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Let $K/F$ be a field extension, I know that if $K/F$ is a finite extension then a simple argument from linear algebra shows that since every homomorphism of fields from $K$ to $K$ that fixes $F$ is 1-1 it is also onto, i.e. an automorphism of $K$.

Can someone please give an example that this is not the case if $K/F$ is not a finite extension ?

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    All field homomorphisms are monomorphisms.2012-06-06
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    @DavidWheeler - indeed, but how is this relavent ?2012-06-06
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    Well, it means you need a field extension of infinite degree, so that you can map injectively to a subfield.2012-06-06
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    @DavidWheeler, I noted this in the post...2012-06-06

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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 $\pi$.

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    Is this because $\pi$ is not in the image ? I think we can use $\mathbb{Q}(x)$ instead and not have to use this fact about $\pi$(that I am told is hard to prove...)2012-06-06
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    In general, one can mimic this same construction for the field of rational functions over any base field of coefficients.2012-06-06
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    You can use any transcendental number you want, or $\mathbb{Q}(x)$ - the argument is exactly the same.2012-06-06
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    Can you show in a similar way that the K-monomorphism (injective homomorphism) is not necessarily K-automorphism? Moved the question [here](http://math.stackexchange.com/questions/1711451/k-monomorphism-that-is-not-k-automorphism).2016-03-24
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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.

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    How did yoy get $\mathbb{Q}(x)$ as a quotient ?2012-06-06
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    @Belgi: $\Bbb Q(x)$ is the fraction field of $\Bbb Q[x]$.2012-06-06
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    yes, I think I remember the build of the fraction field and I think it is not a quotient...2012-06-06
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    I used "quotient field" as a synonym of "field of fractions" which, according to Wikipedia, is the same.2012-06-06