Field theory and polynomials Proposition

Ring homomorphisms from a field are injective

Builds on Ring homomorphism Field A ring homomorphism is injective iff its kernel is zero

Statement

Let $K$ be a field, let $S$ be a nonzero ring and let $f \colon K \to S$ be a ring homomorphism. Then $f$ is injective.

Remarks

The proof: if $a \neq 0$ and $f(a) = 0$ , then $1 = f(1) = f(a) f(a^{-1}) = 0$ in $S$ . This contradicts $S \neq 0$ . The hypothesis $S \neq 0$ is necessary, because the map from $K$ to the zero ring is a ring homomorphism. A consequence: a homomorphism of fields makes the first field a copy of a subfield of the second.

Reading path

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Group concept
Ring definition
Commutative ring definition
Ring homomorphism definition
Unit of a ring definition
Field definition
Kernel of a ring homomorphism definition
A ring homomorphism is injective iff its kernel is zero statement
Ring homomorphisms from a field are injective statement

Tag	`09N7DZ7SDB`
URI	`m2://09N7DZ7SDB`
Kind	statement, proposition
Writing	W0 stub
MSC	12E99
Status	This copy of the site has no build data. Run `m2 status 09N7DZ7SDB`.
Uses	Ring homomorphism, Field, A ring homomorphism is injective iff its kernel is zero

Ring homomorphisms from a field are injective

Statement

Remarks

Reading path

Further reading

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