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So I have a field homomorphism between two fields. I'm trying to show that this induces an isomorphism of their respective prime subfields. Then I'm supposed to show that their characteristics are the same. Assuming the isomorphism, showing the last part felt straightforward, and methinks I've proven surjectivity okay, but I'm having some trouble showing injectivity. Any help and/or hints would be much appreciated!

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    I still can't understand: what kind of problem about fields homomorphism does *not* assume knowledge of rings?! For me It's like trying to do a problem about integers without assuming a knowledge of the naturals...! Anyway, I can't see a way now to approach your problem that doesn't use, one way or another, knowledge about rings and ideals.2012-09-07

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Well, we can construct the prime subfield very explicitly. Let $k$ be any field, we have a unique map $\mathbb{Z} \rightarrow k$, if it's an injection, we induce a map $\mathbb{Q} \rightarrow k$, if it's not, we have a map $\mathbb{F}_p \rightarrow k$. The definition of prime subfield is intersection of all subfields $k'$, and since the map from $\mathbb{Z}$ into $k$ factors through any given $k'$, we deduce either (i) all contain $\mathbb{Q}$, or (ii) all contain $\mathbb{F}_p$, now letting $k' = \mathbb{Q}, \mathbb{F}_p$ gives the result.

With this, if $l \xrightarrow{i} k$ is any map, since the map from $\mathbb{Z} \rightarrow k$ factors through $l$, both have the same characteristic, equivalently prime subfield.


edit: in light of your comment, let's see that $l \xrightarrow{i} k$ must be an injection. Suppose $x \in l$ nonzero, consider its multiplicative inverse $x^{-1}$. We have $1 = i(1) = i(xx^{-1}) = i(x) i(x^{-1})$, hence $i(x)$ has a multiplicative inverse, and hence isn't 0.

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    what is $l$? An ideal?2015-09-04