If a ring is Noetherian, then every subring is finitely generated?

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.

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

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."