Let $R=\mathbb{C}[X,Y], I=(X,Y)$ and $K$ the fraction field of $R$, where we view $K$ as an $R$-module. Show that
$$ R/I\otimes_R I\cong R/I\oplus R/I\quad\text{and}\quad K\otimes_R I\cong K.$$
Conclude that $I$ is not a free $R$-module.
This is the third part of a past exam question I am working on at the moment, and I can't figure out how to show the first isomorphism. The second one seems straightforward since $\frac{p}{q}\otimes f=\frac{pf}{q}\otimes 1\in K\otimes_R I$. The first two parts of the question ask us to show that if $M$ is an $R$-module and
$$0\to N\to E\to N'\to 0 $$
is split short exact sequence of $R$-modules then
$$ 0\to M\otimes_R N\to M\otimes_R E\to M\otimes_R N'\to 0 $$
is a split short exact sequence. We then use this to conclude that $M\otimes_R(N\oplus N')\cong (M\otimes_R N)\oplus (M\otimes_R N')$. So I guess this needs to be used somewhere when showing that $I$ is not a free $R$-module, but I'm not sure how.
Best Answer
For the computation of $R/I\otimes I$, use the fact that $-\otimes I$ is right exact, and apply that to the exact sequence $0\to I\to R \to R/I\to 0$. This implies $$R/I\otimes I\cong I/I^2$$ In the situation of the problem one notes that $I/I^2$ is a vector space over $R/I\cong \mathbb C$ and has basis $\overline X,\overline Y$. More generally, consider the situation where $\mathfrak m$ is a maximal ideal of a ring $A$ with residue field $k:=A/\mathfrak m$. Then, $A/\mathfrak m^2$ is a local ring with maximal ideal $\mathfrak m/\mathfrak m^2$, so that $\mathfrak m/\mathfrak m^2$ can be computed after localizing $A$ with respect to the ideal $\mathfrak m$. Then, by Nakayama's lemma, we have $$\dim_kA/\mathfrak m\otimes_A\mathfrak m = \text{minimal number of generators of $\mathfrak mA_{\mathfrak m}$}$$ Anyway, we have that in the problem statement, $\dim_{\mathbb C}R/I\otimes I = 2$.
For the second isomorphism, consider an exact sequence $$0\to M'\to M \to M''\to 0$$ of $R$-modules. Suppose $\operatorname{ann}_RM''\ne 0$. Since every nonzero element of $R$ is invertible in $K$, we have $M''\otimes K\cong 0$. On the other hand, $-\otimes K$ is exact because it is the same as localizing with respect to the multiplicative set $R\setminus\{0\}$. Hence, we have an exact sequence $$0\to M'\otimes K\to M\otimes K\to 0$$ which is the same as saying $M'\otimes K\cong M\otimes K$. Apply this observation to the short exact sequence $$0\to I\to R\to R/I\to 0$$and you will get the desired isomorphism. The dimension of $M\otimes K$ is called the rank of $M$, for any $R$-module $M$, so the problem has you showing that $I$ is an $R$-module of rank $1$. Can you come up with an equation for the ranks of $M',M,M''$ in the above short exact sequence without making any assumptions on the annihilator of $M''$?
For the conclusion that $I$ is not free, note that if $F$ is a free $R$-module of rank $r$, then for any $R$-module $M$, we have $$F\otimes M\cong M^r$$ by the earlier parts of the problem. In particular, $\dim_{\mathbb C}F\otimes R/I = \dim_{K}F\otimes K$. This shows that $I$ cannot be a free $R$-module.