2
$\begingroup$

Why must every prime ideal of a finitely generated $\mathbb{R}$-algebra (e.g. $\mathbb{R}[X_1,X_2]$) be the intersection of the maximal ideals containing it?

This doesn't follow from the version of the Nullstellensatz I've seen:

  • if $k$ is an algebraically closed field and $S$ is a finitely generated $k$-algebra and $P$ is a prime ideal of $S$, then $P$ is the intersection of all maximal ideals containing $P$

... since $\mathbb{R}$ isn't algebraically closed! But maybe the same method of proof could work?

Many thanks in advance for any help with this!

$\endgroup$
2
$\begingroup$

Let $k$ be a field and $\bar{k}$ its algebraic closure. An algebraic zero of a subset $\Phi$ of $k[x_1,\dots,x_n]$ is an element $(a_1,\dots,a_n) \in \bar{k}^n$ such that $f(a_1,\dots,a_n)=0, \, \forall f \in \Phi$. Then Hilbert's Nullstellensatz says that if $g \in k[x_1,\dots,x_n]$ vanishes at every algebraic zero of $\Phi$, then $g$ is inside the radical of the ideal generated by $\Phi$ (Matsumura, Theorem 5.4).

The next key thing to observe is that given an ideal $I$ of $k[x_1,\dots,x_n]$, there is a $1-1$ correspondence between algebraic zeros of $I$ and maximal ideals of $k[x_1,\dots,x_n]$ that contain $I$. To see that, note that if $m$ is a maximal ideal that contains $I$ and we define $a_i$ to be the class of $x_i$ mod $m$, then $(a_1,\dots,a_n)$ is an algebraic zero of $I$ by the Zariski Lemma. Conversely, if $(a_1,\dots,a_n)$ is an algebraic zero, then $k[a_1,\dots,a_n] = k(a_1,\dots,a_n)$ and the kernel of the $k$-algebra homomorphism $k[x_1,\dots,x_n] \rightarrow k[a_1,\dots,a_n]$ that sends $x_i$ to $a_i$ is a maximal ideal (Matsumura, Theorem 5.1).

Finally, we clearly have that $I \subset \cap_{m \supset I} m$. Conversely, let $f \in \cap_{m \supset I} m$. Then $f$ vanishes at every algebraic zero of $I$ and by Hilbert's Nullstellensatz $f \in \sqrt{I}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.