Determine all ring homomorphism from $\mathbb{Z}\oplus \mathbb{Z}$ into $\mathbb{Z}\oplus \mathbb{Z}$

I understand there is a duplicate question floating around. However, it gives the answer without making explicit reference to the motivation and reasoning. In Group homomorphism, the map of the homomorphism can be specified by the image of the multiplicative identity 1. In a ring homomorphism, there are two binary operation.

How should I THINK about this problem?

Thanks in advance.

  • $\begingroup$ You should specify which convention you mean by "ring": the version where the multiplicative unit is part of the structure or the version where it is not. $\endgroup$ – Hurkyl Jul 15 '17 at 4:57
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    $\begingroup$ Aside: if you are using the convention where the multiplicative unit is part of the structure, then in my opinion, it is somewhat weird to use $\oplus$ here; $\mathbb{Z} \times \mathbb{Z}$ is better notation. $\endgroup$ – Hurkyl Jul 15 '17 at 4:59
  • $\begingroup$ (by "part of the structure", I mean that a ring is required to have a unit and that a homomorphism $f: R \to S$ is required to have $f(1_R) = 1_S$) $\endgroup$ – Hurkyl Jul 15 '17 at 5:03
  • $\begingroup$ @Hurkyl I am using the convention where the unit is not necessarily included. $\endgroup$ – Mathematicing Jul 15 '17 at 5:40

Hint: A ring homomorphism $f: \Bbb Z \oplus \Bbb Z \to \Bbb Z \oplus \Bbb Z$ is determined by the image of the elements $(1,0)$ and $(0,1)$. From there, we have $$ f(a,b) = af(1,0) + bf(0,1) $$ Of course, we have the additional constraint that $f(1,1) = (1,1)$, and of course $(1,0) + (0,1) = (1,1)$. Perhaps you could take it from there.

  • $\begingroup$ Can you explain why the ring homomorphism is determined by the 2-tuples in your hint? If you could make reference to a concept/ or theorem, that would be even better. $\endgroup$ – Mathematicing Jul 15 '17 at 4:45
  • $\begingroup$ To repeat myself: from there, we have $f(a,b) = af(1,0) + bf(0,1)$ $\endgroup$ – Omnomnomnom Jul 15 '17 at 4:46
  • $\begingroup$ Notably, $(1,0)$ and $(0,1)$ are generators of the ring $\Bbb Z \oplus \Bbb Z$. $\endgroup$ – Omnomnomnom Jul 15 '17 at 4:47
  • $\begingroup$ But unless a ring homomorphism is onto, it does not map generators to generators, does it? $\endgroup$ – Mathematicing Jul 15 '17 at 4:52
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    $\begingroup$ You also have constraints $f(1,0)^2=f(1,0)$ etc. $\endgroup$ – Lord Shark the Unknown Jul 15 '17 at 4:58

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