Show that it doesn't exist any homomorphism $\phi: \mathbb{C} \to \mathbb{R}^{\mathbb{R}}$

Question

1. Considering the ring of all of the real functions $ \mathbb{R}^{\mathbb{R}}$. Show that it doesn't exist any homomorphism $\phi: \mathbb{C} \to \mathbb{R}^{\mathbb{R}}$
2. Let $R$ be an integral domain and consider $f:R\to R\,\,\,\,$ s.t. $\,\,\,\, f(a)=a^2,\,\,\,\,\forall a \in R$. Show that $f$ is injective $\iff$ $f$ is a homomorphism.

About 1. any hint on how to show that?

About the point 2.: I have been thinking about that $\forall a \in R, \,\,\, f(a) = a^2 = (-a)^2 = f(-a)$ and, assuming $f$ injective, it means $a=-a \Rightarrow a + a = 2a =0_R\,\,\,\,$ Then: $\,\,\,\,\,a^2+b^2 = a^2+ (2a)b +b^2=(a+b)^2=f((a+b))=f(a)+f(b)=a^2+b^2\,\,\,\,\,$

The fact that $R$ is an I.D. gives the product because we can say that $(ab)^2=a^2b^2$, so $f$ is homomorphism. To show that homomorphism implies injectivity do I need only to retrace this demonstration?

Answer 1

1. You must have $\phi(i)^2=-1$ and it is impossible.

Answer 2

For your other question: if $-1 \neq 1$, then $f $ can't be injective since $f(-1) = f(1) = 1$. And if $-1 = 1$, then $2=0$ and so $f$ respects addition as per your calculation.