# Tensor products of two domains

Let $R$ be an integral noetherian regular local ring. Let $S$ be a noetherian integral domain such that $S/R$ is finite.

That is, $R \subset S$ and the surjection $R^{\oplus n} \twoheadrightarrow S$ between two $R$-modules holds for a certain $n < \infty$.

Consider a ring $T$ which is integral and contains $R$ as its subring, i.e. $R \subset T$.

Let we denote by $k(R)$, $k(S)$, and $k(T)$ fractional fields of $R$, $S$, and $T$, respectively.

## (What if, especially, $T$ is a non-noetherian domain?)

• What is the meaning of $k(R)=k(S)\cap k(T)$?
– abx
Oct 10, 2016 at 10:12
• @abx. Perhaps the OP wants to assume that $k(S)$ and $k(T)$ are linearly disjoint as field extensions of $k(R)$. At any rate, this question does not seem to be appropriate for MO. Since the fraction field of an integral domain is a flat ring extension, the question reduces to the question whether $k(T)\otimes_{k(R)} k(S)$ is integral, which it is when $k(S)$ and $k(T)$ are linearly disjoint extensions of $k(R)$. Oct 10, 2016 at 10:17
• Dear Jason, then could you please explain to me how you use the fraction field of an integral domain is a flat extension in order to reduce the setting into fields case? Thanks. Oct 10, 2016 at 12:31
• Actually, there are counterexamples . . . Oct 10, 2016 at 13:04

First of all, as abx says, let us assume that both $S$ and $T$ are contained in some larger field $L$, in order to make the notation $k(S)\cap k(T)$ well defined.

In the above answer by Jason Starr, it seems that only flatness of $S$ over $R$ is used. To guarantee this, we need not require $S$ to be regular; assuming $S$ to be Cohen-Macaulay is sufficient.

• That is a good point. Somehow when I read the statement of the problem, "regular" stuck in my head. However, as you say, it is sufficient for $S$ to be Cohen-Macaulay. Oct 10, 2016 at 14:38
• Great! Thank you very much. You helped me a lot. Pierre Matsumi Oct 10, 2016 at 16:19

I made a mistake in the comment above. The comment is valid if both $R$ and $S$ are regular. However, as stated, there are counterexamples.

For instance, let $R$ be $k[x,y^{-1},x,y^{-1}]_{\mathfrak{m}}$ where $\mathfrak{m} = \langle x-1, y-1\rangle$. The fraction field of $R$ is $k(x,y)$. Let $k(S)/k(R)$ be the extension field $k(x,y)[z]/\langle z^2 -xy\rangle$. Begin with the integral closure $\widetilde{S}= R[z]/\langle z^2-xy \rangle$ of $R$ in $k(S)$. Then $\widetilde{S}$ is regular, and hence $R\to \widetilde{S}$ is flat. However, instead of $\widetilde{S}$, consider the $R$-submodule $S\subset \widetilde{S}$ generated by $1$ and $\mathfrak{m}\widetilde{S}$. Explicitly, $S$ equal $R[a,b]/I$ for $a=(x-1)z$, $b=(y-1)z$ and the ideal $$I=\langle (y-1)a-(x-1)b, a^2 - xy(x-1)^2, ab-xy(x-1)(y-1), b^2 - xy(y-1)^2 \rangle.$$ Next let $T \subset k(R)$ be the $R$-algebra, $$T = R[(y-1)/(x-1)] \cong R[w]/\langle (x-1)w - (y-1) \rangle.$$

The point is that the element $wa-b$ is a nonzero element of $S\otimes_R T$, yet $(x-1)(wa-b)$ does equal zero. Indeed, consider $S/\mathfrak{m}S$ and $T/\mathfrak{m}T$ as algebras over $R/\mathfrak{m}=k$. First, $S/\mathfrak{m}S$ equals $k[a,b]/\langle a^2,ab,b^2\rangle$. Second, $T/\mathfrak{m}T$ equals $k[w]$. Thus, the quotient ring, $$(S\otimes_R T)/\mathfrak{m}(S\otimes_R T) = (S/\mathfrak{m}S)\otimes_{R/\mathfrak{m}}(T/\mathfrak{m}T)$$ equals $k[a,b,w]/\langle a^2,ab,b^2\rangle$. In particular, the image of $wa-b$ is nonzero. Yet $(x-1)(wa-b)$ equals $(y-1)a-(x-1)b$, which already equals $0$ in $S$.

On the other hand, if $S$ is regular, then the homomorphism $R\to S$ is flat. In that case, since $T\to k(T)$ is injective, also $S\otimes_R T \to S\otimes_R k(T)$ is injective. Since also $k(T)$ is flat over $R$, similarly $S\otimes_R k(T) \to k(S)\otimes_R k(T)$ is injective. Since $k(S)$ and $k(T)$ are linearly disjoint over $k(R)$, $k(S)\otimes_{k(R)} k(T) = k(S)\otimes_R k(T)$ is an integral domain. Therefore the subring $S\otimes_R T$ is also an integral domain.