Find all automorphisms of the ring $ \mathbb{Z} $.

Question

I know that a automorphism is a isomorphism of the ring $\mathbb{Z}$ in $\mathbb{Z}$.

So, f: $\mathbb{Z}$ $\to$ $\mathbb{Z}$ is an isomorphism then the following statements hold true:

i) f is bijective

ii) f is an homomorphism of rings such that:

1. f(x+y) = f(x) + f(y)

2. f(xy) = f(y)f(x) $\forall$ x, y $\in$ $\mathbb{Z}$

1 and 2 are valid for all integers x, y. And f: $\mathbb{Z}$ $\to$ $\mathbb{Z}$ is bijective, so, all integers are isomorphism ?therefore all integers are automorphisms of the ring $ \mathbb{Z} $? Or I understood this wrong?

Answer 1

If $f$ has to be unital, that is, if $f(1)=1$, then there is only one.

If not, then because of condition $2.$, we have that $f(1)^2=f(1)$. That is, $f(1)$ is idempotent. There are only two idempotent elements of $\Bbb Z$, namely $1$ and $0$. So there are (at most) two. But actually only one, because we know $f(0)=0$, so if $f(1)=0$, then $f$ isn't bijective.