For an table filled with natural numbers, we say it is a divisor table if:
- the numbers in the -th row are exactly all the divisors of some natural number ,
- the numbers in the -th column are exactly all the divisors of some natural number ,
- for every .

A prime number is given. Determine the smallest natural number , divisible by , such that there exists an divisor table, or prove that such does not exist.

Proposed by Pavle Martinović

Fix a circle , a line to tangent , and another circle disjoint from such that and lie on opposite sides of . The tangents to from a variable point on meet at and . Prove that, as varies over , the circumcircle of is tangent to two fixed circles.

Find all sequences of natural numbers such that for every pair of natural numbers and , the following inequality holds:

Let be a triangle and let be its incircle. Denote by and the points where is tangent to sides and , respectively. Denote by and the points on sides and , respectively, such that and , and denote by the point of intersection of segments and . Circle intersects segment at two points, the closer of which to the vertex is denoted by . Prove that .

Theorem. Let be a field, let be a degree- irreducible polynomial in , and let be a degree- extension of . If and are coprime, then is still irreducible over .

(The condition is tight; if , then has the factor in .

Proof. Suppose not, and let be a (proper, nontrivial) irreducible polynomial factor of over . Let be the degree of . Adjoin a root of to , producing the field , which has degree over and thus degree over .

Then note that, since is a factor of , we have that is also a root of , which means has degree . Furthermore, . But since the former has degree and the latter has degree over , contradicting the claim that is a proper nontrivial factor.

Let be a vector space over some field . Then we can define a new vector space whose elements look like for . The thing is called the wedge product or exterior product of and . The vector space is called the exterior algebra or Grassmanian algebra, but that's not important. Wikipedia doesn't call it the wedge algebra. Sad.

In fishy terms, is a product — in more formal terms, it's something you get by quotienting out stuff from a tensor algebra — so you have these properties for all :
where in the last equation, is the element of the field . These relations are something you need to travel two Wikipedia articles away from "wedge product" to figure out, but they're not that hard; they're the properties you expect multiplication to have, even if it's hard to visualize what you really get when multiplying two vectors so freely. If you replace with , these properties completely define tensor products; but for the wedge product in addition, we have these two properties:

[list=1]
[*]Alternatingness(sp?): for all
[*]Anticommutativity: for all
[/list]

Alternatingness implies anticommutativity: you can get this by expanding . Anticommutativity almost implies alternatingness: by setting , you get which implies ... unless has characteristic 2.

Today's lessons:
[list=1]
[*]Alternatingness is a better axiom, and
[*]Fields of characteristic 2 are weird.
[/list]
(But is an amazing field, so it's worth it. Maybe that's material for a future post.)

Several relevatory experiences later, I am considering making one short and probably silly math post here every day. I don't know if this is eventually too much commitment, but since today is a holiday, it's a great time to start and do whatever. I just won't be as strict about it as I was when I tried this on the Other Blog.

This post does not do anything interesting except give me an excuse to start a blog post title with the word "monoid". Monoids are underrated.

Definition. Let be a field. Then a -algebra is a possibly noncommutative (!) ring with a copy of inside it (more formally, a homomorphism from into , which has to be injective because fields are nice, except maybe for silly algebras where ).

Examples of -algebras:
[list]
[*] itself
[*]a polynomial ring
[*]the set of linear operators ("endomorphisms of ") for some -vector space (where the copy of is the scalings of the identity)
[*]these things below:
[/list]

Definition. Let be a monoid (a set with an associate binary operation and identity). Then the monoid algebra is the -algebra obtained by considering finite formal sums of the form where , the different are considered linearly independent (over ) elements of , the product of and is the monoid operation on them, and multiplying two of these sums in general is expanding everything and moving scalar coefficients out. (The copy of is given by formally multiplying by the identity of the monoid: .)

In practice we will choose to be a group (a monoid with inverses), which is why I grudgingly accepted the letter from Wikipedia, but I wanted to define this for monoids because it lets you see how other -algebras also kinda fit in the same framework. In particular, the monoid algebra is the polynomial ring (element corresponds to ), and in general, the monoid algebra is the polynomial ring .

We're defining group algebras at all because v_Enhance says this is the right way to do representation theory, so let's go there.

Definition. A representation of a -algebra is a way for to act on some -vector space "nicely": -on- action distributes both ways, -on- action and -on- multiplication associate, and copy-of--on- action is scalar multiplication in .

Equivalently, and more simply but perhaps also more abstractly, a representation is a homomorphism from to .

The driving motivation behind all this is that we want to take abstract groups, which we don't know very much about in general, and turn them into linear transformations, which we do know quite a bit about and have lots of tools to deal with: dimension, trace, and so on.

As traditionally defined, note that a representation of a group (or heck, even a monoid) is still also a homomorphism into , but this time only considering the group structure of , with multiplication being composition, while ignoring the additive and scalar-multiplicative structure. We can naturally and uniquely extend any such representation to a representation of , just by multiplying and summing.

We haven't actually done anything interesting yet in this post, but I'm tired and don't want to set the bar too high for further posts, so let's stop.

I procrastinated posting this :( It's probably because the ending is so trippy.

First, what is Hilbert's Nullstellensatz?

Let be an algebraically closed field (a thing where you can add, subtract, multiply, divide by everything except zero, and find roots of all nonconstant polynomials), let be the polynomial ring you get from adding indeterminates to , and let be an ideal of this polynomial ring. We dumped some of the terminology in the last post already so I won't repeat that.

Then , where:
[list=1]
[*] means the set of zeroes of , i.e. the points of that give 0 when you plug them into any element of
[*] means the ideal of polynomials that are zero on all the points of the
[*] is the radical of , which means all elements that are some th root of an element of .
[/list]

It is "easy" to see that : If , we want to prove is zero wherever all of is zero. But some is in , and wherever is zero, is also zero, since fields are nice enough that you can't multiply nonzero things to get zero.

The other direction is more annoying. But having proven Zariski's lemma, we're already more than half done. Our last intermediate step is Weak Nullstellensatz: If is an algebraically closed field and is a proper ideal in the polynomial ring , then , the set of points at which all elements of are zero, is not empty.

Proof: Since is a field, it's Noetherian, so is Noetherian, so we can pick a maximal ideal that includes .

Then (via ideal preliminary) is a field. Furthermore we can view as a subfield of this. Now we apply Zariski's lemma to see that is module-finite over .

In fact, we claim that this means is isomorphic to . This part relies on the algebraic closedness of .

Suppose that we have some (treated as a field extension of ); we want to prove it is actually in . Then the module-finiteness also implies is integral over (by Lemma 2 of the Zariski post darn I'm forgetful, although in my defense, that lemma was optional for my cobbled-together proof). But if is the root of a polynomial with coefficients in , then we can factor into linear factors merely in because is algebraically closed. So we have is zero and is a field, which means one of those linear factors is zero and , as desired.

Okay. So , so in the field , for each there's a residue that it's in the same class as. So .

But the ideal is maximal, because by "plugging in" in any element of you can reduce it to something in , which differs from it by an element of . Therefore, adding any other element to would reduce it to cover the entirety of . So is maximal, yet and is proper, so .

This means that and we are done.

— end proof —

Now we can tackle the hard direction of Hilbert's nullstellensatz. Quick recap, it states that if be an algebraically closed field and is an ideal of the polynomial ring , then . We now want to prove that , or that if then .

To do this, we add another indeterminate to the polynomial ring to get . And we consider the ideal generated by . Yes, this is trippy. The star doesn't mean anything, I just haven't found an excuse to mean a star superscript in a while.

But this is useful because at any point where are all zero, that means is also zero which means is , so never completely evaluates to zero at any point. Therefore is zero and by the contrapositive of the weak nullstellensatz, . Therefore contains 1.

In other words we have an equation where each is an element of .

Divide everything by a high power of until there's no more of it in the numerator so you get something like:

where each is an element of , where is a magical indeterminate.

Then set to get the result:

And magically, we're done!

Okay, I don't know if you feel like these plugging in variables is fishy the same way I do, but we can do it unfancily like this:
[list]
[*]treat as a subring of
[*]take the homomorphism of that to where or , and supposing
[*]clearing denominators in the equation to make it involve and feature on the right side
[*]saying "And magically, we're done!"
[/list]

Really.

I should be several chapters ahead of where I am right now.

Okay so as dedicated readers (which I still suspect don't really exist) may recall, I was working on the proof of Hilbert's Nullstellensatz after refactoring out the part where I forgot how to prove Hilbert's basis theorem and walked through it on my own because the whole post was too long.

Well, I did that and worked on the shortened proof and patched holes in the draft I forgot about and wrote a bunch of short posts about tangents I ran into and, in spite of all of that, it turns out it was still too long. What a world.

I can't figure out any neat pieces to move out any more, so let's go through Wikipedia's steps. Just in the reverse order (which is actually the logical order of deduction so I don't know what's with Wikipedia). We start with Zariski's lemma. Zariski's lemma states that if you have two fields and and is ring-finite over , then is module-finite over . Hopefully you remember the definitions from the last post and their differences, and that it is easy to prove that module-finiteness is stronger than ring-finiteness.

But let's explore the similarities and differences with some examples. You can take the big field where is a 2015th root of unity, and the small field . The big field is ring-generated* from the small ring plus the single element . It is module-generated* from the small ring plus the elements .

* this is kind-of-not-really a made up term

Another example, consider , which is the big field of rational functions of , over the small field of reals, .

It's certainly not module-finite since you'd at least need the infinitely many elements , which are linearly independent — you can't pick some, slap on nonzero coefficients from , and get the zero polynomial. It's not ring-finite either, however, which is trickier to show. The most obvious "bottleneck elements" that can't be ring-finitely-generated are but to prove that these elements can't be ring-finitely generated isn't so obvious, and this exact idea doesn't generalize to finite fields since you'd only have finitely elements of the form . Fortunately, we'll need the proof of this for general fields, so we can detour to prove it without feeling like we wasted any effort.

Lemma 1. Let be a field. Then the field of rational functions in one free variable, , is not ring-finite over .

Proof. What we have to prove is that for any finite list of elements , the generated ring can't be the entirety of .

To do so, write each as a fraction of polynomials, take all the denominators, multiply them together, and multiply the result by an extra term of just to be safe to get a big polynomial . We claim:

[list=1]
[*]For each element , the product is a polynomial (i.e. element of ) for sufficiently large .
[*]There exists an element such that is not a polynomial (i.e. element of ) for any nonnegative integer .
[/list]

1 is true because for any , you expand it into a polynomial of and pick to be the largest degree of the terms.

2 is true by taking . For any positive integer , we recognize from elementary factoring tricks that which is not a polynomial because it is equal to a polynomial plus the obviously not-a-polynomial . (This is where the extra helps — it guarantees that the denominator is not just a constant.)

Since both are true, is in but not in . And we're done.

— end proof —

Note: Now, using this and the weird results we've built up over the last few posts, we can already prove Zariski's lemma straightforwardly. But since I already wrote all of this down, I'm going to include the requisites for the original proof in Fulton too, and then give both proofs.

Although this seems the natural start and endpoint for writing the lemma, it turns out what we eventually want to use is actually Point 2 above:

Corollary 1. Let be a field and let be a polynomial in . Then there exists an element such that is not a polynomial (i.e. element of ) for any nonnegative integer .

As it turns out we also need to understand some miscellaneous facts about integral elements over rings.

Integral elements are abstractified algebraic integers, as seen in 2013 Winter OMO #48. Also, TIL "omo" is a substring of "automorphism". They are defined thus:

Let be a subring of a commutative ring . Then is integral over if it's the root of a monic polynomial with coefficients in .

We assume is nice below — an integral domain, so we can divide by everything nonzero. Wow, "integral" is really a way overused word. Anyway, we have an interesting alternate definition of integral elements that will turn out useful later.

Lemma 2. Let . These three conditions are equivalent:

[list=1]
[*] is integral over .
[*] is module-finite over .
[*]There exists a subring of containing that is module-finite over .
[/list]

(It looks silly that we need both conditions 2 and 3 in this lemma, but remember, if is replaced with a general ring between and , 3 does not imply 2 unless is Noetherian.)

Proof. : Easy. If the monic polynomial that is a root of has degree , it means that we can express as a linear combination of , so it provides an inductive way to express every for as a linear combination of by factoring out , plugging in, and inducting. Therefore is generated as an -module by .

: Left as an exercise for the reader. Really, it's obvious enough that you should have figured it out in the time it took you to read this unnecessarily long sentence. :)

: This part of the proof is very weird. The difficulty here is that module-finiteness makes it easy to make up a nontrivial linear combination of (a.k.a. a polynomial in ) that sums to zero, but you desperately need the polynomial to be monic.

So, how? Take a finite bunch of things that module-generate the ring that contains .

Then, since it's a ring containing , if you multiply each by , the result stays in , and thus can be expressed as another linear combination with coefficients-in- of all the . So you can write equations about the , each of which has coefficients-in- for all but one ; that has, as a coefficient, plus something from .

The fact that these equations have the nontrivial solution of the themselves means the matrix of the coefficients of these equations has zero determinant. (It's OK to detour to the field of fractions of to prove this clearly.) But if you expand the determinant, you see it's a (degree-) monic polynomial of , since appears only on the main diagonal and nowhere else, so this monic polynomial is what we want.

— end proof —

A wild determinant appearing looks very odd indeed, but apparently the proof sketched on Wikipedia also involves determinants and is even weirder, so I guess I'll have to be satisfied with it. Onward, then:

Corollary 2. (still assuming that is a subring of a commutative ring ) The set of elements of integral over is a subring of containing .

Proof. Clearly it contains because for every , it's a root of the silly polynomial . So we just need to prove that it's closed under addition and multiplication. But this follows quickly from lemma 2: if are integral over then even is module-finite over , because module-finiteness is transitive, as we proved last post. And it's easy to see that each of is contained in . So we activate Lemma 2 using condition 3 to see that are all integral over . Thus the set of elements of integral over is closed under ring operations, and since it inherits those operations from , it's a ring.

(For the nice special case of , you can proved the closedness under ring-operations straightforwardly without lemma 2; if you have monic polynomials with degrees , you just keep using them to express powers of or or as linear combinations of , and then look at the ideals of the coefficients of those linear combinations. But you need the no-infinite-chain-of-descending-ideals property — Noetherian-ness, as you probably recall from the last post — of , so unfortunately this line of argument doesn't generalize to arbitrary rings, or at least, I don't know how to generalize it. Of course, there are a lot of things I don't know here.)

Corollary 2.2. Let be a field. The set of elements of integral over is just .

Proof. Suppose otherwise, that there's a rational function in integral over and is not a multiple of .

Perform the polynomial division to get , where is a polynomial and the remainder satisfies but is not zero.

Then and is trivially integral over , so is also integral over since integral elements over are a subring.

Then by the corollary is module-finite over ; suppose it's generated by the elements .

But consider the "rate-of-growth degree" of rational functions (which is apparently not the usual degree of rational functions), which satisfies .

We have so has elements of the form with arbitrary low rate-of-growth-degrees. But by Lemma 2, must be module-finite over ; if its module generators are , then these module generators getting multiplied by polynomials with nonnegative rate-of-growth degrees and then added together can never produce an element with denominator having a larger degree than the degree of the product of the denominators of all the . So we have a contradiction, as desired.

— end proof —

There. At long last, we return to Zariski, which states, recall, that if you have two fields and , where is bigger and contains , and if is ring-finite over (that is, for some finite bunch of ), then is module-finite over .

Proof. Suppose . We induct on .

The intuitive way to induct is to use the induction hypothesis on over . There's a problem, though, because it's not obvious that will be a field. This can be proved, but only by invoking a lot of prerequisites (which I spent the last few posts randomly proving), which is why I suppose Fulton's book doesn't do it that way. But it turns out that, even using all the stuff I proved, it's easier to induct on the pair of fields over . Of course, to do that we still have to prove that is a field. The result is that the two proofs are really similar:

First, we use the induction hypothesis on , the field generated by and that is a field by construction, and , which is just another way to write the field since it does no harm by letting field-generate stuff in something that's already a field.

So, the induction hypothesis gives us a bunch of that module-generate from .

Now, the goal is just to prove that is a field, so that . To prove this, we just need to prove that is algebraic over and then cite the field preliminary corollary.

The alternative is that is transcendental over , so that we just need to prove that this leads to a contradiction:

< begin proof by contradiction >

If is transcendental over , then would be isomorphic to the field of rational functions . Here the proofs diverge.

My proof using the claim from the last post:

By Lemma 1, is not ring-finite over . But we apply the claim:

[quote]Claim. If we have three rings , is Noetherian, is module-finite over , and is ring-finite over , then is ring-finite over .[/quote]

We take

is clearly Noetherian because it's a field. is module-finite over by the induction hypothesis as we applied it above. is ring-finite over ; that was a given condition for the theorem we're proving. So the claim holds, and is ring-finite over , a contradiction! (That was fast.)

The proof using the results above in this post:

Since is module-finite over , the conditions of Lemma 2 hold for each , so they're integral over , i.e. for each you can write a monic polynomial with coefficients in such that is a root.

Let be the product of ALL the denominators of coefficients (considered as rational functions = elements of ). Then you can turn each polynomial into a monic polynomial in with coefficients in , the ring of polynomials of . Basically you multiply the polynomial by a huge power of and absorb some of it into the variable and the rest of it into the coefficient to get a new polynomial. Write it out and see.

So each is integral over . Because the set of integral elements forms a subring, integralness is closed under ring-generation, so every element of is integral over , which means that for each element of , sufficiently large means is integral over .

But is given to be a field and contains . This means that, for each element of , sufficiently large means is integral over , which, by Corollary 2.2, is the same as saying that that is simply in . But since is transcendental over , this contradicts Corollary 1!

< end proof by contradiction >

Either way, we've proven that is in fact algebraic over . This is the same thing as being integral over , since is a field and you can make any polynomial monic by multiplying by a suitable constant.

To reiterate how the proof ends: invoking Lemma 2 again, we have that is module-finite over . And we can invoke the field preliminary corollary to see that is a field so . So is module-finite over is module-finite over , and by the transitivity of module-finiteness from the last post, we have that is module-finite over . As desired. Yay.

So finally we can announce:

— Zariski done! —

It's about time I introduce the two finteness criteria I was going on about, which feature prominently in Zariski's lemma, and investigate. I already sort-of defined them in the last post, but I should do it formally.

If we have commutative rings and where , then:
[list]
[*] is ring-finite over if you can get from by adding finitely many elements and letting them generate elements until you get a ring. Notationally, you have to be able to find elements so that .

Even more explicitly, this means that for every element , you can write it as a sum of elements of the form where and are nonnegative integers.
[*] is module-finite over if you can get from by adding finitely many elements and letting them generate elements as an -module, which is essentially a vector space over . Notationally, you have to be able to find elements so that .

Even more explicitly, this means that for every element , you can write it in the form where .
[/list]

The difference is that in the former case you can multiply the finitely many elements by themselves and by each other, whereas in the latter case you can only form linear combinations of the finitely many elements you choose, where "linear combinations" means using coefficients in . So it is clear that module-finiteness implies ring-finiteness, since you can pick the same finite set of elements.

One of the intuitive things about these properties is that they are transitive. One of these statements is easier to prove than the other.

Let be commutative rings.
[list]
[*]If is ring-finite over and is ring-finite over , then is ring-finite over .

To prove this, just take the union of the two finite sets of ring-generators. If and then . No big deal.
[*]If is module-finite over and is module-finite over , then is module-finite over .

To prove this, take the Cartesian product of the module bases and multiply each pair together.

Explicitly: suppose is module-generated over by and is module-generated over by ; then for every we can write it as for . We can write each as .

Then so is a finite set that is generated by as an -module.
[/list]

Anyway, one of the questions I got sidetracked into while trying to motivate each step in Hilbert's Nullstellensatz, which I thought would be similarly intuitive, was this:

[quote]Suppose we have three (let's say commutative) rings .
If is ring-finite over , is ring-finite over ?
If is module-finite over , is module-finite over ?[/quote]

Note that in both cases it is easy to deduce the related statement that is (ring|module)-finite over , by using the exact same set of generators.

Surprisingly and unintuitively to me, it turns out both statements are false. The first statement was easier for me to disprove: For any ring we can take and .

v_Enhance offered the startling counterexample to both claims at the same time:

where is a nonzero element such that .

After searching up some random websites I did figure out finally that the statement with module-finiteness is true if we assume that is Noetherian. That was essentially Corollaries 2 and 3 of the last post. (So the motivation for constructing the above example is clearer, because has to be not Noetherian.) The claim follows this way:

Claim. If we have three rings , is Noetherian, and is module-finite over , then is module-finite over .

Proof. By Corollary 3 of the last post, is Noetherian as an -module, so any submodule of it is finitely generated as an -module. is one such, which is just another way to say that it's module-finite over . QED. — end proof —

I had no idea how to go about the statement with ring-finiteness, though, at this point I asked my professor for something similar, and I got the theorem:

Claim. If we have three rings , is Noetherian, is module-finite over , and is ring-finite over , then is ring-finite over .

Proof: Let the ring-generators of over be , and let the module-generators of over be .

The module-finiteness means we can write each and , where , are in .

Now, set . This is a Noetherian ring by the "general" Hilbert basis theorem of the last post.

Subclaim: By construction of ,
Why is this true? Every element of is in so we can write it as the sum of stuff like for . We can plug in to reduce this into a sum of stuff like , then as long as the sum of the is more than 1, repeatedly plug in to reduce each such term into even more terms with less total degree, until we finally end with a sum of an insane number of terms of the form . (I just have to have excuses to use rare Greek letters.)

So from equation and Corollary 3, turns out to be a Noetherian -module. Then is a -submodule of and thus is a finitely generated -module. Therefore, is ring-finite over .

— end proof —

As you can see, the conditions are pretty strong. Now, the claim is still true without the condition that be module-finite over ; however, it appears the shortest path to proving that is citing Zariski's lemma. I'll get to that in the next post, but since my original goal with proving this claim was to use it in an intuitive proof of Zariski's lemma, I can only use the above claim right now.

I asked my professor and he even told me: [quote]I think a direct proof without going through the Zariski Lemma might be hard and even not easy to be found.[/quote]

"Hard and even not easy" probably sums it up.

Grrr. At some point I'm going to have transcribed down all the content of my abstract algebra textbooks and then some. I've got a little confession, I don't know what I'm doing, but if you want you can poke holes in these proofs so I can get closer to understanding. What?

I proved Hilbert's basis theorem a few posts ago while sidestepping the alternate definition that "every ideal contained in [a Noetherian ring] is finitely generated as a module". Then I forgot all about that characterization and was unable to understand when it popped up in discussions of finiteness criteria. If you don't know what the finiteness criteria I'm referring to yet, I just mean when rings are generated as modules or rings by adding finitely many elements over other rings, but it doesn't matter.

It turns out furthermore that stuff we prove is more useful if we generalize the idea of Noetherian-ness to modules.

Let be a commutative ring (I am not brave enough to handle noncommutative ones) and let be an -module (which means we have a function satisfying various clear multiplication axioms that I will irresponsibly not list here.)

[list=1]
[*] is Noetherian_1 if any infinite chain of submodules is eventually constant.
[*] is Noetherian_2 if every submodule of is finitely generated as a module.
[/list]

When we speak of Noetherian rings, that means they're Noetherian when we treat them as modules of themselves. Another note to self: in a ring , an ideal and an -module are the same zarking thing and it means the same thing for them to be finitely generated, which is the thing the Noetherian-ness criterion is all about. I am unbelievably obtuse sometimes. So you can just read the proofs below while mentally replacing "module" or "submodule" by "ideal".

First, let's prove the equivalence.

(standard, ~ DF 12.1 Theorem 1) An -module is Noetherian_1 iff it is Noetherian_2.

Proof. () Let be a submodule of . Starting from the zero module, attempt to pick a sequence of elements such that is in but not in the module generated by . This cannot go on forever since the modules generated by are an infinite chain of ascending modules; the only way it stops is if at some point , at which point we have proven that generate .

() Let be an infinite chain of submodules of . Then the union of all of them is a module (verify the axioms), which is finitely generated. Each of the generators appears in one of the modules ; if the maximum of those then the chain is constant after .

— end proof —

Okay, weirder stuff:

Let be an -module, and let be a submodule. Then is Noetherian iff and are both Noetherian.

Proof. () If is Noetherian, then any submodule of is a submodule of and is thus finitely generated, so is Noetherian. As for , any submodule of it corresponds to a submodule of containing . That submodule is also finitely generated, and the image of those generators in finitely generate it, so is Noetherian.

Now assume and are both Noetherian; we want to prove is Noetherian. Let be a submodule of ; we want to prove it's finitely generated. Consider . Then is a submodule of and is finitely generated, say by . Also, observe that the obvious mapping from into is well-defined, a homomorphism, and injective, so is isomorphic to a submodule of and is also finitely generated. Take preimages of those generators in and call them ; those elements and the generators of together generate because you can pick elements from to reach the equivalence class in of any element and then reach the actual element in that class using .

— end proof —

Corollary 1. If is a Noetherian -module, then (the obvious -module of -tuples of ) is Noetherian.

Proof. Induct on . It's trivial when . If you let be the submodule of tuples of with only the first element nonzero, then is isomorphic to and is a Noetherian -module, and is isomorphic to and is a Noetherian -module by the induction hypothesis, so is Noetherian.

Corollary 2. If is a Noetherian -module, then any module of the form for is also a Noetherian -module. (The condition is just what we mean when we say is module-finite over .)

Proof. There's a surjective homomorphism from to sending the tuple with th element 1 and all others 0 to the element . So is isomorphic to where is the kernel of , and is Noetherian by Corollary 1, so is Noetherian by the theorem.

Corollary 3. If is a Noetherian ring, then an -module is Noetherian iff it is finitely generated.

Proof. If is Noetherian then any submodule of , including itself, is finitely generated. On the other hand, if it's finitely generated by , that means , so it's Noetherian by Corollary 2.

— end corollaries —

Separately:

If is a Noetherian ring and is an ideal, then is a Noetherian ring.

Proof. Any ideal in is for some ideal of . Then is finitely generated and the image of the generators in finitely generate . Alternatively, any chain of ascending ideals in likewise corresponds to a chain of ascending ideals in , which is eventually constant. — end proof —

(Note that this does not instantly follow from the above theorem, which proves that is Noetherian if you treat it as a -module. It appears you can deduce the consequence you want from this, but it's so easy by itself anyway that I don't think it's worth it.)

And, finally: If is a Noetherian ring and is ring-finite over , then is Noetherian.

Proof. Let be ring-generated over by elements . There is the obvious homomorphism from the polynomial ring in indeterminates to ; if is the kernel of this then . By the Hilbert basis theorem is Noetherian and by the above theorem is Noetherian. — end proof —

Wait, I finally figured out the misunderstanding that confused me for a long time; I had my criteria for ideals and quotient rings mixed up. Darn. Also, it's so nice to be able to assume rings are commutative...

(again this is easy stuff you should know this already, self)

As you recall, an ideal of a commutative ring is a subset of that is closed under addition (with another element of ) and under multiplication by any element of . Given an ideal and a ring, you can get a new ring — a quotient ring (also called a factor ring among other names) — by taking the quotient , which contains as elements sets of the form for . It is easy/boring to check that ring operations are well-defined, so I won't do that.

is defined to be prime if implies or . Some authors additionally require that be proper, i.e. not equal to itself.

→ is prime iff is an integral domain. This is a straightforward translation of the definition. The additional requirement depends on if you accept the single-element , in which serves as both the additive and the multiplicative identity, to be a ring.

is defined to be maximal if is proper and no ideal exists such that
→ is maximal iff is a field. This is more or less what we proved in the last post, once we note that any ideal of satisfying the above equation gives rise to an ideal in and vice-versa, which is one of those intuitive homomorphism theorems.

I'm just spamming blog posts now.

Let be a commutative ring. is a field iff it has only two ideals, itself and .

Proof. If is a field then any nonzero ideal contains a nonzero element , which generates , which generates the whole field, so . So and are indeed the only two ideals.

If is not a field then there's a noninvertible element and the ideal generated by is neither nor .

— end proof —

(Theorem 2.16 in Jacobson) Let be a field and let be algebraic over with minimal polynomial . Then is:
[list]
[*]a field if is irreducible;
[*]not even an integral domain if is reducible.
[/list]

Proof. If is irreducible, we claim has no ideals other than itself and . So suppose we have an ideal of . Then the preimage of under the obvious homomorphism from the polynomial ring to would be an ideal of . Since is a principal ideal domain, as seen in the last post, is generated by an element . Also contains , so is a multiple of . Since is irreducible we must have or (up to a constant factor), which respectively means or , as desired. Thus is a field by the statement earlier in this post.

If is reducible as then and are nonzero in but is zero, so is not an integral domain.

Corollary. If is a field and is embeddable in an integral domain, then is a field.

Proof. As above, is either a field or contains zero divisors, and the latter case is impossible if is embeddable in an integral domain.

This is easy you should know this already, self.

Let be a commutative ring. Then is a field iff the field of polynomials is a principal ideal domain.

Proof. Suppose is a field. Given an ideal in , take a polynomial with minimal degree in . This generates every element of because you can use the division algorithm to divide any such element by and any nonzero remainder would be a polynomial of less degree than in , contradicting our choice of .

Suppose is not a field because has no inverse. Then the ideal generated by and in is not principal, because if it had a generator then it would have to be a constant to generate , but it can't generate , which means it can't generate .