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Well the question is stated in the title. I dont know much about field theory and i was suprised when i read it on wikipedia

please provide some examples

thanks in advance

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If you don't know any field theory, why do you find this statement surprising? –  Pete L. Clark May 16 '12 at 19:54
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What Wikipedia in fact says is "Field has neither products nor coproducts." This is slightly weaker than the claim that it doesn't have any products or coproducts, and in fact some trivial cases do exist. For example, the coproduct of $\mathbb{Q}$ with $\mathbb{Q}$ is $\mathbb{Q}$. –  Chris Eagle May 16 '12 at 19:55
    
@Chris: good point. I took the question to be an explanation of the quoted fact (for coproducts, but of course the same argument works for products). If the OP wants to know necessary and sufficient for a coproduct of fields to exist, s/he should clarify accordingly. (This is actually a pretty good question, now that I think about it: it has to do with tensor products, composita and so forth...) –  Pete L. Clark May 16 '12 at 20:02
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@PeteL.Clark: I'd quite like to see an answer to that question! (Also, isn't it in principle possible for the coproduct to exist in the category of fields but be different from the coproduct in the category of commutative rings?) –  Zhen Lin May 16 '12 at 20:06
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@Zhen: sure, that is (like most things!) "in principle possible". However, when the coproduct exists in the category of rings and is a field, then I think it is in principle impossible that this is anything else than the coproduct in the category of fields. :) By the way, please go ahead and ask the followup question, if you are interested. –  Pete L. Clark May 16 '12 at 20:10
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2 Answers

up vote 12 down vote accepted

Among other things, a coproduct of objects $F_1$ and $F_2$ in a category is an object $F$ together with morphisms $\iota_1: F_1 \rightarrow F$, $\iota_2: F_2 \rightarrow F$.

In order to have a homomorphism between two fields $K$ and $L$, $K$ and $L$ must have the same characteristic. Thus for instance $\mathbb{Q}$ and $\mathbb{Z}/p\mathbb{Z}$ (for any prime $p$) cannot have a coproduct in the category of fields. (Added: they can't have a product either, for almost exactly the same reasons.)

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Ok i get that we cant find a object with two arrow from $\mathbb{Q}$ and $\mathbb{Z/pZ}$ but doesn't this simply mean that the pathcategory $(\triangle\downarrow T)$, (T is the functor with the field above) is empty, which inturn should imply that the coproduct is the 0 field, or am i missing something and perhaps there is no 0 field? –  user25470 May 16 '12 at 20:18
    
well... –  anon May 16 '12 at 20:25
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@user25470: Yes, there is no $0$ field. (@anon: the "field with one element", whatever it is, is not an element of the category of fields!) –  Pete L. Clark May 16 '12 at 20:33
    
I know, I know. ("if such a field could exist") –  anon May 16 '12 at 20:36
    
very well, i accept your answer as true, but i dont really understand why. In the rings we accept 0 as multiplicative identity, why can't we do the same for fields? –  user25470 May 16 '12 at 22:46
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As Pete said, two fields $F_1$ and $F_2$ can only have a coproduct if they have the same prime field $F$ ($F=\mathbb Q$ or $F=\mathbb F_p$).

a) If the $F$-algebra $F_1\otimes_F F_2$ is a field, then it is a coproduct of $F_1$ and $F_2$ in the category of fields.
The simplest examples are the fields $\mathbb Q(\sqrt 2) \otimes_\mathbb Q (\sqrt 3)=\mathbb Q(\sqrt 2,\sqrt 3)$ and $\mathbb F_{p^2} \otimes_{\mathbb F_p} \mathbb F_{p^3}=\mathbb F_{p^6}$
It is however a delicate question to decide whether $F_1\otimes_F F_2$ is a field: see here for many examples and non-examples.

b) If $F_1\otimes_F F_2$ is not a field and if a coproduct $F_1\sqcup F_2$ of $F_1$ and $F_2$ exists, we have a ring morphism $F_1\otimes_F F_2\to F_1\sqcup F_2$. But I don't know if it really possible that $F_1\sqcup F_2$ exists if $F_1\otimes_F F_2$ is not a field .

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