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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.

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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 $X\mapsto (1,0)$, $Y\mapsto (0,1)$.

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  • $\begingroup$ Thank you for your answer! If possible, could you also comment on $\mathbb{Z}_{(2)}$ as well? I have been trying to show that it is not a homomorphic image but to no avail (or maybe it is a homomorphic image..?) $\endgroup$ – dhk628 May 18 '17 at 23:41
  • $\begingroup$ I really don't know. The localisation at an element $\mathbf Z_n$ is of course the quotient $\mathbf Z[X]/(nX-1)$, and the localisation at a prime is a direct limit of such quotients, but being a direct limit doen't prove it's not such a quotient. If I find something, I'll post it. $\endgroup$ – Bernard May 18 '17 at 23:46
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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.

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  • $\begingroup$ I suppose that a homomorphic image of $k[x_1,\dots,x_n]$ contains $k$, and $\mathbf{Z}_{(p)}$ doesn't contain any field. $\endgroup$ – user26857 May 19 '17 at 6:35
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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.

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