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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 $1$.2012-05-05

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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?2012-05-05
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    @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 $f=g$.2012-05-05
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    @ArturoMagidin I see. Thanks a lot!2012-05-05
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    @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.2012-05-05
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    @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.2012-05-05
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    @ymar: a ring homomorphism must map idempotents to idempotents. An idempotent must satisfy $x^2-x$; so in a field, the only idempotents are $1$ and $0$. So any ring homomorphism between two fields must map $1$ to $1$, or everything to $0$. If $f$ and $g$ are homomorphisms and one is equal to $0$, then $f+g$ is still a homomorphism. If both homomorphisms are different from zero, then $f(1)+g(1)=1+1$; since $1+1\neq 1$ in any field, if $f+g$ is a homomorphism then we must have $1+1=0$, so the characteristic is $2$. And then as above, we conclude that $f(r)=g(r)$ for all $r$, so $f=g$ and $f+g=0$2012-05-05
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    @ymar: Which means that the sum of two ring homomorphisms between fields is a ring homomorphism if and only if at least one of them is the zero map, or else the characteristic is $2$ and the two homomorphisms are equal (in which case their sum is $0$, which I *am* considering a homomorphism). In particular, if you consider "field homomorphism" to *always* map $1$ to $1$, then the above analysis shows that the sum of two field homomorphisms is never a field homomorphism, and will be a *ring* homomorphism if and only if we are in characteristic $2$ and the two maps are equal.2012-05-05
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    @ArturoMagidin Thank you for your patience. I got confused by what looked like a correction to your previous comment. I'm still not sure if you wanted to make a correction to something or not, but I _do_ understand what you are saying.2012-05-06
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    @ymar: In my first comment, I wrote: `It can happen if one of them is the zero map or the two maps are equal and we are in characteristic 0;` but if the two maps are equal and we are in characteristic $0$, then you don't get a ring homomorphism. What that second clause should say is "or the two maps are equal and we are in characteristic 2."2012-05-06
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    @ArturoMagidin Ah, I'm sorry. I missed that completely. I read it as you intended it.2012-05-06