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Let $R$ be a commutative ring with $1$, and let $K$ be a field. We know that $R$ is Noetherian iff every ideal of $R$ is finitely generated as an ideal.

Question 1: If $R$ is Noetherian, is every subring of $R$ finitely generated as a ring?

Is there a simple (counter)example, preferrably in $R:=K[x_1,\ldots,x_n]$?

Question 2: if $f:K[x_1,\ldots,x_n]\rightarrow K[y_1,\ldots,y_n]$ is a ring homomorphism, can $\mathrm{Im}(f)$ be nonfinitely generated (as a ring)?

$K$ is any field that can be implemented in computer algebra systems, such as finite fields and $\mathbb{Q}$.

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Dear Leon: If you don't follow Jack's suggestion, you could take $R=K=\mathbb Q,n=0$. –  Pierre-Yves Gaillard Dec 29 '11 at 16:55
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@Leon: I'm not sure we made ourselves clear. You can also take n=1 in these examples. For instance take K to be the function field Z/2Z(x,y) as is common in CAS, n=1, R=K[t], but then the subring S=Z/2Z[x,y/x,y/x^2,y/x^3,...] of R is not noetherian. However K is not contained in S, so S is not a sub K-algebra of R, merely a subring of R. I believe magma has no trouble creating k, K, or R, but I'm not it can represent S easily. magma.maths.usyd.edu.au/magma/handbook/text/395#4071 –  Jack Schmidt Dec 29 '11 at 18:31
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@Leon: any finite subset of S has a maximum degree (say n) denominator, which is only degree 0 in x if the numerator is degree 0 in y, so y/x^(n+1) is not in the ring generated by that finite subset. Zev's example is even easier to verify, it just took more characters to type it as a comment. –  Jack Schmidt Dec 29 '11 at 20:39
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You still need to be careful about "ring" or "K-algebra". More or less nothing is going to work for "ring", and I think almost everything works for "K-algebra". For instance, take K=Q to be the rational numbers, R=Q[t], and S=Q. Then S is a finitely generated K-algebra, but S is not finitely generated as a ring. (This is Pierre-Yves Gaillard's comment again). –  Jack Schmidt Dec 29 '11 at 20:43
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@Pierre: You are certainly not slow; but I am :). You are right, I think my argument above is flawed. I get a little confused by your arguments also, so I rewrote my argument. If I am not mistaken again, we have $$K[x,\frac{y}{x},\frac{y}{x^2},\frac{y}{x^3},\ldots]\!=\!\{f(x)\!+\!\frac{yg(x,‌​y)}{x^k};f\!\in\!K[x],g\!\in\!K[x,y],k\!\in\!\mathbb{N}\}.$$ Now for any finite subset $\{f_i(x)\!+\!\frac{yg_i(x,y)}{x^i};1\!\leq\!i\!\leq\!k\}$, the subring generated by it is contained in $K[x,\frac{y}{x},\ldots,\frac{y}{x^k}]$, but the latter does not contain $\frac{y}{x^{k+1}}$. $\blacksquare$ –  Leon Lampret Jan 1 '12 at 20:25

3 Answers 3

up vote 7 down vote accepted

The purpose of this post is to prove

(A) If $K$ is an infinite field, then $K$ is not a finitely generated ring.

The main ingredients are

(Z) (Zariski Lemma) Let $k$ be a field and $B$ a finitely generated $k$-algebra. If $B$ is a field, then it is a finite degree extension of $k$;

and

(B) let $A$ be a unique factorization domain with infinitely many association classes of prime elements, let $K$ be its field of fractions, and let $L$ be a finite degree extension of $K$. Then $L$ is not a finitely generated $A$-algebra.

Recall that these association classes correspond bijectively to the nonzero principal prime ideals. The assumptions in (B) are satisfied in particular by $\mathbb Z$ thanks to Euclid's observation that there are infinitely many prime numbers, and Euclid's argument also applies to polynomial rings in one indeterminate over a field.

Note that (Z) and (B) imply immediately (A). Indeed, to be finitely generated over its prime ring, $K$ must at least be finitely generated over its prime field $K_0$. So, by (Z), we can assume that $K$ is of finite degree over $K_0$, and thus, that $K$ is a finite degree extension of $\mathbb Q$. But then we can apply (B) to $\mathbb Z$. QED

Consider a fourth statement:

(C) Let $A$ be a subring of a domain $B$, and $B$ be integral over $A$. Then $A$ is a field if and only if $B$ is a field.

We'll see that, roughly speaking, (C) $\implies$ (B) $\implies$ (Z). All this is essentially contained in Atiyah and MacDonald's wonderful book Introduction to Commutative Algebra.

Proof of (C). Assume $B$ is a field, and let $x$ be a nonzero element of $A$. We have $$ x^{-n}+a_{n-1}\ x^{1-n}+\cdots+a_0=0,\quad a_i\in A, $$ and thus $$ -x^{-1}=a_{n-1}+\cdots+a_0\ x^{n-1}\in A. $$ We won't need the converse, but let's prove it anyway. Assume $A$ is a field, and let $y$ be a nonzero element of $B$. We have $$ y^n+a_{n-1}\ y^{n-1}+\cdots+a_0=0,\quad a_i\in A. $$ and thus $$ y\ (y^{n-1}+a_{n-1}\ y^{n-2}+\cdots+a_1)=-a_0. $$ Assuming, as we may, that $n$ is minimum, we have $a_0\neq0$, and we see that $y$ is invertible. QED

Proof of (B). Assume by contradiction $$ L=A[x_1,\dots,x_n]. $$ Let $a$ be the product of the denominators of the coefficients of the minimal polynomials of the $x_i$ over $K$. [Sorry for that sentence...] Then $L$ is integral over $A':=A[a^{-1}]$. In view of our assumptions on $A$, the ring $A'$ is not a field, contradicting (C). QED

Proof of (Z). We argue by induction on $n$. The result being clear if $n=1$, assume $n\ge2$. Form the ring $A:=k[x_1]$ and its fraction field $K:=k(x_1)$. By the inductive hypothesis, $B$ is of finite degree over $K$, and we only need to show that $x_1$ is algebraic over $k$. But if $x_1$ were transcendent over $k$, we would get a contradiction by observing that $A$ would satisfy the assumptions of (B). QED

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This is false. Consider the ring $$R= k(x_1,x_2,\ldots)$$ where $k$ is a field and the $x_i$ are indeterminates. Then, as a field, $R$ is certainly noetherian, but the subring $k[x_1,x_2,\ldots]$ is not finitely generated as a ring.

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Because if $K[x_1,x_2,\ldots]\leq K(x_1,x_2,\ldots)$ were finitely generated as a ring, it would contain only finitely many variables. Thanks. Is there still an easy example of a non-finitely generated subring of $K[x_1,\ldots,x_n]$? –  Leon Lampret Dec 29 '11 at 16:21
    
What does $k(x_1, k_2 \ldots )$ mean? –  user38268 Dec 29 '11 at 16:24
    
@Ben: It means the fraction field of $k[x_1,x_2,\ldots]$, the polynomial ring over $k$ in infinitely many indeterminates $x_i$. –  Zev Chonoles Dec 29 '11 at 16:25
    
@ZevChonoles Thanks I was going to ask some other question then realised the difference between the brackets. –  user38268 Dec 29 '11 at 16:27
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The moral of the story is that sub*rings* don't inherit very many nice properties of the rings that they are sitting in. Except having no zero-divisors, for instance, so a subring of an integral domain still is an integral domain. –  Lubin Dec 30 '11 at 5:53

No.

From Wikipedia:

"However, a non-Noetherian ring can be a subring of a Noetherian ring:

The ring of rational functions generated by $x$ and $y/x^n$ over a field $k$ is a subring of the field $k(x,y)$ in only two variables.

Indeed, there are rings that are left Noetherian, but not right Noetherian, so that one must be careful in measuring the "size" of a ring this way."

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