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Let $k$ be an algebraically closed field and $A = k[X_1, \ldots, X_n]$. In Hartshorne's Algebraic Geometry, Example 1.4.4, it is stated that a maximal ideal $\mathfrak{m}$ of $A$ corresponds to a minimal irreducible closed subset of $k^n$, which must be a point $(a_1, \ldots, a_n)$. I get why this holds from the discussion before.

What I'm trying to understand is why we can conclude from this that every maximal ideal is of the form $\mathfrak{m} = (X_1 - a_1, \ldots, X_n - a_n)$ only using the information we have so far - namely, a version of Hilbert's NSS which says that for any ideal $\mathfrak{a}$, $I(Z(\mathfrak{a})) = \sqrt{\mathfrak{a}}$.

I've reasoned that since $\mathfrak{m}$ is represented by a point $(a_1, \ldots, a_n) = Z(X_1 - a_1, \ldots, X_n - a_n)$ have that $$ \mathfrak{m} = IZ(X_1 - a_1, \ldots, X_n - a_n) = \sqrt{(X_1 - a_1, \ldots, X_n - a_n)} $$ where the last equality is by Hilbert's. So now I would just have to show that the ideal $\mathfrak{a} = (X_1 - a_1, \ldots, X_n - a_n)$ is radical, but I'm unable to do that. Is this something that should follow easily just from the information given so far in the book?

rosecabbage
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  • What's $A/\mathfrak a$? – anankElpis Oct 25 '24 at 19:45
  • @anankElpis If $\phi:A \rightarrow k$ is evaluation in $(a_1, \ldots, a_n)$, then $\mathfrak{a} \subset \ker \phi$, but how do I show the other inclusion? Otherwise I don't know how to prove the obvious answer that $A/\mathfrak{a} = k$ using just the info I have. – rosecabbage Oct 25 '24 at 19:53

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