Suppose $k$ is algebraically closed, $A$, $B$ are $k$-algebras and $A$ is an affine $k$-algebra. It is known that then $A\otimes_k B$ is a domain if $A$ and $B$ are domains. This can be found in Milne's Algebraic Geometry notes as Proposition 4.15(b). I do not see where the assumption $k$ algebraically closed is used.

He gives an example that the above is not true if $k$ is not algebraically closed. But i dont see where this assumption is being used in the proof.

I think that for an affine $k$-algebra the Jacobson radical is the nilradical, so here we do not need $k$ to be algebraically closed.

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    $\begingroup$ It is being used when he notes that the quotient $A/\mathfrak{m}$ must be $k$, which is what the other entries are lindearly independent with respect to. If $k$ isn't algebraically closed then the quotient need only be some finite extension of $k$, where the $b_i$ may be linearly dependent. $\endgroup$ Commented Dec 9, 2013 at 6:44
  • $\begingroup$ I recently struggled with the same kind of statement (replacing "domain" by "reduced", but the proof is essentially the same). In order to make things clear, I then wrote the proof down with full detail : the document if you are interested. (Remark the hypothesis $A$ finitely generated is not needed.) $\endgroup$
    – Pece
    Commented Dec 9, 2013 at 9:10

1 Answer 1


I like this question a lot. While he does give a counterexample when $k$ is not algebraically closed, it is hard to see in his proof where this property is used. Where Milne uses algebraic closure is in the lines

For each maximal ideal $\mathfrak{m}$ of $A$, we know $(\sum\overline{a}_ib_i)(\sum \overline{a}_i~'b_i')=0$ in $B$, and so we either $(\sum \overline{a}_ib_i)=0$ or $(\sum \overline{a}_i'b_i')=0$. Thus either all the $a_i\in \mathfrak{m}$ or all the $a_i'\in \mathfrak{m}$.

We know that $\{b_1,b_2,\dots\}$ and $\{b_1',b_2',\dots\}$ are linearly independent over $k$ (emphasis on $k$). The elements $\overline{a}_i$ and $\overline{a}_i'$ live in $A/\mathfrak m$, which is a priori only an algebraic field extension of $k$. It is possible for elements of a $k$ algebra to be linearly independent over $k$ but not over some extension (consider $1,i\in \mathbb C$ over $\mathbb R$ as opposed to $1,i\in \mathbb C$ over $\mathbb C$). The fact that $k$ is algebraically closed forces $A/\mathfrak{m}=k$, so we can apply the linear independence condition.

Great Question! I think Milne could use to write a line or two more for clarity, especially in an introductory treatment.


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