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Let $R=\mathbb{C}[x,y]$ and let $\phi: R^2 \to R^2$ be the $R$-module homomorphism given by the matrix $A= \left( \begin{array}{ccc} a & b \\ c & d \\ \end{array} \right) $ where $a,b,c,d \in \mathbb{C}[x,y]$.

What are the necessary and sufficient conditions for $\phi$ to be an isomorphism?

I know $\phi$ will be bijective iff $A$ has an inverse iff $ad-bc \neq 0$.

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    what exacly do you mean by «is this all?» – 2017-02-11
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    It is not true that $A$ has an inverse iff $ad-bc\neq0$. The matrix $\begin{pmatrix}x&0\\0&1\end{pmatrix}$ does not have an inverse. – 2017-02-11
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    $A$ has an inverse iff $det A$ has inverse in $R=\mathbb C[x,y]$. Which elements have inverse in $R$? – 2017-02-11

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Your matrix $A$ will have inverse iff $\det A$ has inverse in $R=\mathbb C[x,y]$. Now, the invertible elements of $R$ are the invertible in $\mathbb C$, that is, $\mathbb C\setminus\{0\}$. Let us see why.

If $p(x,y)$ is invertible in $R$, then exists $q(x,y)$ so that $p(x,y)\cdot q(x,y)=1$. But that would mean $$\deg p(x,y) + \deg q(x,y) = \deg [p(x,y)\cdot q(x,y)]= \deg 1=0$$ so $\deg p(x,y) = \deg q(x,y) = 0$. That means $p(x,y)\in \mathbb C$.

So, to sum up, $\phi$ is an isomorphism iff $\det A \in \mathbb C\setminus \{0\}$.

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    Yes, that would be the case. – 2017-02-11