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Let $K$ be a field. A field extension of $K$ is a morphism $\iota\colon K \hookrightarrow L$ of fields, often simply written as $K \subseteq L$. Now, there are several proves for the existence of an algebraic closure of $K$. One relies on taking the colimit of a certian diagram $$ K=K_0 \hookrightarrow K_1 \hookrightarrow \ldots \hookrightarrow K_n \hookrightarrow \ldots $$ In particular, we have a morphism $j \colon K \hookrightarrow \overline{K}= \varinjlim K_i$. And with respect to this morphism, $\overline{K}$ is an algebraic closure of $K$. To be precise, $j(K) \subseteq \overline{K}$ is an algebraic extension and $\overline{K}$ is algebraically closed.

Now, my intention for writing this question was to ask if we can always find an algebraic closure $L$ of $K$ such that $K$ is actually a subset of $L$. However, while writing this, I realized that this might not make too much sense if we do not specify, in the first place, where $K$ actually lives in as a set or rather as a subfield: That is, we have to indicate (at least) a superset of $K$ before we start searching for an algebraic closure in between this superset and $K$.

For example, consider $K(X_1,\ldots, X_n)$. Then $K \subseteq K(X_1,\ldots, X_n)$ has transcendence degree $n$. (In fact, this isn't really a subset relation, either; but let us understand $K$ as the constants in the field of rational functions). Now, let $\overline{K}$ denote the algebraic closure of $K$ arising from the colimit construction above. I am wondering, if there is a way of identifying $\overline{K}$ with some $L$ satisfying $K \subseteq L \subseteq K(X_1,\ldots, X_n)$. Of course, this identification should be a $K$-homomorphism, i.e. the triangle involving $K \subseteq L$, and $j\colon K \to \overline{K}$ should commute.
Such an identification is probably not canonical, but I think we can get one by using another construction of the algebraic closure involving Zorn's lemma applied to subfields of $K(X_1,\ldots, X_n)$, which then gives us a non canonical $K$-isomorphism $\overline{K} \cong L$.

In the end, this does not really turn out to be a question. But still, in case someone has helpful details/ information to add or corrections to make, I'll be grateful, if that person lets me know.

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  • $\begingroup$ It is irrelevant if (the underlying set of) $K$ is a subset of (the underlying set of) $\overline{K}$ or not. What is important is there is a homomorphism of fields $K \to \overline{K}$. You can do everything you want with this. $\endgroup$ – HeinrichD Oct 15 '16 at 11:28
  • $\begingroup$ (Steinitz 1) All field $K$ has an algebraic closure $K^*$. (Steinitz 2) The algebraic closure $K^*$ of a field $K$ is unique up to $K$-isomorphim. $\endgroup$ – Piquito Oct 15 '16 at 12:52
  • $\begingroup$ Can you please give me a reference where this language (of colimit to prove existence of algebraic closure) is used? Thanks. $\endgroup$ – Landon Carter Jan 24 '18 at 4:09
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Let $\phi: R \rightarrow S$ be a monomorphism of rings, so $S$ is the disjoint union of $\phi(R)$ and $S - \phi(R)$. Assume that $R$ and $S$ are disjoint as sets. Let $T$ be the set $R \cup [S - \phi(R)]$. Then there is a bijection of sets $f:T \rightarrow S$ given by $f(x) = x$ if $x \in S - \phi(R)$, or $f(x) = \phi(x)$ if $x \in R$.

The bijection $f$ allows you to endow $T$ with the structure of a ring. Now $T$ contains $R$ as a subset, and in fact the operations on $T$ restrict to the existing operations on $R$. Therefore $T$ is a ring which is isomorphic to $S$ and which contains $R$ as a subring.

But nobody ever does this construction. Usually, we just identify $R$ with its image in $S$.

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