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Homomorphism to Continuous function: TIFR GS 2019, Part B Problem 9

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
Topic
Abstarct algebra
Difficulty Level
Medium
Suggested Book
Abstarct algebra, Dummit and Foote

Start with hints

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}).

Hence, the statement is false.

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