Determine all ring homomorphisms of $\varphi:\mathbb{Z}_{10} \to \mathbb{Z}_{10}$

Question

This is what I came up with as a solution and I wanted some suggestions on how to proceed with the last part or perhaps a new direction of thinking.

Given that $m=n=10$ we must have $0=m\cdot a=10a$ which is in $\mathbb{Z}_{10}$ $\Longleftrightarrow$ $10$ divides $10a$ $\Longleftrightarrow$ 1 divides $a$. Clearly this is trivial so $\varphi(1)\in \mathbb{Z}_{10}$. But the elements which are idempotent in $\mathbb{Z}_{10}$ are $0,1,5,6$. According to Gallian and Van Buskirk (1984) there should be $2^{\omega(n)-(\omega(n/(m,n))}$ ring homomorphisms. In other words there should be two. I'm leaning toward $0,5$ but I am unsure how to demonstrate that $1,6$ cannot be included. Any suggestions?

I'm not looking for a response where people can stroke their ego, I am genuinely not sure how to do this other than looking at the group homomorphisms, which I didn't fully understand either.

Answer 1

Usually a ring homomorphism must by definition send the multiplicative identity to the multiplicative identity, so we must have $\phi(1) = 1$.

Answer 2

I was incorrect in assuming there were only two ring homomorphisms. When I calculated $2^{\omega(n)-\omega(n/(m,n))}$ I neglected to realize that it yields 4 instead of 2. The four ring homomorphisms for $\varphi : \mathbb{Z}_{10}\to\mathbb{Z}_{10}$ are in fact:

$$\varphi : \mathbb{Z}_{10}\to\mathbb{Z}_{10};x\mapsto0$$ $$\varphi : \mathbb{Z}_{10}\to\mathbb{Z}_{10};x\mapsto x$$ $$\varphi : \mathbb{Z}_{10}\to\mathbb{Z}_{10};x\mapsto 5x$$ $$\varphi : \mathbb{Z}_{10}\to\mathbb{Z}_{10};x\mapsto 6x$$