Towers of commutative rings and finiteness criteria (edited)

by math_explorer, Apr 20, 2015, 3:07 AM

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 $R$ and $S$ where $S \subseteq R$ , then:

$R$ is ring-finite over $S$ if you can get $R$ from $S$ by adding finitely many elements and letting them generate elements until you get a ring. Notationally, you have to be able to find elements $v_1, \ldots, v_n$ so that $R = S[v_1, \ldots, v_n]$ .

Even more explicitly, this means that for every element $r \in R$ , you can write it as a sum of elements of the form $sv_1^{\alpha_1}v_2^{\alpha_2}\cdots v_n^{\alpha_n}$ where $s \in S$ and $\alpha_i$ are nonnegative integers.
$R$ is module-finite over $S$ if you can get $R$ from $S$ by adding finitely many elements and letting them generate elements as an $S$ -module, which is essentially a vector space over $S$ . Notationally, you have to be able to find elements $w_1, \ldots, w_m$ so that $R = Sw_1 + Sw_2 + \cdots + Sw_m$ .

Even more explicitly, this means that for every element $r \in R$ , you can write it in the form $s_1w_1 + \cdots + s_mw_m$ where $s_i \in S$ .

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 $K$ . 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 $A \subseteq B \subseteq C$ be commutative rings.

If $B$ is ring-finite over $A$ and $C$ is ring-finite over $B$ , then $C$ is ring-finite over $A$ .

To prove this, just take the union of the two finite sets of ring-generators. If $B = A[u_1, \ldots, u_n]$ and $C = B[v_1, \ldots, v_N]$ then $C = A[u_1, \ldots, u_n, v_1, \ldots, v_N]$ . No big deal.
If $B$ is module-finite over $A$ and $C$ is module-finite over $B$ , then $C$ is module-finite over $A$ .

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

Explicitly: suppose $B$ is module-generated over $A$ by $w_1, \ldots, w_m$ and $C$ is module-generated over $B$ by $\psi_1, \ldots, \psi_\mu$ ; then for every $c \in C$ we can write it as $c = b_1\psi_1 + \cdots + b_\mu\psi_\mu$ for $b_i \in B$ . We can write each $b_i$ as $b_i = a_{i1}w_1 + \cdots + a_{im}w_m$ .

Then $\begin{align*} c &= (a_{11}w_1 + \cdots + a_{1m}w_m)\psi_1 + (a_{21}w_1 + \cdots)\psi_2 \\ &\qquad + \cdots + (\cdots + a_{\mu m}w_m)\psi_\mu \\ &= \sum_{i=1}^\mu \sum_{j=1}^m a_{ij}w_i\psi_j \end{align*}$ so $\{w_i\psi_j \mid i = 1, \ldots, n;\; j = 1, \ldots, \mu\}$ is a finite set that $C$ is generated by as an $A$ -module.

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 $A \subsetneq B \subsetneq C$ .
If $C$ is ring-finite over $A$ , is $B$ ring-finite over $A$ ?
If $C$ is module-finite over $A$ , is $B$ module-finite over $A$ ?

Note that in both cases it is easy to deduce the related statement that $C$ is (ring|module)-finite over $B$ , 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 $A$ we can take $B = A[x, xy, xy^2, xy^3, \ldots]$ and $C = A[x, y]$ .

v_Enhance offered the startling counterexample to both claims at the same time:
$\begin{align*} A &= \mathbb{Z}[x_1, x_2, x_3, \ldots] \\ B &= \mathbb{Z}[x_1, x_2, x_3, \ldots, \varepsilon x_1, \varepsilon x_2, \varepsilon x_3, \ldots] \\ C &= \mathbb{Z}[\varepsilon, x_1, x_2, x_3, \ldots] \end{align*}$
where $\varepsilon$ is a nonzero element such that $\varepsilon^2 = 0$ .

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

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

Proof. By Corollary 3 of the last post, $C$ is Noetherian as an $A$ -module, so any submodule of it is finitely generated as an $A$ -module. $B$ is one such, which is just another way to say that it's module-finite over $A$ . 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 $A \subsetneq B \subsetneq C$ , $A$ is Noetherian, $C$ is module-finite over $B$ , and $C$ is ring-finite over $A$ , then $B$ is ring-finite over $A$ .

Proof: Let the ring-generators of $C$ over $A$ be $v_1, \ldots, v_n$ , and let the module-generators of $C$ over $B$ be $w_1, \ldots, w_m$ .

The module-finiteness means we can write each $v_i=\sum_{s=1}^m p_{is} w_s$ and $w_iw_j = \sum_{t=1}^m q_{ijt} w_t$ , where $p_{is}$ , $q_{ijt}$ are in $B$ .

Now, set $B'=A[p_{is}, q_{ijt}]$ . This is a Noetherian ring by the "general" Hilbert basis theorem of the last post.

Subclaim: By construction of $B'$ , $C=B'w_1+ \ldots +B'w_m.\qquad(*)$
Why is this true? Every element of $C$ is in $A[v_1, \ldots, v_n]$ so we can write it as the sum of stuff like $a_{\mathrm{junk}}v_1^{e_1}\cdots v_n^{e^n}$ for $a_{\mathrm{junk}} \in A$ . We can plug in $v_i = \sum_{s=1}^m p_{is} w_s$ to reduce this into a sum of stuff like $a_{\mathrm{junk}}w_1^{\eta_1}\cdots w_m^{\eta_m}$ , then as long as the sum of the $\eta_i$ is more than 1, repeatedly plug in $w_iw_j = \sum{t=1}^m q_{ijs} w_s$ 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 $a_{\mathrm{too much junk}}w^i$ . (I just have to have excuses to use rare Greek letters.)

So from equation $(*)$ and Corollary 3, $C$ turns out to be a Noetherian $B'$ -module. Then $B$ is a $B'$ -submodule of $C$ and thus $B$ is a finitely generated $B'$ -module. Therefore, $B$ is ring-finite over $A$ .

— end proof —

As you can see, the conditions are pretty strong. Now, the claim is still true without the condition that $C$ be module-finite over $B$ ; 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.

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

This post has been edited 5 times. Last edited by math_explorer, Apr 27, 2015, 9:26 AM

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Towers of commutative rings and finiteness criteria (edited)

by math_explorer, Apr 20, 2015, 3:07 AM

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