Zariski's lemma (can't think of a witty title ):

by math_explorer, Apr 24, 2015, 3:59 AM

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 $L$ and $K$ and $L$ is ring-finite over $K$ , then $L$ is module-finite over $K$ . 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 $L = \mathbb{Q}[\omega]$ where $\omega$ is a 2015th root of unity, and the small field $K = \mathbb{Q}$ . The big field is ring-generated* from the small ring plus the single element $\omega$ . It is module-generated* from the small ring plus the elements $1, \omega, \omega^2, \ldots, \omega^{2014}$ .

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

Another example, consider $\mathbb{R}(x)$ , which is the big field of rational functions of $x$ , over the small field of reals, $\mathbb{R}$ .

It's certainly not module-finite since you'd at least need the infinitely many elements $x, x^2, x^3, x^4, \ldots$ , which are linearly independent — you can't pick some, slap on nonzero coefficients from $\mathbb{R}$ , 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 $\frac{1}{x}, \frac{1}{x+1}, \frac{1}{x+2}, \cdots$ 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 $\frac{1}{x+k}$ . 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 $F$ be a field. Then the field of rational functions in one free variable, $F(x)$ , is not ring-finite over $F$ .

Proof. What we have to prove is that for any finite list of elements $v_1, \ldots, v_n \in F(x)$ , the generated ring $F[v_1, \ldots, v_n]$ can't be the entirety of $F(x)$ .

To do so, write each $v_i$ as a fraction of polynomials, take all the denominators, multiply them together, and multiply the result by an extra term of $x^{1337}$ just to be safe to get a big polynomial $P$ . We claim:

For each element $w \in F[v_1, \ldots, v_n]$ , the product $wP^m$ is a polynomial (i.e. element of $F[x]$ ) for sufficiently large $m$ .
There exists an element $z \in F(x)$ such that $zP^m$ is not a polynomial (i.e. element of $F[x]$ ) for any nonnegative integer $m$ .

1 is true because for any $w$ , you expand it into a polynomial of $v_1, \ldots, v_n$ and pick $m$ to be the largest degree of the terms.

2 is true by taking $z = 1/(P-1)$ . For any positive integer $m$ , we recognize from elementary factoring tricks that $\frac{P^m}{P-1} = P^{m-1} + \cdots + 1 + \frac{1}{P-1},$ which is not a polynomial because it is equal to a polynomial plus the obviously not-a-polynomial $\frac{1}{P-1}$ . (This is where the extra $x^{1337}$ helps — it guarantees that the denominator is not just a constant.)

Since both are true, $z$ is in $F(x)$ but not in $F[v_1, \ldots, v_n]$ . 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 $F$ be a field and let $P$ be a polynomial in $F[x]$ . Then there exists an element $z \in F(x)$ such that $zP^m$ is not a polynomial (i.e. element of $F[x]$ ) for any nonnegative integer $m$ .

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 $R$ be a subring of a commutative ring $S$ . Then $v \in S$ is integral over $R$ if it's the root of a monic polynomial with coefficients in $R$ .

We assume $S$ 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 $v \in S$ . These three conditions are equivalent:

$v$ is integral over $R$ .
$R[v]$ is module-finite over $R$ .
There exists a subring of $S$ containing $R[v]$ that is module-finite over $R$ .

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

Proof. $(1 \Longrightarrow 2)$ : Easy. If the monic polynomial that $v$ is a root of has degree $d$ , it means that we can express $v^d$ as a linear combination of $1, v, \ldots, v^{d-1}$ , so it provides an inductive way to express every $v^e$ for $e \geq d$ as a linear combination of $1, v, \ldots, v^{d-1}$ by factoring out $v^d$ , plugging in, and inducting. Therefore $R[v]$ is generated as an $R$ -module by $1, v, \ldots, v^{d-1}$ .

$(2 \Longrightarrow 3)$ : 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.

$(3 \Longrightarrow 1)$ : 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 $1, v, v^2, \ldots$ (a.k.a. a polynomial in $v$ ) that sums to zero, but you desperately need the polynomial to be monic.

So, how? Take a finite bunch of things $w_1, \ldots, w_n$ that module-generate the ring $R'$ that contains $R[v]$ .

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

The fact that these equations have the nontrivial solution of the $w_i$ themselves means the matrix of the coefficients of these equations has zero determinant. (It's OK to detour to the field of fractions of $R'$ to prove this clearly.) But if you expand the determinant, you see it's a (degree- $n$ ) monic polynomial of $v$ , since $v$ 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 $R$ is a subring of a commutative ring $S$ ) The set of elements of $S$ integral over $R$ is a subring of $S$ containing $R$ .

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

(For the nice special case of $R = \mathbb{Z}$ , you can proved the closedness under ring-operations straightforwardly without lemma 2; if you have monic polynomials $P(x), Q(y)$ with degrees $n, m$ , you just keep using them to express powers of $x+y$ or $x-y$ or $xy$ as linear combinations of $x^iy^j, 0 \leq i < n, 0 \leq j < n$ , 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 $\mathbb{Z}$ , 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 $F$ be a field. The set of elements of $F(x)$ integral over $F[x]$ is just $F[x]$ .

Proof. Suppose otherwise, that there's a rational function $P/Q$ in $F(x)$ integral over $F[x]$ and $P$ is not a multiple of $Q$ .

Perform the polynomial division $P/Q$ to get $P = KQ + R$ , where $K$ is a polynomial and the remainder $R$ satisfies $\deg R < \deg Q$ but is not zero.

Then $P/Q = K + R/Q$ and $K$ is trivially integral over $F[x]$ , so $R/Q$ is also integral over $F[x]$ since integral elements over $F[x]$ are a subring.

Then by the corollary $F[x][R/Q]$ is module-finite over $F[x]$ ; suppose it's generated by the elements $\{u_i\}$ .

But consider the "rate-of-growth degree" of rational functions $\deg V/W := \deg V - \deg W$ (which is apparently not the usual degree of rational functions), which satisfies $\deg AB = \deg A + \deg B$ .

We have $\deg R/Q < 0$ so $F[x][R/Q]$ has elements of the form $(R/Q)^n$ with arbitrary low rate-of-growth-degrees. But by Lemma 2, $F[x][R/Q]$ must be module-finite over $F[x]$ ; if its module generators are $w_1 \ldots w_n$ , 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 $w_i$ . 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 $L$ and $K$ , where $L$ is bigger and contains $K$ , and if $L$ is ring-finite over $K$ (that is, $L = K[v_1, \ldots, v_n]$ for some finite bunch of $\{v_i\} \subseteq L$ ), then $L$ is module-finite over $K$ .

Proof. Suppose $L = K[v_1, \ldots, v_n]$ . We induct on $n$ .

The intuitive way to induct is to use the induction hypothesis on $K[v_1, \ldots, v_{n-1}]$ over $K$ . There's a problem, though, because it's not obvious that $K[v_1, \ldots, v_{n-1}]$ 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 $K[v_1, \ldots, v_n]$ over $K[v_n]$ . Of course, to do that we still have to prove that $K[v_n]$ is a field. The result is that the two proofs are really similar:

First, we use the induction hypothesis on $K' = K(v_n)$ , the field generated by $K$ and $v_n$ that is a field by construction, and $L' = K(v_n)[v_1, \ldots, v_{n-1}]$ , which is just another way to write the field $L$ since it does no harm by letting $v_n$ field-generate stuff in something that's already a field.

So, the induction hypothesis gives us a bunch of $w_1, \ldots, w_m$ that module-generate $L$ from $K(v_n)$ .

Now, the goal is just to prove that $K[v_n]$ is a field, so that $K[v_n] = K(v_n)$ . To prove this, we just need to prove that $v_n$ is algebraic over $K$ and then cite the field preliminary corollary.

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

< begin proof by contradiction >

If $v_n$ is transcendental over $K$ , then $K(v_n)$ would be isomorphic to the field of rational functions $K(x)$ . Here the proofs diverge.

My proof using the claim from the last post:

By Lemma 1, $K(v_n)$ is not ring-finite over $K$ . But we apply the claim:

Quote:

Claim. If we have three rings $A \subsetneq B \subsetneq C$ , $A$ is Noetherian, $C$ is module-finite over $B$ , and $C$ is ring-finite over $A$ , then $B$ is ring-finite over $A$ .

We take
$\begin{align*} A &= K \\ B &= K(v_n) \\ C &= L = K[v_1, \ldots, v_n] \end{align*}$
$A$ is clearly Noetherian because it's a field. $C$ is module-finite over $B$ by the induction hypothesis as we applied it above. $C$ is ring-finite over $A$ ; that was a given condition for the theorem we're proving. So the claim holds, and $B = K(v_n)$ is ring-finite over $A$ , a contradiction! (That was fast.)

The proof using the results above in this post:

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

Let $c$ be the product of ALL the denominators of coefficients (considered as rational functions = elements of $K(v_n)$ ). Then you can turn each polynomial $P_i$ into a monic polynomial in $cv_i$ with coefficients in $K[v_n]$ , the ring of polynomials of $v_n$ . Basically you multiply the polynomial by a huge power of $c$ 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 $cv_i$ is integral over $K[v_n]$ . Because the set of integral elements forms a subring, integralness is closed under ring-generation, so every element of $K[cv_1, \ldots, cv_n]$ is integral over $K[v_n]$ , which means that for each element $x$ of $K[v_1, \ldots, v_n]$ , sufficiently large $\ell$ means $c^\ell x$ is integral over $K[v_n]$ .

But $K[v_1, \ldots, v_n]$ is given to be a field and contains $K(v_n)$ . This means that, for each element of $K(v_n)$ , sufficiently large $\ell$ means $c^\ell x$ is integral over $K[v_n]$ , which, by Corollary 2.2, is the same as saying that that $c^\ell x$ is simply in $K[v_n]$ . But since $v_n$ is transcendental over $K$ , this contradicts Corollary 1!

< end proof by contradiction >

Either way, we've proven that $v_n$ is in fact algebraic over $K$ . This is the same thing as being integral over $K$ , since $K$ 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 $K[v_n]$ is module-finite over $K$ . And we can invoke the field preliminary corollary to see that $K[v_n]$ is a field so $K(v_n) = K[v_n]$ . So $K[v_1, \ldots, v_n]$ is module-finite over $K(v_n) = K[v_n]$ is module-finite over $K$ , and by the transitivity of module-finiteness from the last post, we have that $K[v_1, \ldots, v_n]$ is module-finite over $K$ . As desired. Yay.

So finally we can announce:

— Zariski done! —

algebra

abstract algebra

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I am a dedicated reader (who hopefully exists)! Just haven't learned any abstract algebra :/

by briantix, Apr 25, 2015, 4:43 PM

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Zariski's lemma (can't think of a witty title ):

by math_explorer, Apr 24, 2015, 3:59 AM

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