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?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $S_2$.

  • $\begingroup$ 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

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.


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.