# Field homomorphism is trivial or injective

Let $$F$$ and $$G$$ be fields, and let $$f: F \to G$$ be a field homomorphism. Then one of the following is the case:

• $$f$$ is the constant $$0$$ map: for every $$x \in F$$, we have $$f(x) = 0_G$$.

• $$f$$ is injective.

# Proof

Let $$f: F \to G$$ be non-constant. We need to show that $$f$$ is injective; equivalently, for any pair $$x,y$$ of elements with $$f(x) = f(y)$$, we need to show that $$x = y$$.

Suppose $$f(x) = f(y)$$. Then we have $$f(x)-f(y) = 0_G$$; so $$f(x-y) = 0_G$$ because $$f$$ is a field homomorphism and so respects the “subtraction” operation. Hence in fact it is enough to show the following sub-result:

Suppose $$f$$ is non-constant. If $$f(z) = 0_G$$, then $$z = 0_F$$.

Once we have done this, we simply let $$z = x-y$$.

## Proof of sub-result

Suppose $$f(z) = 0_G$$ but that $$z$$ is not $$0_F$$, so we may find its multiplicative inverse $$z^{-1}$$.

Then $$f(z^{-1}) f(z) = f(z^{-1}) \times 0_G = 0_G$$; but $$f$$ is a homomorphism, so $$f(z^{-1} \times z) = 0_G$$, and so $$f(1_F) = 0_G$$.

But this contradicts that the image of the identity under a group homomorphism is the identity, because we may consider $$f$$ to be a group homomorphism between the multiplicative groups $$F \setminus \{ 0_F \}$$ and $$G \setminus \{0_G\}$$, whereupon $$1_F$$ is the identity of $$F \setminus \{0_F\}$$, and $$1_G$$ is the identity of $$F \setminus \{0_G\}$$.

Our assumption on $$z$$ was that $$z \not = 0_F$$, so the contradiction means that if $$f(z) = 0_G$$ then $$z = 0_F$$. This proves the sub-result and hence the main theorem.

Parents:

• Algebraic field

A field is a structure with addition, multiplication and division.