# When does a f.g. algebra over a field $F$ make it “look like $F$ is algebraically closed?”

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 $\textrm{Spec } B$.

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

## 1 Answer

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 $F[x, y]/(x^2, xy, y^2)$.

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