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Consider $K$ field and consider the ring $R=K[X^2,X^3]\subset K[X]$. It is clear that $R$ is not a Dedekind domain, since with the element $X$ one immediately see that it is not integrally closed. But $R$ is a Noetherian domain and every non trivial prime ideal is maximal. I have a proof for the last two part, but it is a bit heavy and maybe someone can help me finding a better proof! Thanks in advance!

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    $\begingroup$ Have you tried considering $R = K[X^2, X^3] \cong \frac{K[Y, Z]}{(Y^3 - Z^2)}$? $\endgroup$ – André 3000 Jan 22 at 16:25
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    $\begingroup$ @André3000 so the polynomial is irreducible and the ideal generated by it is prime. Moreover, $K[Y,Z]$ should be Noetherian by Hilbert's Basissatz, so the quotient is Noetherian. Right? The dimension $1$ (i.e., all its prime ideal are maximal) is not so evident for me... $\endgroup$ – Lei Feima Jan 22 at 16:35
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    $\begingroup$ @LeiFeima: $\langle Y^{3}-Z^{2} \rangle$ is a prime ideal of $K[Y, Z]$, and any proper chain of prime ideals of $K[Y, Z]/\langle Y^{3}-Z^{2} \rangle$ lifts to a chain of prime ideals of $K[Y, Z]$ which begins $0 \subset \langle Y^{3}-Z^{2} \rangle \subset \cdots$. Since $K[Y, Z]$ has Krull dimension $2$, it follows that $K[Y, Z]/\langle Y^{3}-Z^{2} \rangle$ has Krull dimension at most $1$. One can see that it has Krull dimension exactly $1$ by (e.g.) considering the projection of the maximal ideal $\langle Y, Z\rangle$ of $K[Y, Z]$. $\endgroup$ – Alex Wertheim Jan 22 at 18:37
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    $\begingroup$ I also have an alternative approach for showing that $K[X^{2}, X^{3}]$ has dimension $1$ which only uses basic facts on integrality, if you're interested. $\endgroup$ – Alex Wertheim Jan 22 at 18:39
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    $\begingroup$ @LeiFeima It depends on how elementary you'd like your proof to be. For instance, $R$ is the coordinate ring of a curve, so its dimension is $1$. Or since the ideal $(Y^3 - Z^2)$ is principal, it has codimension $1$ by Krull's Hauptidealsatz. Or as Alex Wertheim suggests, $R$ is an integral (quadratic) extension of $K[Y]$. Integral extensions preserve dimension by the Cohen-Seidenberg theorems, so $R$ has the same dimension as $K[Y]$. I suspect you may be able to reproduce this result for integral quadratic extensions by hand, if you don't want to appeal to these theorems. $\endgroup$ – André 3000 Jan 22 at 22:26
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Suppose $\mathfrak{p}$ is a nonzero prime ideal of $A := K[X^{2}, X^{3}]$; we want to show that $\mathfrak{p}$ is maximal. Note that $K[X^{2}]$ is a subring of $A$, and $K[X^{2}]$ is a principal ideal domain, since it is isomorphic to $K[X]$ via the morphism of $K$-algebras $K[X] \to K[X^{2}], X \mapsto X^{2}$. Note that $\mathfrak{m} := \mathfrak{p} \cap K[X^{2}]$ is a prime ideal of $K[X^{2}]$, since it is the preimage of $\mathfrak{p}$ under the inclusion morphism $K[X^{2}] \hookrightarrow A$. If $\mathfrak{m}$ is nonzero, then $\mathfrak{m}$ is maximal, since $K[X^{2}]$ is a PID. Moreover, since the inclusion $K[X^{2}] \hookrightarrow A$ is integral, so too is the induced (injective) morphism $K[X^{2}]/\mathfrak{m} \hookrightarrow A/\mathfrak{p}$. Since $K[X^{2}]/\mathfrak{m}$ is a field and $A$ is a domain, $A$ must be a field as well (this is, e.g., Proposition 5.7 in Atiyah Macdonald).

It therefore suffices to show that $\mathfrak{p} \cap K[X^{2}]$ is nonzero for any nonzero prime ideal $\mathfrak{p}$ of $A$. This amounts to showing that any nonzero $\mathfrak{p}$ contains a polynomial whose monomial terms all have even degree. Take $f(X) \in \mathfrak{p}$ nonzero, and write $f(X) = g(X) + h(X)$, where $g$ has monomial terms of even degree only, and $h$ has monomial terms of odd degree only. Then $f(-X) = g(X) - h(X)$, and $f(-X) \in A$, so $f(-X)f(X) = g(X)^{2} - h(X)^{2} \in \mathfrak{p}$, which clearly has monomial terms of even degree only.

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  • $\begingroup$ Thanks a lot! This is exactly the proof that I had in mind when I posed the question, even if I showed that $p\cap K[X^2]$ is nonempty in a slightly different way... $\endgroup$ – Lei Feima Jan 23 at 9:00
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    $\begingroup$ @LeiFeima: great minds think alike. :) I suppose it's good to have it recorded here, if nothing else. $\endgroup$ – Alex Wertheim Jan 23 at 9:04
  • $\begingroup$ Just for the sake of completeness and clarity: I considered the $p$ prime ideal of $R$ and $p':=p\cap K[X^3]$. Since $R$ is integral over $K[X^3]$, I consider a nonzero $c\in p$ such that, for $b_i \in K[X^3]$ with $b_0 \neq 0$, $c^n+c^{n-1}b_{n-1}+...+b_0 =0$. Then $c(c^{n-1}+b_{n-1}c^{n-2}+...+b_1)=-b_0 \in p'$. $\endgroup$ – Lei Feima Jan 23 at 14:05
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Merely because $K[X]$ is an integral extension of $K[X^2,X^3]$.

More precisely, the going up theorem ensures any prime of $K[X^2,X^3]$ is a contraction of $K[X]$, and you must know that an ideal $\mathfrak{m}\subseteq K[X]$ is maximal if and only if $\mathfrak{m}\cap K[X^2,X^3]\subseteq K[X^2,X^3]$ is maximal.

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