Let $I=(xy)$ be an ideal of the ring $R=\mathbb{Z}[x,y]$. I want to find the presentation of $I$ as an $R$-module. This may be very simple, but it's betraying my intuition on the subject, so I want to make sure. Here's my process:
Let $\varphi:\mathbb{Z}[x,y] \to (xy)$ be a module homomorphism such that $1 \mapsto xy$. Then, $\ker(\varphi)=\{0\}$. But this would give us that $\mathbb{Z}[x,y] \cong (xy)$, which can't be true, can it? I mean, $1 \in \mathbb{Z}[x,y]$ but $1 \not \in (xy)$, so they can't be isomorphic. What am I missing here?
You have proven that $\mathbb{Z}[x,y]$ and $(xy)$ are isomorphic as modules.
However, $\mathbb{Z}[x,y]$ and $(xy)$ as not isomorphic as rings!
Your observation that $\mathbb{Z}[x,y]$ has a multiplicative identity whereas $(xy)$ doesn't is testament to the fact that they are not isomorphic as rings. However, if we instead view them a modules, then we lose the notion of multiplying two elements of the set, and we lose the notion of multiplicative identity. So there is no contradiction with $\mathbb{Z}[x,y]$ and $(xy)$ being isomorphic as modules.
Said another way:
Now let's revisit your argument. There is an element $1_{\mathbb Z [x, y]} \in \mathbb Z [x, y]$ such that $1_{\mathbb Z [x, y]}\times p = p$ for all $p \in \mathbb Z [x, y]$. If $f$ is a ring isomorphism, then we would have $f(1_{\mathbb Z [x, y]}) \times f(p) = f(p)$ for all $p \in \mathbb Z [x, y]$; since every $r \in (x, y)$ is equal to $f(p)$ for some $p \in \mathbb Z [x, y]$, this would prove that $f(1_{\mathbb Z [x, y]})$ is a multiplicative identity in $(x, y)$. If $f$ is merely a module isomorphism, then this argument doesn't work.