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.


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.

  • $\begingroup$ 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

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