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 $\mathbb{Q}$.
Best Answer
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.