Example of a Noetherian ring that is not a homomorphic image of $\mathbb{Z}[t_1,\dots,t_n]$ or $F[t_1,\dots,t_k]$ I need to prove or disprove the claim in the title, and I do believe there is indeed a Noetherian ring that is not a homomorphic image of $\mathbb{Z}[t_1,\dots,t_n]$ or $F[t_1,\dots,t_k]$, where $F$ is a field. I'm thinking about local rings, such as $\mathbb{Z}_{(2)}$ or a direct sum of Noetherian rings, such as $\mathbb{Z} \oplus \mathbb{Z}$ will do the trick, but I'm not exactly sure how to show they are not homomorphic images of the above two types of rings.
 A: A ring of formal power series $\,K[[X]]$ or $\;\mathbf Z[[X]]$ is such an example.
$\mathbf Z\times\mathbf Z$ is the homomorphic image of $\mathbf Z[X,Y]$ by the mapping $X\mapsto (1,0)$, $Y\mapsto (0,1)$.
A: Let $R = \mathbf{Z}_{(p)}$ where $p$ is a prime. Let $\frac{a_1}{b_1}, \dots, \frac{a_n}{b_n} \in R$ be a set of finitely many elements. Consider every rational number $\frac{a}{b}$ that can be expressed as a polynomial in $\frac{a_1}{b_1}, \dots, \frac{a_n}{b_n}$. The denominator of such a rational number, in reduced form, must be some monomial in $b_1,\dots,b_n$, or a divisor thereof. Since there are infinitely many primes, $\frac{a_1}{b_1}, \dots, \frac{a_n}{b_n}$ cannot generate $R$ as a $\mathbf{Z}$-algebra. Thus $R$ is not a homomorphic image of $\mathbf{Z}[x_1,\dots,x_n]$ for any $n$.
Also in $R$, the element $p$ is not invertible, thus $R$ cannot be a homomorphic image of $k[x_1,\dots,x_n]$ unless $p$ is also not invertible in $k$. This implies that $\operatorname{char} k$ must be $p$. But in this case $p = 0$ in $R$, a contradiction.
A: Here is a much more general result:
Note that finitely generated $\mathbb Z$- or $F$-algebras are Jacobson rings. In particular any noetherian local ring, which has a non-maximal prime ideal is a an example of a noetherian ring, that is not a quotient of a polynomial ring over a field or the integers.
