Counting the homomorphisms $\mathbb{Z} \rightarrow \mathbb{Z} \times .....\times\mathbb{Z}$ (n times)

Question

How many ring homomorphisms are there from $\mathbb{Z} \rightarrow \mathbb{Z} \times .....\times\mathbb{Z}$ (n times)?

My attempt: I got $2^n$. Is it true?

Answer 1

Let $f:\mathbb{Z} \rightarrow \mathbb{Z} \times .....\times\mathbb{Z}$(ntimes) be a ring homomorphism. Set $f(1)=(a_1,...,a_n)$. Now since $f(1)f(1)=f(1)$, we have $a_i^2=a_i$ for each $i$. Thus, $a_i\in\{0, 1\}$. Hence we have at most $2^n$ ring homomorphis. An easy argument shows that every function defined by $f(n):=n(a_1,...,a_n)$ , where $a_i\in\{0, 1\}$ is a ring homomorphism, so there are exactly $2^n$ ring homomorphism from $\mathbb{Z}$ to $\mathbb{Z} \times .....\times\mathbb{Z}$(ntimes). The only unital of them is the ring homomorphism such that $a_i=1$ for all $i$.