Prove $(X,Y) \subset \mathbb{C}[X,Y]$ is not free as $\mathbb{C}[X,Y]$-module using two isomorphisms

Question

Consider the ring $R = \mathbb{C}[X,Y]$ and let $I = (X,Y)\subset R$ be the ideal generated by $X$ and $Y$. In the exercise I have proven that $R/I \otimes_R I \cong R/I \oplus R/I$ and that $Q(R) \otimes_R I \cong Q(R)$ where $R/I$ is viewed as a quotient module and $Q(R)$ is the field of fractions of $R$ viewed as an $R$-module. Next the exercise asks to conclude that $I$ is not a free $R$-module.

However I do not see how this follows from the two isomorphisms above. Can someone help? Thanks in advance.

By the way, I am aware of different solutions (for example The ideal $(x,y)$ is not a free $K[x,y]$-module). But I am particularly interested in how these isomorphisms help to reach the same conclusion.

Answer 1

Suppose that $I$ is a free $R$-module. Since $I$ is finitely generated there exists a positive integer $m$ such that $I\simeq R^m$ as $R$-modules. From $R/I \otimes_R I \simeq R/I \oplus R/I$ and $I\simeq R^m$ we get $(R/I)^m\simeq (R/I)^2$. Now note that this is not only an isomorphism of $R$-modules, but it is also an isomorphism of $R/I$-modules, and by the invariant basis number property it follows that $m=2$.
From $Q(R) \otimes_R I \simeq Q(R)$ and $I\simeq R^2$ we get $Q(R)^2\simeq Q(R)$. Note that this is also an isomorphism of $Q(R)$-modules (that is, vector spaces), a contradiction.

Answer 2

Question: "Next the exercise asks to conclude that I is not a free R-module. However I do not see how this follows from the two isomorphisms above. Can someone help? Thanks in advance."

Answer: $R/I \otimes_R I \cong I/I^2$ as $k:=R/I$-module and $I/I^2$ has a basis consisting of the elements $\overline{x}, \overline{y}$ as $k$-vector space. Hence

$$R/I\otimes_I I \cong I/I^2 \cong R/I\{\overline{x}, \overline{y} \}.$$

If $I$ is free of rank one on the element $e$ it follows

$$R/I\otimes_R I \cong R/I e$$

is free of rank one the element $e$. This is not true, hence if $I$ is free, it is free of rank greater than one, but this is impossible for the following reason: If $a_1,a_2\in I$ are arbitrary non-zero elements it follows

$$(a_2)a_1+(-a_1)a_2=0$$

hence the elements $a_1,a_2$ cannot be "linearly independent" elements.

In general if $I \subseteq A$ is a non-principal ideal it follows $I$ cannot be a free $A$-module: Assume $a_1,a_2 \neq 0$ are part of a basis of $I$. It follows $(a_2)a_1+(-a_1)a_2=0$ hence $\{a_1,a_2\}$ cannot be linearly independent over $A$ for any elements $a_1,a_2\in I$.