1y, 1y3, 1y5, and 1y7 are associates, since their ratios are units. The extension is totally ramified at 2, and the prime lying over 2 is principal, generated by 1y. Localize about 2 and find a ring wherein the only prime ideal is principal. This is a pid, and a dvr, wherein the valuation of 1y is 1/4. This is integrally closed, hence the original extension is integrally closed, dedekind, and a ufd.
Here is a gcd algorithm for cyclotomic 8. That in turn provides an alternate proof of unique factorization.
Let y be the eighth root of 1 that lies in the first quadrant. Let 1, y, y2, and y3 span the ring as a free Z module. Of course y4 = 1. Let a vector v = a+by+cy2+dy3 span a sublattice in 4 space. The base cell is v times (1, y, y2, and y3). Note that vy = d+ay+by2+cy3 has the same length, in 4 space, as v. (This is not the norm within the cyclotomic extension; just a distance metric in a parallel universe.) In the same way, vy2 and vy3 have the same length. The cell is a 4 dimensional rhombus.
Pull w back into this base cell. If the base cell is a perfect cube, then w is farthest from the corners when it lies in the center. In that case the distance to each corner is the length of v. Write w as v/2 + vy/2 + vy2/2 + vy3/2, or v times (1+y+y2+y3)/2. Remember that i = y2, and write 2 as (1y2)*(1+y2). Now our expression reduces to v/(1y). Since 1y lies over 2, and has norm 2, w is a vector of the same length, but with half the norm. This can't continue forever, so we may assume w is not in the center of the cube. It is nearer to one of the corners, or the rhombus is not a perfect cube. When the rhombus is tilted, w is closer to a corner than the length of v. Travel from w perpendicular to the nearest 3 dimensional lattice, a distance bounded by half of v. Travel perpendicular to the nearest plane, a distance bounded by half of v. Travel to the nearest line of points, then to the nearest lattice point. The distance is less than twice half of v, thus less than v. March along in the quotient remainder fashion and find the gcd.
n = 9 is an extension of dimension 6, thus y has 6 conjugates. The zeta polynomial sets y6 = 1y3. The discriminant is 39 (calculation). Everything off of 3 is integrally closed. The minkowski bound is 4.46. Evaluate the product of the conjugates of 1y and get 3. This is a totally ramified extension, and the one and only ideal over 3 is principal. Since 2 has order 6 mod 9, 2 is an inert prime. The extension is a ufd.
n = 12 has dimension 4. Evaluate the discriminant and get 144. The minkowski bound is 1.82, and that assures a ufd. Technically we need to prove this composite cyclotomic extension is integrally closed and dedekind, but I'll get to that later.
If the prime over p is principal, is there a procedure to find the generator efficiently? (I found the generator for the prime over 23 in cyclo 11, but that was rather a follow your nose fishing expedition.) Ideally, you want to take the gcd of p and anything whose norm has precisely one factor of p, the latter generating the ideal in the local ring.
Assume we are adjoining the m roots of 1, and warm up with a residue degree of 1. Thus p = 1 mod m. Find u primitive mod p, and raise it to an appropriate power so that u becomes a primitive mth root of 1. Then write um1 = a multiple of p. If um1 is divisible by a higher power of p, add p to u. Split um1 in the cyclotomic ring. uy times uy2 times uy3 times … Let the elements generate ideals in the dedekind domain. For Q over P, there is some uw divisible by Q, where w is a power of y. Suppose w is not a primitive mth root of 1. Perhaps it is a primitive kth root of 1. Multiply the k terms together and get uk1. This is an integer not divisible by p. The norm in the composite extension, a power of the aforementioned integer, is not divisible by p either. Yet the norm of uw, containing Q, is divisible by p. This is a contradiction; hence we can use uw where w is a primitive mth root of 1. There are φ(m) such terms, and they are all conjugate. Thus there are φ(m) primes over p, as we would expect. Find one by taking the gcd of uy and p. Conjugate to find the rest.
I don't have a general gcd algorithm that is efficient and/or proven to work; but when you reduce a mob b, and put a in a parallelotope spanned by b, anchored at the origin, there always seems to be a corner v such that av has norm smaller than b. I've implemented a couple of these extensions in software and it seems to work. Of course m has to be modest, as there are 2φ(m) corners to consider.
Now let p have residue degree l. Thus pl = 1 mod m. Reduce the cyclotomic ring mod p, giving a box of size pd, where d = φ(m). Use the chinese remainder theorem to write this as a product of rings mod Qi, for each Qi lying over P. This is a direct product of fields, where each field has size pl. Just about any point you select will be a unit, and will have order pl1. Raise u to the (pl1)/m. Now um1 is divisible by pl, and um1 splits in the cyclotomic ring. Run gcd(uy,p) as before.
Since u is not an integer, uy2 is no longer the conjugate of uy. If you want all the primes over p, you may need to run gcd(uyi,p) for all the powers of y.
Adjoin y to the integers, where y is an nr root of 1. Start by building the cyclotomic polynomial ζ. Adjoin the nth root of 1 first, giving ζ = xn1 + xn2 + … + x + 1. Then replace each x with x to the nr1. The lead term of ζ, and ζ itself, now has degree (n1)×nr1, which is the same as φ(nr). Here are a couple of examples.
ζ25 = x20 + x15 + x10 + x5 + 1
ζ27 = x18 + x9 + 1
ζ16 = x8 + 1
Verify that y is a solution, and so is yi, provided i is not divisible by n. Raising y to the nr1 produces a nontrivial nth root of 1. This is plugged into the cyclotomic polynomial for n, and the result is 0. Thus the roots of ζ are the conjugates of y.
You can also confirm the construction of ζ using the recursive or mobious formulas. In any case, we have built ζ, a polynomial of degree φ(nr), that contains n terms. Each term is a power of x, and exponents are multiples of nr1.
Since ζ is the product of the conjugates of xy, we can evaluate the norm of a+by by replacing x with a, and bringing in powers of b to balance, so that the resulting expression is homogeneous, i.e. each term has degree (n1)×nr1. Set a = b = 1 to find an alternating sum 1  1 + 1  1 … + 1, giving a norm of 1. This confirms 1+y is a unit, the ratio of 1y2 over 1y. (This doesn't work when n = 2.)
In the same fashion, set a = 1 and b = 1, and the norm of 1y is n. (This does work when n = 2.) As elements or principal ideals, the product of the conjugates of 1y is n. Each ideal has index n, and is maximal.
The quotient of two generators, 1yi over 1yj, is a ratio unit. The generators are associates, and the maximal ideals coincide. Once again n is totally ramified.
Localize about n and find a dvr over a dvr. If n has valuation 1, 1y has valuation 1 over (n1)×nr1.
Let 1y generate the maximal ideal M. A natural homomorphism maps the cyclotomic ring onto Z/n, with M as kernel. This homomorphism, called s(x), is the sum of the coefficients of x, reduced mod n, when x is represented as a linear combination of powers of y. The proof is the same as the one given earlier. As a corollary, the norm of x is divisible by n if x lies in M, or it is equal to 1 mod n otherwise.
Here is a lemma that will help determine the discriminant. What is the product of 1yj, as j runs from 1 to nr1? Omit 1yj when j is divisible by n, and we have the product of the conjugates of 1y, which is n. By induction, the product of 1yj, where j is a multiple of n, is nr1. Put this all together and the product over 1yj is nr.
Write the discriminant as the square of the product of differences of conjugates. Thus each yiyj appears twice. Count this once as yiyj, and once as yjyi.
When you look at a term like yiyj, pull yi out, so that it becomes 1 minus a power of y. When I did this for n, I carefully kept track of the powers of y, but there's no real need to do this, and I'm not going to bother this time around. The discriminant comes out a positive number, so the powers of y must clump together to make 1. Let's not worry about them.
I'll use 27 as an example. Imagine a 27 by 27 matrix from which we select i and j, avoiding rows and columns divisible by 3, and avoiding the main diagonal. This is a bookkeeping tool to make sure we have all the differences of the powers of y.
The top row contains yyj, as j runs from 1 to 27. Skip the upper left entry and let j run from 2 to 27. Divide by y and find 1yj as j runs from 1 to 26, skipping the entries that are 2 mod 3.
Divide the second row by y2, start at the third column, and wrap around to the first column. The result is 1yj, as j runs from 1 to 26, skipping the entries that are 1 mod 3.
Put these two rows together, and for j not divisible by 3, we would have the norm of 1y twice, but we skipped the entries 1 mod 3, and 2 mod 3, so that takes one of the norms away. In general the norm of 1y appears n2 times over. With 1y = n, this contribution is nn2.
Then there are the columns that are multiples of 3. This happens for both rows, or n1 rows in the general case. Invoke the previous lemma and get nr1 raised to the n1. The exponent on n becomes nrrn+1. Bring in the contribution from the previous paragraph and the exponent on n becomes nrr1.
After dividing out y4, the fourth row looks just like the first. And the fifth row looks just like the second. And so on. The pattern repeats 9 times, hence the previous expression is raised to the nr1. Put this all together and the discriminant is n raised to the (nrr1) × nr1.
Set r = 1 and get n2 as expected.
Set r = 2 and get (2n3) × n. In the last section we computed the discriminant for 32 by hand, or perhaps with a computer. There are only 6 primitive roots, so the algebra is tractable. The discriminant was 39. Set n = 3 in (2n3) × n and get 9 as expected.
Review this proof when n = 2. Consider only the odd rows and columns of the matrix. With y pulled out, the first row is the product of 1yj for j even. This is 2r1 as per our lemma, And this is raised to the 2r1. The exponent becomes (r1) × 2r1. Surprisingly, you get the same exponent if you substitute n = 2 in the previous formula. We saw this in the last section, when the 8th roots of 1 produced a discriminant of 256.
Finally we can prove the extension is dedekind. The discriminant is always a power of n. Localize about any other prime and the discriminant is a unit. Thus the localization is integrally closed. Localize about n and find a dvr, which is also integrally closed. Thus the extension is integrally closed, and dedekind.
Now for the splitting problem. As shown above, n is totally ramified. In other words, {n} is {1y} raised to the (n1)×nr1.
Let p be a prime other than n. With a unit discriminant, each p is unramified. The ramification degree is 1. The residue degree is l, where l is the order of p mod nr. That leaves (n1)×nr1/l primes lying over p.
I skated past an important detail here. Continuing the 27 example, what if y maps to a cube root of 1 in K? y3 maps to 1, and 1y3 is in the maximal ideal Q lying over P. The norm of 1y3 is n or a power of n, which puts n in Q, which is a contradiction, since Q lies over p. Therefore y must map to a primitive root of 1 in K, and the residue degree is established. That completes the proof.
Let n be composite and let y be the nth root of 1. To illustrate, let n = 9×5×7 = 315. We want to show that this cyclotomic extension is dedekind. You probably know by now that if the extension is integrally closed, we're done. And it is enough to prove this locally.
The cyclotomic polynomial ζn has degree φ(n), which becomes the dimension of the cyclotomic extension. Instead of adjoining y directly, adjoin the fifth root of 1, then the seventh root of 1, then the ninth root of 1. These extensions have dimensions, over Q, of φ(5), φ(7), and φ(9) respectively. Stack them up in a tower and the dimension is at most φ(5)×φ(7)×φ(9), which is φ(315).
Since gcd(5,7) = 1, some linear combination of 5 and 7 yields 1. Hence the fifth root of 1 and the seventh root of 1 conspire to build the 35th root of 1. Bring in the ninth root of 1 and span y. Thus the tower of 3 coprime extensions builds Q[y].
Since dimensions are preserved, these three extensions are linearly disjoint. Intuitively, they run independently of each other. In fact, the galois group associated with the ninth roots of 1 is the same whether we adjoin these roots first or last. All six conjugates are present, and they map onto each other in the same way.
Evaluate the discriminant as the square of the product of the differences between the six conjugates. It doesn't matter whether this is brought in first or last. Whenever the ninth roots come in, the discriminant is 39.
Localize about some prime p apart from 315. Adjoin cyclo 5 to get an integrally closed ring, then cyclo 7 for an extension of that ring via a discriminant that is a unit mod p, hence integrally closed. Finally adjoin the ninth roots of 1 with another unit discriminant mod p. The result is integrally closed. At each level the fraction field is Q; integrally closed means integrally closed in Q. This works for all the off primes.
If p is 3 5 or 7, adjoin that one first, giving a dedekind domain that is integrally closed. Localize about that prime, then pile on the next two extensions, each having a unit discriminant, and each integrally closed over the previous ring. Thus Z[y] is integrally closed over each prime, and integrally closed. This is the integral ring of Q(y), and dedekind.
Assume once again that p does not divide n. Localize about p and see how it splits. Bring in each prime power of n in turn. The discriminant is always a unit, giving a ramification degree of 1. Ramification degrees multiply in a tower of extensions, hence the entire extension is unramified at p. Find the residue degree and we're done.
Suppose y maps to something other than a primitive 315th root of 1 in the residue field K. Push y up until the image is 1. Either a cube root of 1, or a fifth root of 1, or a seventh root of 1, maps to 1. Let z be one of these roots, mapping to 1. Now 1z maps to 0, and is contained in q, lying over p. Yet this generates the prime 3 5 or 7, which is not part of p. This is a contradiction, hence y maps to a primitive nth root in K, and it takes the entire cyclotomic extension along for the ride.
The residue degree is the least l such that pl = 1 mod n. This is the lcm of the order of p under each prime power dividing n.
Once l is established, the number of primes over p is φ(n)/l.
Assume p is 3. Adjoin the ninth roots first, and p is totally ramified. There is but one prime over p, and the residue field remains at Z/3. Now bring in y, where y is the 35th root of 1. If this maps into a nonprimitive root of 1 in its residue field then 5 or 7 is part of the prime lying over 3, and is part of the ideal generated by 3. This is impossible, hence y maps to a primitive root of 1. Find l such that pl = 1 mod 35. This is the residue degree of the extension, and the number of primes over 3 is 24/l. 5 and 7 are handled similarly.
The splitting problem has been solved for all cyclotomic number fields. To summarize, let m be the part of n that is coprime to p. The residue degree is the order of p mod m, denoted l. The ramification degree is φ(n/m), and the number of primes lying over p is φ(m)/l.
When n is a prime power, 1y generates the prime ideal lying over n. However, when n has multiple prime factors, 1y becomes a unit. Remember that ζn is the product over the conjugates of xy. Replace x with 1 to find the product over the conjugates of 1y, or the norm of 1y. Thus we are interested in the sum of the coefficients of ζ.
Review the recursive definition of ζ. Start by dividing xn1 by x1, giving a polynomial with n terms, whose coefficients sum to n. Divide this by ζd for each divisor strictly between 1 and n. Apply the ring homomorphism that replaces x with 1, mapping polynomials into integers. By induction ζd is 1 whenever d has multiple prime factors. When d is a power of p, ζd has coefficients that sum to p. (This is how we determined 1y = p.) Thus each power of p reduces n by p. Do this for all primes dividing n, and the result is 1. The norm of 1y is 1, and 1y is a unit.
If n is odd, 1+yi is a unit, specifically, the ratio (1y2i)/(1yi).
What is the product of 1yi for all the powers of y? If n is a prime power, an earlier lemma shows that the product is n, so assume n has multiple prime factors. When the exponents are coprime to n, the product over 1yi is the norm of 1y, or 1. These drop away.
Let g be a factor of n strictly between 1 and n. Consider all the exponents i such that gcd(n,i) = g. The product of 1yi, across this set, is the norm of 1z, where zn/g = 1. This is 1 when n/g contains multiple primes, and it is p when n/g is a power of p. As g ranges over the proper divisors of n, n/g ranges over the proper divisors. All the powers of p appear, for each prime p dividing n, and their product is n. In summary, the product of 1yi, as i ranges from 1 to n1, is n, for any n.
Let y be the 8th root of 1, generating a 4 dimensional extension of the rationals and the integers. The cyclotomic polynomial is x4 + 1 = 0. The basis of the extension is 1, y, y2, and y3. Perform polynomial arithmetic as usual, but replace y4 with 1 wherever it appears.
The dirichlet unit theorem tells us there is one fundamental unit, along with the 8 roots of 1. Let's see if we can find it.
Let u = 1+yy3. As a real number this is 1+sqrt(2). Its inverse v is 1+yy3. As a real number this is sqrt(2)1. Multiply u times v and get 1. u lies outside the unit circle and v lies inside the unit circle. Powers of u head out along the x axis to infinity, while powers of v head in towards the origin.
Complex conjugation doesn't change u, and replacing y with y3 or y5 swaps u and v. This holds in general. Let g be a unit outside the unit circle, with inverse h. The product of the four conjugates of g is 1. Both g and its complex conjugate, reflections through the x axis, lie outside the unit circle. The other two conjugates must lie inside the unit circle. Each unit g outside the unit circle has a specific distance d(g) from the origin, that does not change with conjugation, or is inverted if that conjugate lies inside the unit circle.
If u is not fundamental then there is some generator g outside the unit circle with d(g) < d(u). Write g = a+by+cy2+dy3. If a is negative then picture a as a vector pointing left along the y axis. If c is negative then picture c as a vector pointing down the y axis. Similarly let b and d slant up, or down, along 45 degree lines, as they are positive or negative respectively. Now we can think of all four vectors as positive integer lengths.
If a points right, and b and d both slant to the right, then a b and d conspire to build a real component that is at least as large as u. c, and the imaginary contributions from b and d, can only make g larger. Therefore this configuration is impossible.
In general, 3 consecutive vectors cannot point into the same half plane. Spin the picture around (multiplying by a power of y) so that a points right and c points up. b has to point down and to the left. If d points down and to the right, spin the picture 135 degrees counterclockwise, so that a points right, c points up, b points down and to the left, and d points up and to the left. Now replace y with y3 to find a conjugate of g. This unit has 3 vectors heading into the right half plane, hence d(1/g) is at least d(u). That puts g inside the unit circle. Therefore g cannot comprise four nonzero vectors.
If g has three nonzero vectors, you can still spin it around so that a points right and c points up. Suppose b is 0 and d is nonzero. a and c contribute at least sqrt(2) up the 45 degree line. In fact, if either is greater than 1 then they contribute at least 2.12 up the 45 degree line, and the actual distance to the origin can only increase from their as a result of d and any transverse contributions from a and c. Granted this is less than d(u), but not less than the square root of d(u), as is required if u is some power of g. We need d(g) < 1.554. So a = c = 1, and if d is 1 then the length is sqrt(3), which is still to large.
If b is the third nonzero vector then it points down and to the left. Replace y with y3 and find three adjacent nonzero vectors in the fourth quadrant.
Step back to two nonzero vectors. If the angle is 45 degrees, and the lengths are 1, then d(g) is precisely the square root of d(u). This looks promising, but 1+y lies over 2, and is not a unit. Lengthen either vector, or increase the angle to 90 or 135 degrees, and d(g) is too large. That finishes the case of two nonzero vectors. Of course g does not consist of a single vector, else it lies on the unit circle. Therefore u is fundamental.
I suggest, without proof, that the fundamental unit for cyclo 5 is y2+y3, and the fundamental unit for cyclo 12 is y3+y+1. Perhaps these yield to a geometric analysis similar to the one shown above.