Let M be any module, or a left module if you prefer. A chain H of submodules is ascending if Hi ⊂Hi+1, and descending if Hi ⊃ Hi+1. In other words, the submodules are getting larger (ascending) or smaller (descending).

Emmy Noether investigated modules with no infinite ascending chains, and these modules are now called noetherian in her honor. Emmy Noether has been called "The most important woman in the history of mathematics." She rebuilt abstract algebra from the ground up, and her work in theoretical physics is equally impressive. She accomplished all this as a Jew in early 20th century Germany, often working without pay, since she was not permitted to be a professor.

At approximately the same time, Emil Artin explored modules with no infinite descending chains, and these modules are now called artinian in his honor.

Another synonym for noetherian is "ascending chain condition", or acc. As you might guess, an artinian module has the "descending chain condition", or dcc. I will usually use the words noetherian and artinian, but the brevity of acc and dcc is compelling.

If R is a ring, R is a left R module, and the submodules are the left ideals. Thus a noetherian ring has no infinite ascending chain of left ideals, and an artinian ring has no infinite descending chain of left ideals.

A division ring is a rather trivial example. It only has two left ideals, 0 and the entire ring. This is noetherian and artinian.

The matrices over a field have at most n+1 left ideals in a chain, corresponding to subspaces from 0 through n dimensions. This is a noetherian and an artinian ring.

If R is a pid, such as the integers, and p is a prime element, the powers of p generate smaller and smaller ideals, thus R is not artinian. To go the other way, let H be a proper ideal of R, generated by some x. An ideal contains H iff it is generated by some divisor d of x. Factor x uniquely into primes. Combine these prime factors in finitely many ways to produce a finite number of divisors of x. H cannot belong to an infinite chain of ideals, thus R is noetherian.

There is another definition of noetherian and artinian that is more general. Let S be a collection of subsets of a given universal set U. These could be submodules, or ideals, or anything else for that matter.

Let W be an arbitrary nonempty subset of S, a collection of submodules for instance, and assume each collection W includes a maximal element. This set is maximal with respect to the sets of W. This means W cannot be an infinite ascending chain, and since W was arbitrary, S is noetherian. Conversely, assume some W has no maximal member. There is always another set in W that is larger, so build an infinite ascending chain, whence S is not noetherian.

Similarly, the collection S is artinian iff every subcollection W of S includes a minimal member. The proof is the same as above, just run the chains down instead of up.

A module (or left module) is noetherian iff every submodule is finitely generated. This theorem has no analog for artinian modules.

First assume every submodule of M, including M itself, is finitely generated. Let H1 H2 H3 … be an infinite ascending chain of submodules, with H the union over Hi. Now H is a submodule of M, and is spanned by a finite set of generators. All these generators live in some submodule Hn, and beyond that, all submodules in the chain are equal to Hn. The infinite ascending chain cannot exist, and M is noetherian.

Conversely, assume H cannot be spanned by a finite set of generators. Select x1 in H and let H1 be the submodule generated by x1. This is not all of H, so select any x2 in H-H1 and let H2 be the submodule generated by x1 and x2. Once again these two generators cannot span all of H, so select any x3 in H-H2 and let x1 x2 and x3 span H3. This process continues forever, building an infinite ascending chain. Therefore M is noetherian iff every submodule is finitely generated.

Every ring with finitely generated ideals, such as a pid, is noetherian.

If all the prime ideals in a commutative ring R are finitely generated, then R is noetherian.

Consider the set of infinitely generated ideals in R, and show that the union of an ascending chain of ideals in this set cannot be finitely generated. Those generators would appear in one of the ideals, and ideals could not increase beyond this point.

Use zorn's lemma to establish a maximal infinitely generated ideal. Such an ideal is prime, hence it is finitely generated after all.

Any infinitely generated ideal seeds an ascending chain, and leads to a maximal infinitely generated prime ideal, which is a contradiction. Therefore all ideals are finitely generated, and R is noetherian.

Let M be a noetherian module, or left module if you prefer, and let Q be a quotient module of M with kernel K. Both K and Q are noetherian. An infinite chain in Q lifts to an infinite chain in M by correspondence, and an infinite chain in K belongs to M. Thus acc or dcc propagates down to both quotient and kernel.

It is possible to put this in reverse, but we need a lemma. Assume, for a moment, that each submodule in a chain corresponds uniquely to its image in Q and its intersection with K. Map an ascending chain in M onto an ascending chain in Q, and intersect with K to find an ascending chain in K. At each step the image in Q or the intersection with K must increase. Thus K and Q noetherian implies M is noetherian, and similarly for artinian. As a bonus, the length of the chain in M, ascending or descending, is no longer than the sum of the lengths of the chains in K and in Q. The length of M, if it has a well defined length across all possible chains, is bounded by the length of K plus the length of Q.

And now for the all important lemma. Assume an image Q0 and an intersection K0, and build the submodule from the ground up. I am not saying there is a unique submodule with quotient Q0 and intersection K0. Rather, given two such modules, one cannot contain the other. If you want to increase the module, you have to increase the kernel or the quotient. A chain can't have two modules exhibiting the same Q0 and K0.

The submodule includes K0, and at least one coset of K0 for each coset of K representing Q0. But the submodule cannot contain anything beyond K0 in K, and if it contains x+K0 and y+K0 in any other coset of K, it contains x-y in K, which is a contradiction. Exactly one coset of K0 in each dictated coset of K; that's it. There is no room for M to grow, without increasing K0 or Q0.

M

 K K0
 K0
 K0
 K0
 K0
Q

In the following short exact sequence, M is noetherian iff K and Q are noetherian. This is merely a restatement of the kernel quotient theorem using different notation. The arrows represent module homomorphisms from the previous module into the next, wherein the image of one homomorphism is the kernel of the next. Thus K embeds into M, because its kernel is 0; and then the kernel of M is K, with quotient Q. We'll encounter this notation again in upcoming chapters, so please take a moment to assimilate it.

0 → K → M → Q → 0

Let M be the direct product of two modules. Let the first component act as kernel, while the second component is the quotient. Thus M is noetherian/artinian iff the same holds for both components. Use induction to generalize this to a finite direct product of modules.

If R is a noetherian ring and Q is the quotient ring, Q is a noetherian R module. (If R is not commutative, assume R is left noetherian, and work with left modules.) But Q is also a Q module. A chain of left ideals in Q becomes a chain of R modules in Q. Thus Q is a noetherian Q module, and a noetherian ring. Similarly, the quotient of an artinian ring is artinian.

Let A and B be submodules of M, such that M/A and M/B are noetherian. If H = A ∩ B, is M/H noetherian?

To simplify notation, mod out by H. By correspondence, M/A = (M/H)/(A/H), so M/A and M/B are still noetherian; but now their intersection is 0. In other words, A and B are disjoint.

If two distinct elements u and v are in A, and are in the same coset of B, then their difference gives an element that is in both A and B, which is a contradiction. Every element in A represents a unique coset of B, and by symmetry, every element of B represents a unique coset of A.

Suppose S is an infinite ascending chain of submodules in M. Divide through by A and the sequence eventually becomes constant. Divide through by B and the sequence eventually becomes constant. Beyond some point, both image sequences are constant. In other words, Sj and Sj+1 have the same image mod A, and the same image mod B.

Suppose x is in Sj+1, but not in Sj. This because Sj+1 properly contains Sj in the ascending chain. Now x does not create any new elements in the image M/A. Thus Sj already contains some y, in the same coset as x. Subtract to find z, an element in the kernel A. Clearly z is a new element, just as x is a new element. When Sj+1 is intersected with A, we find a larger submodule, containing z. This goes on forever, hence A is not noetherian.

Let Tj = Sj ∩ A. We just showed that T is an infinite ascending sequence of submodules inside A. However, each Tj is its own submodule in M/B. This because each element in Tj represents a unique coset of B. Therefore M/B is not noetherian, and that is a contradiction. Our original sequence S cannot exist, and M is noetherian after all.

Let R be noetherian and let M be finitely generated over R. Write M as the homomorphic image of a free R module F spanned by those generators. Since F is a finite direct product of R, F is noetherian. M is the quotient of F, and is noetherian. Finitely generated over a noetherian/artinian ring remains noetherian/artinian.

Beyond this, finitely generated over a noetherian ring is finitely presented. Write M as the quotient of F, which is noetherian. The kernel K is a submodule of F, and is also noetherian, hence finitely generated, which makes M finitely presented.

The n×n matrices over R are noetherian or artinian, according to R. After all, the matrix ring is a free R module of rank n2.

According to jordan holder, the composition series of a module is unique. But we really need this chapter to put a bow on it.

If M has a composition series, then each quotient is simple, and thus noetherian and artinian. Apply the kernel quotient theorem, and after n steps M is noetherian and artinian.

Conversely, assume M is noetherian and artinian (from the left or the right as you prefer). Start with the trivial chain of M and 0. Give M the rational number 1, and map 0 to 0. If M is not simple, refine it by a submodule that is assigned the number ½. Make a further refinement, above or below this module, if possible, and give the next submodule the number ¼ or ¾. If this process goes on forever then rational numbers approach some real number s from above or from below. This violates acc or dcc. Thus the process terminates, and leaves a composition series.

In summary, M has a composition series iff M is both noetherian and artinian, and that composition series is unique.

Let M be an artinian module and let f be an endomorphism that is injective. We will show that f is also surjective, hence an automorphism.

Let M1 = f(M), let M2 = f(M1, let M3 = f(M2, and so on. Suppose M1 ⊂M. In other words, f is not onto.

Proceed by induction on j. At the jth step, Mj is contained in Mj-1. Apply f, and all of Mj-1 winds up in Mj, so when f is restricted to Mj, that image is certainly contained in Mj. Therefore Mj+1 is contained in Mj. This is a chain of descending modules.

M → M1 → M2 → M3 → M4 → …

Since M is artinian, the chain stops. Let j be the first index where Mj+1 = Mj. Remember that f is one to one, and can be reversed. The set Mj+1 has Mj as its preimage, and both sets are the same. Reverse f again to show Mj-1 = Mj. This is impossible, unless j = 0 and M = M1. Therefore f maps M onto M, and f is a module automorphism.

Next let M be noetherian and let f be an endomorphism that is onto. Suppose f is not injective. Find x and y such that f(x) = f(y). Subtract, and f maps x-y to 0. Of course f maps 0 to 0 as well.

Let M0 = 0 and run backwards, running f in reverse. Thus M1 is the preimage of 0 under f. Now M1 is a submodule of M that properly contains M0.

Assume Mj properly contains Mj-1. Let Mj+1 be the preimage of Mj under f. Since Mj includes Mj-1, taking the preimage includes the preimage of Mj-1, which is Mj. Each submodule (moving left) contains the previous.

… → M4 → M3 → M2 → M1 → 0

If Mj+1 = Mj then f(Mj) should also equal Mj, yet Mj-1 is smaller. Therefore Mj+1 is larger. This builds an infinite ascending chain of submodules. Since M is noetherian, this is impossible. Therefore f is injective, and an automorphism on M.

Let M be a finite dimensional vector space, hence both noetherian and artinian. This theorem reaffirms something we already know - a linear transformation from M into itself is injective iff it is surjective.

Let R be a left artinian domain. Given a nonzero x in R, consider the descending chain of principal left ideals spanned by the powers of x. At some point, xn and xn+1 generate the same ideal. Thus yxn+1 = xn, and yx = 1. Since every x is left invertible, R is a division ring.

Let R be a noetherian ring and let S be the polynomial extension R[x]. We will prove S is also noetherian. Obviously S isn't a noetherian R module, since it has an infinite basis, namely the powers of x. But S is a noetherian S module, and hence a noetherian ring.

S is not artinian, since the powers of x generate an infinite descending chain of principal ideals. This is one of those noetherian only theorems.

R need not be commutative. If you like, think of R as left noetherian, and replace the word "ideal" with left ideal throughout. Right noetherian is handled similarly.

Consider any ideal W in R[x] and let H be the set of extant lead coefficients. If W includes 5x3+x+1 and 3x2, then W includes 3x3, being an ideal, and W includes 8x3+x+1. The presence of 5 and 3 in H brings in 8. Verify that H is an ideal in R, hence finitely generated by a1 a2 a3 … an.

For each generator ai of H, pick a polynomial pi(x) in W with the lead coefficient ai. These will act as the first n generators of W.

Let m be the maximum degree of these polynomials, and let q be any polynomial in W with degree at least m. Write its lead coefficient as a linear combination of a1 through an, e.g. c1a1 + c2a2 + c3a3 + … cnan. Apply c1 through cn, times various powers of x, to p1 through pn, and subtract from q. The result is a polynomial of lesser degree, that is spanned by p1 through pn iff q is spanned. In other words, the generators p1 through pn span, and remove, the leading term of q, whatever it may be. Repeat this process until you are left with a polynomial in W of degree less than m. If this is spanned, then our original polynomial q is spanned.

Notice that none of the intermediate expressions cixkpi exceeds the degree of q. The polynomials line up at the left edge of q, combine, and strip off the leading term.

The polynomials of degree m-1, living in W, have lead coefficients that form a finitely generated ideal. This implies a finite set of polynomials capable of killing off the lead term. Do the same for degree m-2, m-3, and so on down to the constant polynomials of W. This builds a finite set of generators that span W as a left ideal in R[x], and that completes the proof.

Let R be noetherian and let S be a finitely generated R algebra. This isn't a finitely generated module; it's a finitely generated ring. Adjoin elements to R to build a larger ring, just as we adjoined i to the integers to get the gaussian integers. Also, the generators adjoined to R commute with each other and with R.

Repeatedly adjoin indeterminants to R and apply hilbert's basis theorem at each step. The resulting polynomial ring, in several variables, is noetherian. Since S is a quotient ring, with certain polynomials generating the kernel, it too is a noetherian ring.

If R is itself finitely generated, and commutative, it is noetherian. This because R is the quotient of Z adjoin x1 x2 x3 … xn, which is noetherian.

A variant of Hilbert's proof shows R[[x]] is noetherian. These are the power series in x, with coefficients in R. The powers of x go on forever.

R need not be commutative. If you like, think of R as left noetherian, and replace the word "ideal" with left ideal throughout. Right noetherian is handled similarly.

Build H using the lead coefficients in the power series of an ideal W. In this case the lead coefficient has the lowest degree, not the highest. Let H0 be the ideal of constant coefficients of all series in W. Let H1 be the ideal of coefficients on x when the constant term is 0. This will include H0, for multiplying those series by x still leaves them in W, but H1 could include more values of R. Build H2, H3, H4, and so on, so that the union of Hi is H.

By acc the chain is finite, so that Hm = H.

Select generators a1 a2 a3 etc, and corresponding power series p1 p2 p3 etc, spanning H0, then H1, then H2, and so on up to Hm. These are all the generators we will need.

Given a series q in W, subtract a linear combination of generators to kill off the constant term. Then kill off the linear term, then the squared term, and so on. We only need consider series in W that start beyond xm.

How might we span a series q that starts with xm? A linear combination of generators of H reproduces the lead coefficient of q. Subtract this away, leaving a series that starts with xm+1. Once again a linear combination of generators builds the lead coefficient. Multiply this linear combination by x, and subtract from q. This leaves a series that starts with xm+2. Again, a linear combination of generators yields the desired coefficient, and when this is multiplied by x2, q loses another term. Repeat this forever, and each generator is multiplied by an infinite series, and their sum spans q. W is finitely generated, and that completes the proof.

By induction, the formal power series in several variables is noetherian, and so is any quotient ring thereof.

If x and y do not commute, R[x,y] is not noetherian; nor R[[x,y]].

Let Z be the base ring, and consider the polynomials, or power series, in x and y. I'll build an ascending chain of ideals; that will cover left ideals.

Start with the ideal generated by yxy. Every string in every polynomial in this ideal contains yxy somewhere in it. There are no strings with at least x2 between every pair of y's. So bring in yx2y. Now there are no strings with at least x3 between each pair of y's. Bring in yx3y, and so on, building an infinite ascending chain of ideals.

This ring is a domain. Sort strings by length (expanding x3 to xxx etc), then lexicographically. Constants have degree 0. The longest string in pq comes from the longest string in p times the longest string in q. Of these, the last alphabetically comes from the last in p times the last in q. This string stands alone in the product, and pq is nonzero.

Since the ring is a domain, 0 is a prime ideal.

Assume f cannot be written as a product of irreducibles, and build complementary chains of ascending and descending ideals as follows.

Let f0 = f. Since f0 is not irreducible, write it as x1f1, where f1 cannot be written as a product of irreducibles. Then write f as x1x2f2, where f2 cannot be written as a product of irreducibles. Next write f as x1x2x3f3, then x1x2x3x4f4, and so on.

Let yj be shorthand for the product of x1 through xj. Clearly yj spans yj+1. As principal ideals, the former contains the latter. If the latter contains the former then yj = cyj+1, hence yj = cxj+1yj, and xj+1 becomes a unit. This is a contradiction, hence the ideals generated by yj form an infinite descending chain. At the same time, the ideals generated by fj form an infinite ascending chain. Each fj+1 properly spans fj; the proof is essentially the same.

An integral domain that is either noetherian or artinian is a factorization domain. This includes any ring that is finitely generated over the integers, such as the cyclotomic extensions.

Let R be a noetherian ring with respect to its ideals. (It need not be left or right noetherian.)

Suppose S is the set of ideals that do not contain a finite product of prime ideals. By acc, let H be a maximal ideal in S. Clearly H is not prime, nor is it R, since R contains a maximal prime ideal. Since H is not prime, find ideals A and B with A*B in H, but A and B not in H. Since A+H and B+H properly contain H, they each contain a finite product of primes. Multiply these two products together to get a longer, but still finite product of primes. This is contained in (A+H)*(B+H), which is H. Thus H contains a finite product of primes, which is a contradiction. Therefore S is empty.

In particular, S does not contain 0, hence 0 is a finite product of prime ideals. This product lies inside any minimal prime ideal, and hence every minimal prime is one of the factors in this product. Thus R contains finitely many minimal prime ideals.

Write 0 as a finite product of prime ideals, including the aforementioned minimal prime ideals. Each prime in this product descends to a minimal prime ideal. This can only make the product ideal smaller. Thus R contains finitely many minimal prime ideals, and some product of these minimal prime ideals is 0.

Some of the minimal prime ideals may be repeated in the product. Let K be a field and recall that K[x], the polynomials over K, is a pid. Let H be the ideal generated by x2. A prime ideal that contains H is generated by a prime element (irreducible polynomial) that divides x2. That can only be x. Mod out by H and the quotient ring has but one prime ideal, generated by x. This is the minimum prime ideal and the maximum prime ideal. This ideal, squared, is 0.

Let's build a ring R that is left noetherian, but not right noetherian.

Let K be a field, and let F/K be the transcendental field extension produced by adjoining infinitely many indeterminants to K.

Let R consist of 2×2 matrices, with elements of F in the top row, 0 in the lower left, and something from K in the lower right.

 F F 0 K

Let L be any intermediate field between K and F. The matrices that are 0, except for L in the upper right, form a right ideal. Adjoin more and more indeterminants, and L steps through a tower of field extensions. This builds an ascending chain of right ideals, hence R is not right noetherian. Start with L = F and take away indeterminants, and build a descending chain of right ideals. Hence R is not right artinian either.

Now view R as a left R module. Build a submodule of matrices that are 0, except for F in the upper right. This is a simple R module, generated by any of its elements.

Bring in K, in the lower right, to build the next submodule of R. Mod out by the previous submodule, and the quotient looks like K. Again, the quotient is a simple R module.

Finally step up to R. Once again the quotient is a simple left R module. This builds a composition series, hence M is left noetherian and left artinian.

Transpose matrices to reverse the rolls of left and right in the above example.

If R is a factorization domain, a nonzero prime ideal inside a proper principal ideal becomes the entire principal ideal. This theorem was proved earlier.

Let's approach this from another angle. Let R be a commutative noetherian ring. If R is an integral domain then it is a factorization domain, but perhaps we don't know that. Let g be a nonunit that generates a principal ideal, and let P be a nonzero prime ideal that is properly contained in g*R. For every y in P, y = gx for some x. Since P is prime, x belongs to P. Thus P is contained in gP, and of course, gP is contained in P, hence P = gP.

Let H0 be an ideal that lives inside P. Let H1 be the conductor ideal that drives g into H0. This is sometimes denoted [g:H0]. Of course H1 includes all of H0, and since g generates H0, g*H1 = H0. Since g lies outside of P, H1 lies in P.

Let H2 = [g:H1], let H3 = [g:H2], and so on. The ascending chain cannot continue forever, so say H6 = H7. Now gH7 = H6 and gH6 = H5. This forces H6 = H5, even though H6 is larger than H5. The only way around this contradiction is to assume H0 = H1. In other words, gH0 = H0. Multiplication by g maps each ideal in P onto itself.

 R g*R P H3 H2 H1 H0

Next assume R is a noetherian integral domain. Select g and P as above. Let z generate a principal ideal inside P. Thus gz generates z, or gxz = z, making g a unit. This is a contradiction, hence there is no prime ideal strictly between 0 and g*R. A nonzero prime ideal inside g*R becomes g*R.