An example for a homomorphism that is not an automorphism

Solution 1:

Consider the extension $\mathbb Q(X)$ over $\mathbb Q$. The homomorphism $\mathbb Q[X] \to\mathbb Q[X]$, $X\mapsto X^2$ induces an homomorphism of the quotient field $\mathbb Q(X) \to \mathbb Q(X), f/g \mapsto f(X^2)/g(X^2)$. To show that this is not surjective assume $X = f(X^2)/g(X^2)$ for some $f,g \in \mathbb Q[X]$ with $g\neq 0$. Then $f(X^2) = Xg(X^2)$. But the left hand side contains only even powers of $X$, the right hand side contains only odd powers of $X$. Comparing coefficients shows $f = g = 0$ which is a contradiction since $g$ has to be non-zero.

We had to choose an extension that is not only infinite but also transcendent since any endomorphism is an automorphism if the extension is algebraic.

Solution 2:

Consider $\mathbb{Q}(\pi) / \mathbb{Q}$ and the field homomorphism induced from mapping $\pi$ to $\pi^2$.

The fact that this is not an automorphism follows from the transcendence of $\pi$.