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!
Prime Subfield Isomorphism
-
0Hint: any *ring* homomorphism from a field to some ring (that maps 1 to 1, to avoid trivial homomorphism) is automatically injective... – 2012-09-07
-
0Further hint: a commutative unital ring is a field iff its only ideals are $\,\{0\}\,$ and it itself. – 2012-09-07
-
0@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. – 2012-09-07
-
0At 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... – 2012-09-07
-
0@DonAntonio I do know! But the problem assumes I don't! – 2012-09-07
-
0I 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
1 Answers
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.
-
0Ah, very cool! Very helpful! Thankya sir. – 2012-09-07
-
0haha thanks man, but I haven't been knighted yet lmao :p – 2012-09-07
-
0?? 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... – 2012-09-07
-
0@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... – 2012-09-07
-
0what is $l$? An ideal? – 2015-09-04