Note: In this class, a ring homomorphism must map multiplicative and additive identities to multiplicative and additive identities. This is different from our textbook's requirement, and often means there are fewer situations to consider.

I always have a pretty hard time answering these types of questions:

Let $\phi: \mathbb{Z~ \times ~Z} \rightarrow \mathbb{Z~ \times ~Z}$ be a ring homomorphism. We know, then, by definition of a ring homomorphism, that $\phi(1,1) = (1,1)$ (because $(1,1)$ is the multiplicative identity of $\mathbb{Z~ \times ~Z}$). Any ring homomorphism must then have the form $\phi(a,b) = (a,b)$ or $\phi(a,b) = (b,a)$. Any addition/multiplication to elements would cease to send $(1,1)$ to $(1,1)$.

Is... this correct? It seems too simple, but I'm pretty sure it covers the possibilities.


2 Answers 2


You know that $(1,0) \cdot (0,1)=(0,0)$ then $f(1,0) \cdot f(0,1)=f(0,0)=0$. This implies that $f(1,0)=(a,0)$ or $(0,a)$, for some $a\in\mathbb{Z}$ and similarly for $f(0,1)=(b,0)$ or $(0,b)$. Since $f(1,1)=(1,1)$ you get that $a=b=1$ and that only two of the four options are valid. In this way you get exactly your two functions.

  • 2
    $\begingroup$ The number of homomorphisms has to be a square because of the universal property of products, so there can't be only two functions. $$ \hom(\mathbb{Z} \times \mathbb{Z}, \mathbb{Z} \times \mathbb{Z}) \cong \hom(\mathbb{Z} \times \mathbb{Z}, \mathbb{Z}) \times \hom(\mathbb{Z} \times \mathbb{Z}, \mathbb{Z})$$ $\endgroup$
    – user14972
    Aug 15, 2018 at 10:14
  • $\begingroup$ True, the cases where f(1,0) or f(0,1) are (0,0) are missing but could be handled similarly. $\endgroup$
    – Quimey
    Aug 17, 2018 at 22:34

I have a manual that goes through the solutions to this problem. Although Quimey is on track, there is actually 9 possibilities, and they all describe a ring homomorphism.

Think of it this way. Let $f\colon \mathbb Z\times\mathbb Z \to \mathbb Z\times\mathbb Z$ be the function. Then suppose that $f(1,0) = (m,n)$. Well, $f(1,0) = f( (1,0)(1,0) ) = f(1,0)f(1,0) = (m,n)(m,n) = (m^2,n^2)$. So when is it true in $\mathbb Z$ for $m^2 = m$ and $n^2 = n$? only when $m$ and $n$ are $1$ or $0$.

This means that $f(1,0) = (1,0) , (0,1) , (1,1) , (0,0)$

Notice though, that they have to be in certain combinations with each other because of the reason Quimey stated.

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    $\begingroup$ You say 9 possibilities then list 4 answers.... $\endgroup$
    – user14972
    Aug 15, 2018 at 10:15
  • $\begingroup$ Also, not all preserve the multiplicative identity as required by the OP. $\endgroup$
    – badjohn
    Aug 15, 2018 at 10:53

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