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I have the following homework question:

Let $X \subseteq \mathbb{A}^n$ be an irreducible algebraic subset, and let $\mathbb{K}$ be algebraically closed. Show that every maximal ideal in $\mathbb{K}[X]$ determines a unique point $p \in \mathbb{A}^n$ with $p \in X$.

I've tried a proof along the following lines. Let $M$ be a maximal ideal of $\mathbb{K}[X]$. Then there is a corresponding maximal ideal in $\mathbb{K}[x_1, \dots ,x_n]$, which must be of the form $\langle x_1 - p_1, \dots x_n - p_n \rangle$ for some $p=(p_1, \dots,p_n) \in \mathbb{A}^n$ by the Nullstellensatz. Then $p$ must be in $X$, since otherwise $M$ would be empty, contradicting the definition of a maximal ideal (obviously, some details have been omitted).

However, this doesn't make use of the fact that $X$ is irreducible, leading me to believe that I'm missing something. Am I working along the right lines? Where have I gone wrong?

lokodiz
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1 Answers1

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The coordinate ring is the quotient $\mathbb{K}[X]:= \dfrac{ \mathbb{K}[x_1,\ldots, x_n]}{I(X)}.$ By the correspondence theorem for rings, the maximal ideals of $\mathbb{K}[X]$ are uniquely of the form $\mathfrak{m} + I(X)$ for some maximal ideal $\mathfrak{m}\subset \mathbb{K}[x_1,\ldots, x_n]$ containing $I(X).$ Hilbert's Nullstellensatz tells you that $\mathfrak{m}$ is of the form $\mathfrak{m}_p = (x_1-p_1,\ldots, x_n-p_n)=I(\{ p\})$ for some $p=(p_1,\ldots, p_n) \in \mathbb{K}^n.$ Then $I(X)\subseteq I( \{p \} )$ gives $\{ p\} \subseteq X$ i.e. $p$ is in $X.$ The irreducibility does not change anything I just wrote.

Ragib Zaman
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