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# Understand the problem

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 function from $\Bbb R$ to $\Bbb R$.
##### Source of the problem
TIFR GS 2019, Part B Problem 9
Abstarct algebra
Medium
##### Suggested Book
Abstarct algebra, Dummit and Foote

Do you really need a hint? Try it first!

In the ring $\Bbb{R}\oplus\Bbb{R}$ you have $(1,0)\times(0,1)=0$ and $(1,0)+(0,1)=1$.

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
$0=fg=f(1-f)=f-f^2.$ Now can you use this information?

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 $f\equiv 0$ or $f\equiv 1$. 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 homomorophism from $\Bbb{R}\oplus\Bbb{R}$ to $C(\Bbb{R})$.

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