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In the text by Dummit and Foote, the authors introduce the idea of an algebraic closure, and also prove that one always exists for every field $F$. They then comment that we can "...speak sensibly of the composite of any collection of algebraic extensions by viewing them as subfields of an algebraic closure."

But I'm a bit confused by this. Suppose we have two fields $K_1$ and $K_2$. Then we have the two algebraic closures $\overline{K_1}$ and $\overline{K_2}$, but it doesn't necessarily follow that either one of these algebraic closures contains both $K_1$ and $K_2$...

So I have two questions:

  1. How do we construct the big field that contains both $K_1$ and $K_2$?

  2. The quoted comment said that $K_1$ and $K_2$ must be algebraic extensions...why is this?

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  • $\begingroup$ I don't know much field theory, but don't you need more conditions on $K_1,K_2$? Choose $K_1 = \mathbb F_2$, and $K_2 = \mathbb Q$. How can you have a mutual extension of them? $\endgroup$ – Mark Apr 11 '17 at 23:39
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    $\begingroup$ If you know about tensor products then, for a field $F$ contained in both $K_1$ and $K_2$, the ring $K_1 \otimes_F K_2$ leads to a construction you seek: pick a maximal ideal $\mathfrak m$ in this ring (an $F$-algebra) and the composite $K_1 \rightarrow (K_1 \otimes_F K_2)/\mathfrak m$ where $x \mapsto x \otimes 1 \bmod \mathfrak m$ is a ring homomorphism of fields and hence injective, and likewise for $K_2 \rightarrow (K_1 \otimes_F K_2)/\mathfrak m$ by $y \mapsto 1 \otimes y \bmod \mathfrak m$. $\endgroup$ – KCd Apr 11 '17 at 23:59
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The first thing to note here is that $K_1$ and $K_2$ must be algebraic extensions of the same base field. There's no way we can find a field $L$ such that both $\mathbb{F}_9$ and $\mathbb{Q}(\sqrt{2})$ are subfields of $L$.

So let's assume that $K_1$ and $K_2$ are both algebraic extensions of some base field $F$. The general idea here is that $K_1 = F(S_1)$ and $K_2 = F(S_2)$, where $S_1$ and $S_2$ are (possibly countably infinite) collections of roots of polynomials in $F[x]$. Because the algebraic closure of $F$ is constructed by adjoining all the roots of every possible polynomial in $F[x]$ to $F$, we must have $F(S_1) \subset \overline{F}$, and likewise for $F(S_2)$. In many cases, we do not have to go all the way to the algebraic closure; notice that both $F(S_1)$ and $F(S_2)$ will be contained in $F(S_1 \cup S_2)$. For a concrete example, $\mathbb{Q}( \sqrt{2})$ and $\mathbb{Q}( \sqrt{3})$ are both contained in $\mathbb{Q}( \sqrt{2}, \sqrt{3})$.

Technically, it seems that one can find a larger field containing both $K_1$ and $K_2$ as long as these fields have the same characteristic -- i.e. both have $\mathbb{Q}$ or both have $\mathbb{Z}_p$ (for the same $p$) as a subfield. Even if $K_1$ and $K_2$ are not strictly algebraic extensions, they will nevertheless be generated by (i.e. have a basis of) some collection of algebraic elements together with some collection of transcendental elements. One can find a field containing both $K_1$ and $K_2$ by considering the field generated by the union of all of the algebraic / transcendental generators for both fields.


Recall that we can think of extensions of a field $F$ as a vector space over $F$. Because of this, we are able to make analogies with vector spaces we're comfier working with to help grasp the concept:

Suppose we have two vector spaces $S_1$ and $S_2$, which are subspaces of $\mathbb{R}^{10}$. Let's denote the "standard basis vectors" as $\mathbf{e}_k$, which have all zeros except for a $1$ in the $k^\text{th}$ slot. Let $S_1$ be the subspace of $\mathbb{R}^{10}$ with the basis $\mathbf{e}_1, \mathbf{e}_5, \text{ and } \mathbf{e}_9$, and let $S_2$ be the subspace with basis $\mathbf{e}_2$ and $\mathbf{e}_3$. That is to say:

$$S_1 = \text{Span} \Big( \mathbf{e}_1, \mathbf{e}_5, \mathbf{e}_9 \Big)$$ $$S_2 = \text{Span} \Big( \mathbf{e}_2, \mathbf{e}_3 \Big)$$

One can ask: "What is the smallest subspace of $\mathbb{R}^{10}$ that contains both $S_1$ and $S_2$?". It isn't too hard to convince ourselves that the answer will be the subspace spanned by $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3, \mathbf{e}_5, \text{ and } \mathbf{e}_9$. In other words, the subspace requested has as its basis elements the union of the basis elements from the two original subspaces $S_1$ and $S_2$.

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  • $\begingroup$ thanks for this answer. Two questions: (1) If $K_1$ is an algebraic extension, how do you know there exists a set $S_1$ of roots with $K_1 = F(S_1)$? If this is some fundamental result that I am just not aware of, I apologize. And (2) What is your definition of $F(S_1)$? I would like to say it is the smallest subfield of _____ containing $F$ and $S_1$, but the whole point of this discussion seems to be about finding ______...? $\endgroup$ – Sam Y. Apr 12 '17 at 0:34
  • $\begingroup$ @SamY., to answer your first question, the definition of an algebraic extension is that every element is a root of a polynomial in $F[x]$ (with elements of $F$ being roots of linear polynomials). We can simply start choosing elements in $K_1 \setminus F$ to adjoin to $F$ so as to create a tower $F \subset F(a_1) \subset F(a_1, a_2) \subset \cdots \subset F(S_1) = K_1$. There are two ways to think of the notation $F(a)$, and one can show that they are equivalent. First, $F(a)$ can be thought of as the smallest field containing both the element $a$ and $F$ as a subfield. $\endgroup$ – Kaj Hansen Apr 12 '17 at 3:35
  • $\begingroup$ Second, $F(a) = \{ f(a) \ | \ f(x) \in F[x] \}$ - i.e. the set of all polynomial evaluations at $a$. Sorry if it feels like I'm rambling, but I think this is all potentially relevant to get a good feel for what's going on. You can read further discussion I have on this point here: math.stackexchange.com/questions/1080336/… $\endgroup$ – Kaj Hansen Apr 12 '17 at 3:39
  • $\begingroup$ And with that, it seems I have answered your second question: $F(S_1)$ can indeed be thought of as the smallest subfield of the algebraic closure of $F$ containing both $F$ and $S_1$. This is sort of going back to the above point: it can also be thought of in terms of a set of polynomial evaluations. What's more, there is a set $T \supseteq S_1$ such that $F(T)$ is the algebraic closure of $F$. It seems to me the whole point of this discussion is, given two algebraic extensions $K_1$, $K_2$ of $F$, to construct an extension $K \subseteq \overline{F}$ such that $K$ contains both $K_1$ and $\endgroup$ – Kaj Hansen Apr 12 '17 at 3:43
  • $\begingroup$ $K_2$. The best way of seeing this construction is in terms of a "basis" for the extension (in the sense that every extension of $F$ can be thought of as a vector space over $F$). The "basis" for the field you want to construct is simply the union of the two bases for the original extensions $K_1$ and $K_2$. It's all about spanning sets. At any rate, I think the above link might be key to understanding the idea I'm struggling to convey here: indeed, we can think of $F(S_1)$ as "the smallest field containing $S_1$ and $F$", but that's not being satisfactorily descriptive unless we think $\endgroup$ – Kaj Hansen Apr 12 '17 at 3:48
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After some thought, I don't think my question is too complicated. The first thing to note (as was pointed out in the comments section and by Kaj Hansen) is that the fields $K_1$ and $K_2$ must both be extensions of the same base field $F$.

Now to answer the two questions I originally asked:

  1. How do we construct the big field that contains both $K_1$ and $K_2$?

Since $K_1 / F$ is algebraic over $F$, we must have $K_1 \subseteq \overline{F}$, where $\overline{F}$ is one choice of algebraic closure for $F$ (there are many choices, but they are all isomorphic). If we stick with this same choice of algebraic closure, then $K_2 \subseteq \overline{F}$. So $\overline{F}$ is the "big field".

  1. The quoted comment said that $K_1$ and $K_2$ must be algebraic extensions...why is this?

I think it is still possible to construct a big field extension of $F$ containing $K_1$ and $K_2$ if $K_1$ and $K_2$ are not both algebraic over $F$. But it is more straightforward if $K_1$ and $K_2$ are algebraic over $F$, since we can use $\overline{F}$ as our big field.

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