If $\mathfrak p_0 \subsetneq \mathfrak p_1\subset k[x_0,...,x_n]$ are homogeneous prime ideals such that there is no homogeneous prime ideal strictly between them, could there be a (non-homogeneous) prime ideal strictly between them?

Part of me thinks the answer is probably no, but I couldn't come up with a counterexample. Part of me thinks the answer might be yes since the dimension of a projective variety seems like it should be one less than that of its cone, although idk if that makes any sense or if I'm even using any of those terms correctly.

Entertaining the possibility that it's true, my only progress has been to note that if such a prime ideal $\frak q$ exists, then the set $\mathfrak q - \mathfrak p_0$ must not contain any homogeneous elements, since if one such $f$ exists, then $\mathfrak q$ is a minimal prime over $(f)+\mathfrak p_0$, which implies $\mathfrak q$ must be homogeneous.


1 Answer 1


If $\mathfrak p_0 \subsetneq \mathfrak q \subsetneq \mathfrak p_1$, then the codimension of $\mathfrak p_1$ in $R/\mathfrak p_0$ is at least 2. Now let $f\in \mathfrak p_1 - \mathfrak p_0$ with $f$ homogeneous and let $\mathfrak p_1'$ be a minimal prime over $\mathfrak{p}_0+(f)$ such that $\mathfrak p_1'\subseteq\mathfrak p_1$. By Krull's PIT, the codimension of $\mathfrak p_1'$ in $R/\mathfrak{p}_0$ is not greater than $1$. But $\mathfrak{p}_0 +(f)$ is homogeneous, and therefore $\mathfrak p_1'$ is homogeneous. Therefore $\mathfrak p_1' = \mathfrak p_1$. Contradiction.

  • $\begingroup$ I think you mean to also require that $\mathfrak{p}_1'\subseteq\mathfrak{p}_1$ when you first define it. $\endgroup$ Commented May 27, 2021 at 18:23
  • $\begingroup$ @EricWofsey yes, thank you $\endgroup$ Commented May 27, 2021 at 20:53

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