3
$\begingroup$

Let $F$ be a field, and let $A$ be a finitely generated algebra over $F$. If $\mathfrak m$ is a maximal ideal of $A$, then $A/\mathfrak m$ is an algebraic extension of $F$, although it is in general not equal to $F$.

Are there f.g. algebras $A$ for which $A/\mathfrak m = F$ for all maximal ideals $\mathfrak m$ of $A$?

There is an obvious situation where this is true, when $A$ is a finite product of copies of $F$. I was wondering whether there were some nontrivial examples and whether they were of interest to anyone.

My motivation is as follows: If $B$ is another f.g. generated $F$-algebra, then there is an injection of $\textrm{Max } A \otimes_F B$ into $\textrm{Max } A \times \textrm{Max } B$, since every maximal ideal of $A \otimes_F B$ takes the form $\mathfrak m \otimes B + A \otimes \mathfrak n$ for $\mathfrak m, \mathfrak n$ maximal ideals of $A, B$. Not all ideals of this form are maximal though, since $$(A \otimes_F B)/[\mathfrak m \otimes B + A \otimes \mathfrak n] \simeq A/\mathfrak m \otimes_F B/\mathfrak n$$ need not be a field. So if $A,B$ satisfy the property above, then as in the algebraically closed case, the closed points of the scheme $\textrm{Spec } A \otimes_F B$ can be identified with the cartesian product of the closed points of $\textrm{Spec } A$ and $\textrm{Spec } B$.

$\endgroup$
  • $\begingroup$ Did you forget to mention the case when $F$ is algebraically closed? $\endgroup$ – user26857 May 12 '16 at 5:37
  • $\begingroup$ Lol, I guess I thought that would fall under "trivial" examples $\endgroup$ – D_S May 12 '16 at 5:38
6
$\begingroup$

By Noether normalization, $A$ is a finite extension of $F[x_1, \dots x_n]$ for some $n$. By the going up theorem, the induced map $\text{Spec } A \to \mathbb{A}^n$ is surjective (on prime ideals). Now, $F[x_1, \dots x_n]$ clearly has maximal ideals $m$ with quotient a nontrivial algebraic extension of $F$ (meaning any such ideal in $\text{Spec } A$ mapping to $m$ has the same property) unless either $F$ is algebraically closed or $n = 0$.

If $n = 0$, then $A$ is a finite-dimensional $F$-algebra, and any maps from $A$ to a field extension of $F$ factor through the quotient $A/N(A)$, where $N(A)$ is the nilradical. This quotient is semisimple, hence a product of finite extensions of $F$.

So if $F$ is not algebraically closed, then the algebras $A$ which satisfy your property are precisely the finite-dimensional algebras such that $A/N(A)$ is a finite product of copies of $F$. More explicitly, these are finite products of finite-dimensional local rings over $F$ with residue field $F$. Examples include $F[x]/x^n$ or $F[x, y]/(x^2, xy, y^2)$.

$\endgroup$
  • $\begingroup$ Great answer. I thought there might be some way to reduce to the case where $A$ is a polynomial ring. $\endgroup$ – D_S May 12 '16 at 15:14

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.