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Let $F$ be a field of characteristic $p$, and define $\phi: F \to F$ by $\phi(a) = a^p$, i.e. let $\phi$ be the Frobenius endomorphism of $F$.

In the text by Dummit and Foote, there is a result that states that $\phi$ is surjective if $F$ is finite. The reason for this is because $\phi$ is injective (the kernel is $\{0\}$) and $F$ is finite, which thus implies surjectivity.

I'm wondering why I can't drop the finiteness assumption by using the first isomorphism theorem for rings. We have $F/ker(\phi) = F/\{0\} \cong \phi(F)$, which gives $\phi(F) \cong F$. This would mean that $\phi$ is surjective, right?

Sam Y.
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  • Consider this analogy: $\phi : \mathbb Z \to \mathbb Z$ is a group homomorphism sending $x \mapsto 2x$. Here, $ker(\phi) = 0$, so $\mathbb Z \cong Z/ker(\phi) \cong \phi(\mathbb Z)$, yet $\phi$ is not surjective. – Kenny Wong Mar 15 '17 at 17:52
  • You may wish to take a look at my answer here: http://math.stackexchange.com/q/91688 – Zev Chonoles Mar 15 '17 at 17:54
  • @KennyWong Great example. So I guess the error in my "proof" was that $\phi(R) \cong R$ implies $\phi$ is surjective. But if these two rings are isomorphic, then there is a bijection between them, right? – Sam Y. Mar 15 '17 at 18:00
  • @ZevChonoles Very comprehensive answer you posted there... – Sam Y. Mar 15 '17 at 18:01
  • @SamY. There clearly exists a bijection between the two rings $R$ and $R$: the identity bijection will do nicely. But this doesn't help you. Your task is to show that $\phi$ is a bijection. – Kenny Wong Mar 15 '17 at 18:15
  • @KennyWong Sorry, I should have been more clear. When I said "but if these two rings are isomorphic", I was referring to the rings $\phi(R)$ and $R$ ... not the rings $F$ and $F$. – Sam Y. Mar 15 '17 at 19:58
  • @SamY. ...and yes, in my example, there is indeed a bijection between $\mathbb Z$ and $2 \mathbb Z$. – Kenny Wong Mar 15 '17 at 20:14
  • @KennyWong I agree there is a bijection between $\mathbb{Z}$ and $\mathbb{2Z}$. It seems like I am not understanding some pretty basic stuff here...by the First Isomorphism Theorem for Rings, we should have $\mathbb{Z}/ ker(\phi) \cong Im(\phi)$ or $\mathbb{Z} \cong \mathbb{2Z}$. But these rings are clearly not isomorphic, since the former has a $1$. – Sam Y. Mar 16 '17 at 00:13
  • @ My homomorphism $\phi : \mathbb Z \to 2 \mathbb Z$ is only a group homomorphism. It is not a ring homomorphism, since it fails to send $1 $ to $1$. – Kenny Wong Mar 16 '17 at 00:14
  • @KennyWong...I think I see the issue. The map in your example is an isomorphism of groups but not of rings. It is not an isomorphism of rings because $8 = \phi(2 \cdot 2) \neq \phi(2) \cdot \phi(2) = 16$. So we only have $\mathbb{Z}$ isomorphic to $\mathbb{2Z}$ as groups...So I'm not sure that the example you gave at the beginning applies, because the Frobenius endomorphism is an isomorphism of rings... – Sam Y. Mar 16 '17 at 00:17
  • @KennyWong You beat me to it! You type faster than me. I guess I would like to see a "counterexample" where $\phi: R \to R$ is an injective homomorphism of rings, but fails to be surjective...(this is what your example is for group homomorphisms) – Sam Y. Mar 16 '17 at 00:19

1 Answers1

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“Right?” No, fortunately, not right. You can see very clearly by noting that even in characteristic zero, if $x$ is an indeterminate, $$ \Phi:\Bbb Q(x)\to\Bbb Q(x), \qquad \frac{P(x)}{Q(x)}\mapsto\frac{P(x^2)}{Q(x^2)} $$ is a field morphism, so with zero kernel, so one-to-one, but the image (range) is $\Bbb Q(x^2)$, a proper subfield.

Lubin
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  • This is exactly what I wanted! Thank you. One question...why do you say at the beginning "No, fortunately, not right"? Why would it be "unfortunate" if every injective homomorphism of rings was surjective? – Sam Y. Mar 16 '17 at 00:22
  • Well, I guess I was being a wise-guy there. All sorts of theorems that I really like would be false otherwise. For instance, in my example, if $p(x)$ is a polynomial expression in $x$, then $\Bbb Q(p)\subset\Bbb Q(x)$, and the degree of this field extension is the degree of $p$. If instead of a polynomial, you use a rational function $p(x)=g(x)/h(x)$, then the field extension degree is the maximum of the degrees of $g$ and $h$. – Lubin Mar 16 '17 at 00:28
  • OK, I understand. I just spent a few minutes thinking about this example and it is just so simple and beautiful! It would have taken me much much longer to provide a counterexample to the claim that every injective endomorphism is surjective. So thanks. – Sam Y. Mar 16 '17 at 00:51