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.

  • $\begingroup$ 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

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 $S$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.