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Let $k$ be a commutative ring with total quotient ring $K$, and let $A$ be a commutative $k$-algebra such that the multiplication map $A \otimes_k A \longrightarrow A$ is an isomorphism. EDIT: Assume also that $k \longrightarrow A$ is injective. Then: Must $A$ be isomorphic to a $k$-subalgebra of $K$? If necessary, assume also that $k$ is an integral domain and/or is of characteristic zero.

Conversely: For which $k$-algebras $A$ contained in $K$ is the multiplication map $A \otimes_k A \longrightarrow A$ an isomorphism? Equivalently, for which $k$-algebras $A$ contained in $K$ is $A \otimes_k A$ $k$-torsion-free?

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I think the missing word is flat : $A$ should be a flat finitely presented $k$-algebra. Indeed put $Y=\mathrm{Spec}\,k$ and $X=\mathrm{Spec}\,A$. The condition on $A/k$ is that the diagonal morphism $X\rightarrow X\times _YX$ is an isomorphism. This means exactly that $f:X\rightarrow Y$ is unramified and universally injective ("radiciel"). If we add that $f$ is flat, then it is an open embedding (EGA IV, 17.9), which means $A\subset K$.

It is easy to give conter-examples if for instance we ask only for $k\rightarrow A$ to be injective.

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  • $\begingroup$ That makes sense to me. $\endgroup$ – Jason Starr Jan 25 '14 at 20:30
  • $\begingroup$ Flatness is sufficient for the given conclusion, but is it necessary? And what is a counterexample when $k \longrightarrow A$ is injective? $\endgroup$ – Jesse Elliott Jan 25 '14 at 23:23
  • $\begingroup$ "Which means $A\subset K$": yes assuming e.g. that $k$ is a domain and $A\neq0$. In fact, the zero algebra is a counterexample to many claims. Also, "finitely presented" is irrelevant, I think. $\endgroup$ – Laurent Moret-Bailly Jan 26 '14 at 0:07
  • $\begingroup$ Please add $A \neq 0$ to my assumptions. $\endgroup$ – Jesse Elliott Jan 26 '14 at 0:13
  • $\begingroup$ @JesseElliot: "Please add $A\neq 0$ to my assumptions." Why don't you add that to your assumptions, yourself? In fact, I suggest that you just accept one of the two correct answers. If you want to ask another question, you can always do that. $\endgroup$ – Jason Starr Jan 26 '14 at 0:18
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Perhaps there is a hypothesis missing. There are counterexamples as stated. For instance, if $k$ is $\mathbb{C}[t]$ and $A$ is the quotient ring $\mathbb{C}[t]/\langle t \rangle$, then the multiplication map, $$ \mathbb{C}[t]/\langle t \rangle \otimes_{\mathbb{C}[t]} \mathbb{C}[t]/\langle t \rangle \to \mathbb{C}[t]/\langle t \rangle,$$ is an isomorphism.

Regarding your second question, there certainly are counterexamples. For instance, let $k$ equal $\mathbb{C}[x,y]/\langle y^2-x^3 \rangle$ with quotient ring $K=\mathbb{C}(t)$, where $t=y/x$. Since $x$ equals $t^2$ and $y$ equals $xt = t^3$, then the subring $A=\mathbb{C}[t]$ of $K$ is a $k$-algebra. In fact, $A$ equals $k[t]/\langle x-t^2,y-t^3 \rangle$. Thus, we have $$A\otimes_k A = k[t_1,t_2]/\langle x-t_1^2,y-t_1^3, x-t_2^2,y-t_2^3 \rangle.$$ In particular, $t_2-t_1$ is a zerodivisor in $A\otimes_k A$. Therefore the multiplication map cannot be an isomorphism.

To guarantee that the multiplication map is an isomorphism for every $k$-subalgebra $A$ of $K$, you could assume that $k$ is a Dedekind domain. Then, automatically, $A$ is a flat $k$-algebra. That is the only general hypothesis that works for all $A$ simultaneously of which I am aware.

Edit. I meant to add the words "of which I am aware" in my original answer (but I got distracted by something else). I added them now.

Second edit. In the comments below, the OP clarifies that he wants a counterexample where, not only is $A$ not isomorphic to a $k$-subalgebra of $K$, indeed $A$ is not isomorphic to a subring of $K$ (ignoring the $k$-algebra structure). Of course there are equally simple examples of this: let $k$ be $\mathbb{Z}$, and let $A$ be $\mathbb{Z}/2\mathbb{Z}$.

Third edit. The OP has changed his question. Let $k$ be $\mathbb{Z}$. Let $A$ be the product ring, $(\mathbb{Z}[1/2])\times (\mathbb{Z}/2\mathbb{Z})$, with its unique structure of $\mathbb{Z}$-algebra, i.e., $$A=k[x,y]/\langle 2x+y-1,xy,2y\rangle.$$ Of course the unique ring homomorphism, $$\mathbb{Z}\to (\mathbb{Z}[1/2])\times (\mathbb{Z}/2\mathbb{Z}),$$ is injective. Distributing out the tensor product, and using that $\mathbb{Z}[1/2]\otimes (\mathbb{Z}/2\mathbb{Z})$ is zero, the multiplication homomorphism for $A$ is an isomorphism. Yet $A$ contains nonzero zerodivisors, hence it is not isomorphic to a subring of $\mathbb{Q}$.

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  • $\begingroup$ I guess algebras are meant to be augmented, which would exclude this kind of counterexamples. $\endgroup$ – user43326 Jan 25 '14 at 17:54
  • $\begingroup$ @user43326: "I guess algebras are meant to be augmented ..." That is certainly possible, but that would also rule out most examples where $A$ is a subalgebra of the fraction field $K$. So I am not certain that is what the OP meant. Perhaps the OP wanted $k$ to be a subalgebra of $A$. $\endgroup$ – Jason Starr Jan 25 '14 at 18:37
  • $\begingroup$ $k$ is a Pr\"ufer domain iff every $k$-torsion-free $k$-module is flat, so that is a more general hypothesis that guarantees it for all $k$-subalgebras of $K$. $\endgroup$ – Jesse Elliott Jan 25 '14 at 22:38
  • $\begingroup$ And, yes, I would like an example where the map $k \longrightarrow A$ is injective, which abx in his or her answer claims exists. $\endgroup$ – Jesse Elliott Jan 25 '14 at 23:29
  • $\begingroup$ I'm confused, probably about something silly. In your first example, isn't $A \cong \mathbb C$? So then $A$ is isomorphic to a subring of $K$? $\endgroup$ – Peter Samuelson Jan 25 '14 at 23:35

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