5
$\begingroup$

Let $k$ be an algebraically closed field, and let $A$ be a finitely generated commutative $k$-algebra. Given any maximal ideal $\mathfrak{m}\subset A$, we can form the quotient to obtain a map $A\to A/\mathfrak{m}\cong k$. Conversely, the kernel of any map $A\to k$ is a maximal ideal of $A$. This shows that maximal ideals of $A$ are in one-to-one correspondence with maps of $k$-algebras $A\to k$.

I'd like to know if there is a similar correspondence when $A$ is a finitely generated graded $k$-algebra such that $A_0=k$. Specifically, is there a graded $k$-algebra $B$ such that every homogeneous ideal of $A$ that is maximal among homogeneous ideals properly contained in $\bigoplus_{i>0}A_i$ is the kernel of some surjective map of graded $k$-algebras $A\to B$?

Initially, my guess is that $B$ should be a polynomial ring in one variable, but I'm having trouble working out the details. Any help is much appreciated. Thanks.

$\endgroup$
4
$\begingroup$

Your guess is right with the following additional assumption: $A$ is generated by $A_1$ as a $k$-algebra. Without loss of generality, we can further assume that $A$ is reduced (since every maximal ideal contains the nilradical). Therefore we have a surjective morphism of graded $k$-algebras $k[x_0,\dots,x_n] \to A$ where the $x_i$ are sent to elements of degree one. The kernel of this map is a homogeneous radical ideal and $A$ is the homogeneous coordinate ring of the zero set of this ideal in $\mathbb{P}^n$. Thus if we divide out a maximal homogeneous ideal of A properly contained in $A_+$, we get the homogeneous coordinate ring of a point in $\mathbb{P}^n$, which is the polynomial ring in one variable over $k$.

If you want $B$ to have a fixed grading, namely the standard grading, then the additional assumption is necessary. Consider for example the polynomial ring $k[T]$ where $T$ has degree two.

$\endgroup$
  • $\begingroup$ Can we write $k$ as the quotient of $A$ by some maximal homogeneous ideal? $\endgroup$ – R. Singh Apr 25 '17 at 10:03
  • $\begingroup$ Sure. Divide out the ideal generated by $A_1$. $\endgroup$ – Hans Apr 29 '17 at 6:58

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.