- #1

Icebreaker

"Let F be a field and let f:F->R be a ring homomorphism satisfying f(0) != f(1). Show that f is necessarily injective."

Assume f(a)=f(b), then f(a)-f(b)=0R => f(a-b)=0R. f(0F)=0R and therefore a=b.

But this implies that every homomorphism is injective. How can that be?

Assume f(a)=f(b), then f(a)-f(b)=0R => f(a-b)=0R. f(0F)=0R and therefore a=b.

But this implies that every homomorphism is injective. How can that be?

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