# 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.