True or False:
There exists injective ring homomorphism from $\Bbb R \oplus \Bbb R$ to $C(\Bbb R)$, where $C(\Bbb R)$ is the set of continuous functions from $\Bbb R$ to $\Bbb R$.
I was trying to think that I can exploit the fact $\Bbb R \oplus \Bbb R$ is not an integral domain but we can't because $(1,0) \to f$ and $(0,1) \to g$ both non zero continuous function s.t their product is zero.
Hint: In the ring $\Bbb{R}\oplus\Bbb{R}$ you have $(1,0)\times(0,1)=0$ and $(1,0)+(0,1)=1$.
Details: Let $f$ and $g$ denote the images of $(1,0)$ and $(0,1)$ in $C(\Bbb{R})$, respectively. Then $fg=0$ and $f+g=1$, so $g=1-f$, meaning that $$0=fg=f(1-f)=f-f^2.$$ It follows that $f=f^2$, and so $f(x)\in\{0,1\}$ for all $x\in\Bbb{R}$. Because $f$ is continuous we have either $f\equiv0$ or $f\equiv1$. But the elements $0,1\in R$ are already mapped to the functions $0,1\in C(\Bbb{R})$. This shows that there is no injective ring homomorphism from $\Bbb{R}\oplus\Bbb{R}$ to $C(\Bbb{R})$.