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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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Hint: any ring homomorphism from a field to some ring (that maps 1 to 1, to avoid trivial homomorphism) is automatically injective... –  DonAntonio Sep 7 '12 at 4:22
    
Further hint: a commutative unital ring is a field iff its only ideals are $\,\{0\}\,$ and it itself. –  DonAntonio Sep 7 '12 at 4:23
    
@DonAntonio So I see that the kernel of the homomorphism is an ideal, and so it must be the trivial one by hints 1 and 2, but I'm supposed to be able to do this without knowing what a ring is yet. Is there a more direct computational way I could go about it? I feel like my dumb is showing. –  AsinglePANCAKE Sep 7 '12 at 4:31
    
At most what is showing is your ignorance (as it happened once upon a time to us all), not your dumb. Anyway, you know what field is but not a ring? That's odd, as you seem to know what an ideal is... –  DonAntonio Sep 7 '12 at 4:33
    
@DonAntonio I do know! But the problem assumes I don't! –  AsinglePANCAKE Sep 7 '12 at 4:37

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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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Ah, very cool! Very helpful! Thankya sir. –  AsinglePANCAKE Sep 7 '12 at 4:45
    
haha thanks man, but I haven't been knighted yet lmao :p –  uncookedfalcon Sep 7 '12 at 4:46
    
?? How the above does not use rings, @AsinglePANCAKE? What do you think the map $\,\Bbb Z\to k\,$ is? A ring homomorphism! Otherwise you cannot deduce all that stuff... –  DonAntonio Sep 7 '12 at 4:47
    
@DonAntonio The edit watered it down enough (or technically, the opposite of that). I mean, i can just be thought of as a field homomorphism... –  AsinglePANCAKE Sep 7 '12 at 4:53

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