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Every prime ideal $\mathfrak{p}$ of the ring $\Bbb C[x_1, \dots x_n]$ is the intersection of maximal prime ideals containing $\mathfrak{p}$. Prove the same result for the coordinate ring $\Bbb C[V]$, where $V \subset \Bbb A^n$ is an affine variety.

The coordinate ring $\Bbb C[V]$ is isomorphic to $\Bbb C[x_1, \dots, x_n]/I(V)$ so what I need to show is that if I take a prime ideal $\mathfrak{p} \subset \Bbb C[x_1, \dots, x_n]/I(V)$, then $\mathfrak{p} = \bigcap_{\mathfrak{p} \subset \mathfrak{m}} \mathfrak{m}$ where $\mathfrak{m}$ is a maximal prime ideal.

Is this a proof where I need to show both inclusions $\subset$ and $\supset$ or is there some clever way to approach this?

I couldn't get anywhere by taking $f \in \mathfrak{p}$. The only properties I know for prime ideals are that if the product $fg \in \mathfrak{p}$, then either $f \in \mathfrak{p}$ or $g \in \mathfrak{p}$, but I don't think this is something I should consider here.

Jonas
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  • Have you learned about the Nullstellensatz yet? If I recall, this is an equivalent formulation. See this post for how to do it over $\mathbb{R}$, the case of $\mathbb{C}$ should be easier. – morrowmh Apr 11 '22 at 18:42
  • I have! Does the result follow from the fact that the radical of and ideal $J$ is the intersection of all primes containing $J$? – Jonas Apr 11 '22 at 19:49
  • A ring having this property is called a Jacobson ring: https://en.wikipedia.org/wiki/Jacobson_ring – Viktor Vaughn Apr 11 '22 at 20:10

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I'll give the argument for $\mathbf{C}[x_1, \ldots, x_n]$; see if you can modify it for the $\mathbf{C}[V]$ case.

The $\subset$ inclusion is immediate. Conversely, let $f \in \bigcap_{\mathfrak{m} \supset \mathfrak{p}}\mathfrak{m}$. We have that $f \in \mathfrak{m}$ for all maximal ideals containing $\mathfrak{p}$. What does $f \in \mathfrak{m}$ mean? It implies that $V(f) \supset V(\mathfrak{m})$, i.e. $f$ vanishes at the point $V(\mathfrak{m})$. What does $\mathfrak{m} \supset \mathfrak{p}$ mean? It implies that $V(\mathfrak{m}) \subset V(\mathfrak{p})$, i.e. each $V(\mathfrak{m})$ is a point lying on the affine variety $V(\mathfrak{p})$. Putting the two together, we have that $f$ vanishes at every point on $V(\mathfrak{p})$, and therefore $V(f) \supset V(\mathfrak{p})$, which means that $I(V(f)) \subset I(V(\mathfrak{p}))$, or $\sqrt{(f)}\subset \sqrt{\mathfrak{p}} = \mathfrak{p}$ by the Nullstellensatz. Since $f \in \sqrt{(f)}$, the reverse containment $\supset$ follows.

gf.c
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  • Thanks for this! I got a hint that for $\Bbb C[V]$ I should use the correspondence between ideals of $\Bbb C[x_1, \dots x_n]/\mathfrak{p}$ and ideals of $\Bbb C[x_1, \dots, x_n]$ containing $\mathfrak{p}$, but I still am confused about how I should use this correspondence. – Jonas Apr 12 '22 at 10:26
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    @Jonas: Try using the fact that this correspondence preserves primality/maximality of ideals. The result should then follow from what I've written above. – gf.c Apr 12 '22 at 17:05
  • So in a sense I can just map $\mathfrak{p} \longmapsto \varphi(\mathfrak{p})$ where $\varphi$ is the bijection giving the correspondence. Now $\varphi(\mathfrak{p})$ is also prime and I need to show the equality $\varphi(\mathfrak{p}) = \bigcap_{\varphi(\mathfrak{m}) \supset \varphi(\mathfrak{p})} \varphi(\mathfrak{m})$? Each of these maximal and prime ideals under $\varphi$ are also maximal and prime so bascially the exact argument you gave should work here? Should I be concerned about the varieties $V(\varphi(\mathfrak{p}))$ and $V(\varphi(\mathfrak{m}))$? – Jonas Apr 12 '22 at 17:30
  • Be careful. You start with a prime ideal $\overline{\mathfrak{q}} \in \mathbf{C}[V]$ and lift this via the correspondence to a prime ideal $\mathfrak{q} \subset \mathbf{C}[x_1, \ldots, x_n]$ containing $\mathfrak{p}$. Invoke the argument in my answer, and then pass back to the quotient $\mathbf{C}[V]$. – gf.c Apr 12 '22 at 20:33