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Recall that a homomorphism $f:G\longrightarrow G$ is called idempotent if $f\circ f=f$.

What are idempotent homomorphisms $f:\mathbb{Z}\times \mathbb{Z}\longrightarrow \mathbb{Z}\times \mathbb{Z}$ exactly?

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There's the zero endomorphism, the identity, and all endomorphisms given by matrices of the form

$$ \begin{bmatrix} a & b \\ a(1-a)/b & 1-a \end{bmatrix} $$

where $ a, b \in \mathbb Z $ and $ b \mid a(1-a) $. We can get this by noting that the minimal polynomial of such an operator must divide $ x^2 - x $, so there are three choices; these three cases correspond to the three possible choices of minimal polynomial. (To derive the above form, compute the characteristic polynomial and compare coefficients.)

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Hint: a homomorphism ${\mathbf Z}^2\to {\mathbf Z}^2$ is represented by $2\times 2$ matrix of integers. Matrices with nonzero determinant have cancellation.

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  • $\begingroup$ A group homomorphism or a ring homomorphism? $\endgroup$ – M.Ramana May 3 '18 at 4:57
  • $\begingroup$ @M.Ramana: Group homomorphism. There's no reason for it to preserve multiplication. $\endgroup$ – tomasz May 3 '18 at 7:33
  • $\begingroup$ I understood. Thank you very much. $\endgroup$ – M.Ramana May 3 '18 at 10:28
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Let $R$ be an arbitrary commutative ring and $V$ a $R$-module. If $f$ is an endomorphism of $V$ with $f^2=f$, then $V=\mathrm{Ker}(f)\oplus\mathrm{Im}(f)$ and $f$ is the projection on the second factor, in this decomposition.

Now suppose that $R$ is a PID (or more generally that every projective $R$-module is free, e.g., $R$ is a polynomial algebra or a local ring). Take $V=R^n$. Then every decomposition of $R^n$ is conjugate, by some element of $\mathrm{GL}_n(R)$, to a standard decomposition $R^k\oplus R^{n-k}$.

This means that any matrix $u$ (square of size $n$) with $u^2=u$ is conjugate to the corresponding diagonal matrix, which can always be chosen of determinant 1. For $R$ additionally required to be a domain, $n=2$ and $u\neq 0,1$, this yields the form given in Starfall's answer.

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