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Is it true that the direct limit of a Directed system in category of fields is just the union of fields in the system with obvious maps?

I think its true but i am confused because i am new to category theory.Any ideas?

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    Not only category theory but also Universal Algebra can provide an answer to your question, the latter being "less general" in a sense, so that your question would be easier to answer from this point of view. In Grätzer's *Universal Algebra*, the direct limit of any directed system of algebras (fields are a certain type of algebras) is described. It is in fact the quotient space of the union, with a certain equivalence relation, namely $x\sim y$ if and only if there exist $i,j,k,z$ such that $i, j\leq k$, $x\in A_i, y\in A_j, z \in A_k$ with $f_{ik}(x) = f_{jk}(y) = z$, with obvious notations – 2017-02-08
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    @Max I always thought that _universal algebra_ was about _equational theories_ (what is modelled by Lawvere theories in the categorical style). How are fields models of an equational theory ? How do you handle the axiom "$x \neq 0 \implies \exists y. xy=yx=1$" with only equational ones ? – 2017-02-08
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    Pece : field aren't models of any equational theory (it follows from Birkhoff's theorem that equational classes, aka varieties, are exactly those classes that are stable by product, homorphic images, and subalgebras; whereas fields obviously aren't stable by products. But Universal Algebra studies algebras in general, not only the equationally definable ones (otherwise, how could there be such a theorem ?). To stress this point, in Grätzer's book, there's barely been any mention of equations when limits are introduced – 2017-02-08

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Yes, this is correct. If $\{K_\alpha\}$ is a directed system, we can show that $K = \bigcup_\alpha K_\alpha$ has the universal property of a direct limit: if $L$ is any field, then a collection of compatible morphisms $f_\alpha : K_\alpha \to L$ lifts uniquely to a morphism $f: K\to L$.

To see this, note that any $a\in K$ comes from some $a_\alpha \in K_\alpha$, so we must have $f(a) = f_\alpha (a_\alpha)$. Once we have shown that this is really a well-defined (independent of $\alpha$) homomorphism of fields with the desired property, this gives us uniqueness. (Exercise: Fill in the details!)

One important note: It is easy to casually write $K = \bigcup_\alpha K_\alpha$, but we have to understand that this is a very specific kind of union, a directed union. We cannot just throw together the various $K_\alpha$, since they are not given as part of a larger set. Rather, we need to take the disjoint union and identify pairs that are compatible in the directed system. (This is the direct limit in the category of sets.)

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    Slade's answer overlooks the fact that the disjoint union of fields cannot in general be given a field structure. – 2017-02-08
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    @SixWingedSeraph I took a disjoint union of _sets_, not fields. – 2017-02-08
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    @Slade: I am facing trouble in verifying the well definedness,could you please add some more details? – 2017-02-10
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    @MathLover Hint: If $a_\alpha = a_\beta$ in the directed union, then $a_\alpha \in K_\alpha$ and $a_\beta\in K_\beta$ are compatible in the directed system. What does "compatible" mean? – 2017-02-10