Can a Field lie "strictly" inside itself?

Question

Is there a field $F$ and a non-zero ring homomorphism $f:F \rightarrow F$ such that $f$ is not surjective?

In other words, is it possible for a field to strictly lie inside itself? I couldn't come up with an example, but I don't see this couldn't be the case. Clearly this is impossible for finite fields and for $\mathbb{Q}$.

Consider $t \longmapsto t^p$ for $\mathbb{F}_p(t)$, or any morphism $x \longmapsto R(x)$ of $k(x)$ where $R$ is a rational fraction which isn’t the quotient of two polynomials of degree at most one. — Aphelli, Mar 13 '21 at 00:10
You could also consider $F = k(x_1, x_2, \ldots)$ with $x_1 \mapsto x_2, x_2 \mapsto x_3, \ldots$. — Daniel Schepler, Mar 13 '21 at 00:18
You may find the Wikipedia article perfect field helpful to find fields which are not perfect. — Somos, Mar 13 '21 at 00:54
@DanielSchepler That should be an answer, it's certainly the simplest construction I can think of. — Noah Schweber, Mar 13 '21 at 01:34

score 4 · Accepted Answer · answered Mar 13 '21 at 01:53

4

Start with an arbitrary field $k$, and let $F=k(x)$ be the field of rational functions in one variable $x$ over $k$. There is a homomorphism $f:k(x)\to k(x)$ that acts on any rational function by substituting $x^2$ for $x$ (leaving the coefficients unchanged). It's clearly a non-zero homomorphism, and $x$ is not in its range.

answered Mar 13 '21 at 01:53

Andreas Blass

75,557

This would have been my answer. Thanks. – Lubin Mar 13 '21 at 03:11

score 1 · Answer 2 · answered Mar 13 '21 at 01:31

$\mathbb{C} \rightarrow \overline{\mathbb{C}(t)}$ are both algebraically closed fields, and have transcendence degree $2^{\aleph_0}$ over $\mathbb{Q}$.

Therefore, $\mathbb{C} \simeq \overline{\mathbb{C}(t)}$ (see here), but the inclusion is not surjective.

Can a Field lie "strictly" inside itself?

2 Answers2

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