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Let $\mathbb{K}$ be an algebraically closed field. A version of Hilbert Nullstellensatz for polynomials rings says that,

Every proper ideal $\mathfrak{a}$ in $\mathbb{K}[X_1,\dotsb,X_n]$ has a zero in $\mathbb{K}^n$, i.e., there exists some $x\in\mathbb{K}^n$, such that $x\in\mathcal{V}(\mathfrak{a})$.

I was wondering if such a statement is true for a general co-ordinate ring of an irreducible affine variety, i.e.,

Let $V$ be an irreducible affine variety. Then, every proper ideal $\mathfrak{a}$ in $\mathbb{K}[V]$ has a zero in $V$, i.e., there exists some $x\in V$, such that $x\in\mathcal{V}(\mathfrak{a})$.

It'sMe
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1 Answers1

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There is a corresponding statement for varieties, and it's exactly what you intuit.

To see why this is the case, recall we have an (inclusion reversing) correspondence between radical ideals and subvarieties of $V$. So asking if

$$\mathcal{V}(\mathfrak{a}) \supseteq \{ x \}$$

is asking if the variety of $\mathfrak{a}$ contains a point $x$ (that is, a minimal nonempty subvariety $\{ x \}$). By the inclusion reversing correspondence, this is the same as asking if

$$\mathfrak{a} \subseteq \mathcal{I}(\{x\})$$

that is, if $\mathfrak{a}$ is contained in the ideal associated to a point $x$ (that is, a maximal proper ideal).

Of course, we know that every proper ideal is contained in some maximal proper ideal (by Zorn's Lemma), and now by the nullstellensatz (since $k$ is algebraically closed) we know that maximal ideals of $k[V]$ are all of the form $(x_1 - a_1, \ldots, x_n - a_n)$ for some point $(a_1, \ldots, a_n) \in k^n$.

If you're not sure why we can apply the nullstellensatz here, it's because ideals in $k[V] = k[x_1, \ldots, x_n] / I$ are exactly the ideals of $k[x_1, \ldots, x_n]$ containing $I$, so we get the claim by the "ordinary" nullstellensatz followed by the correspondence principle.

So, to put all the pieces together, how would you actually write up a proof of this fact? You should try it yourself (everything you need is in this answer) and check against the version I've written up under the fold:

Let $\mathfrak{a}$ be a proper ideal in $k[V]$. Then by Zorn's Lemma it is contained in a maximal ideal $\mathfrak{m}$. By the nullstellensatz (and the correspondence theorem) we know $\mathfrak{m} = (x_1 - a_1, \ldots, x_n - a_n)$ for some $(a_1, \ldots, a_n) \in k^n$. Now taking $\mathcal{V}$ of the relation $\mathfrak{a} \subseteq \mathfrak{m} = (x_1 - a_1, \ldots, x_n - a_n)$, we find $\mathcal{V}(\mathfrak{a}) \supseteq \mathcal{V}(\mathfrak{m}) = \mathcal{V}((x_1 - a_1, \ldots, x_n - a_n)) = \{ (a_1, \ldots, a_n) \}$, so the point $(a_1, \ldots, a_n) \in \mathcal{V}(\mathfrak{a})$. As desired.

I hope this helps ^_^

HallaSurvivor
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