# In what vector space does Dedekind's lemma live?

Dedekind's lemma in field theory says this:

Let $E$ and $L$ be fields, and $\sigma_1,\ldots,\sigma_n:E\longrightarrow L$ be distinct field homomorphisms. Then $\sigma_1,\ldots,\sigma_n$ are $L$-linearly independent, that is $$\sum_{i=1}^na_i\sigma_i=0,\;a_i\in L\implies (\forall i)\;a_i=0,$$

Is there an $L$-vector space in which this linear independence takes place? All field homomorphisms from $E$ to $L$ don't constitute a vector space because there is no neutral element in this set. Also, the sum of two homomorphisms may not be a homomorphism. If $\operatorname{char}L=2,$ and $\sigma:E\longrightarrow L$ is a field homomorphism, then $\sigma+\sigma=0$ isn't a field homomorphism. I'm not really sure when the sum of two homomorphisms is again a homomorphism. Also, I'm not sure when the additive inverse of a homomorphism is again a homomorphism. Could you please help with these questions?

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The zero map can be considered a homomorphism between fields. Any ring homomorphism between fields is either maps $1$ to $1$, or is the zero map, so if $\mathrm{char}(L)\neq 2$, then the sum of two nonzero ring homomorphisms cannot be a field homomorphism, since $1$ is idempotent but it would be mapped to $2$; the only two idempotents in a field are $0$ and $1$. –  Arturo Magidin May 5 '12 at 23:11

## 1 Answer

The space of all additive homomorphisms from $E$ to $L$ ($E$ and $L$ are considered as abelian groups w.r.t. addition). If $f$ is such homomorphism and $a\in L$, then $af$ maps $x$ to $a\cdot f(x)$.

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Thanks! Could you also say something about the sum of two field homomorphisms being a field homomorphism? When can it happen? –  user23211 May 5 '12 at 19:44
@ymar: It can happen if one of them is the zero map or the two maps are equal and we are in characteristic $0$; note that $f(1)=g(1)=1$ implies $(f+g)(1) = 1+1$; since any field homomorphism must either map $1$ to $1$ or everything to zero, we need $1+1=0$, so the characteristic is $2$. Then for every $a,b$, $b\neq 0$, expanding $(f+g)(ab)$ two ways we obtain $f(a)g(b)=g(a)f(b)$, so $f(\frac{a}{b}) = g(\frac{a}{b})$, so taking $a=rb$ we conclude $f(r)=g(r)$ for all $r$, so $f=g$. –  Arturo Magidin May 5 '12 at 22:00
@ArturoMagidin I see. Thanks a lot! –  user23211 May 5 '12 at 23:26
@ymar: Above, the first sentence should say "and we are in characteristic 2". I'm considering the zero map to be a "homomorphism", which you are not. If you don't consider the zero map to be a homomorphism, then the sum of two field homomorphisms is never a field homomorphism. –  Arturo Magidin May 5 '12 at 23:29
@ArturoMagidin I understood that about the zero map being a homomorphism. I'm not attached to the notion that it's not. But now I'm not sure I understand what you mean by the first sentence in your last comment. I can't see a mistake in your previous one. –  user23211 May 5 '12 at 23:47