Let $f$ be the ring automorphism of $\mathbf{C}$. Prove $f(1) = 1$.
I am a little stuck on this proof. I know by definition of ring isomorphism that the unity gets mapped to the unity, but that's not a proof that just me stating a property of ring homomorphisms.
Here is my attempt at a proof looking for feedback as proof writing is a struggle for me:
Suppose we have ring R and ring S where $f(x_R) \rightarrow x_S$, with R having unity 1.
and $f$ is onto.
Since $f$ is onto we always have $x \in R$ that maps to a element in $S$.
Note that then $f(x)\times f(1) = f(x \times 1)= f(x)$
That is $f(1)$ is the unity of $S$
Since the problem states that "$f$ be the ring automorphism of $\mathbf{C}$",meaning $f$ maps $\mathbf{C} \rightarrow \mathbf{C}$ and in our example $R, S =\mathbf{C}$. Then this implies 1 is the unity of $S$. Therefore $f(1)=1$