# Questions about basic field theory from Lang's book

I began to study Field Theory from Lang's book and I would be happy to discuss some questions which I am going to write down below:

Question 1. Suppose that $$\mathbb{k}$$-field, $$E$$ extension field of $$\mathbb{k}$$ and $$\alpha_1,\alpha_2\in E$$.

Am I right that $$\mathbb{k}(\alpha_1,\alpha_2)=\mathbb{k}(\alpha_1)(\alpha_2)$$?

The LHS is the smallest subfield in $$E$$ containing $$\mathbb{k},\alpha_1$$ and $$\alpha_2$$. The RHS is the smallest subfield in $$E$$ containing $$\mathbb{k}(\alpha_1)$$ and $$\alpha_2$$.

Proof: Since $$\mathbb{k}(\alpha_1,\alpha_2)$$ contains $$\mathbb{k},\alpha_1,\alpha_2$$ then it also contains $$\mathbb{k}(\alpha_1)$$ and $$\alpha_2$$ then it contains also $$\mathbb{k}(\alpha_1)(\alpha_2)$$. Hence $$\mathbb{k}(\alpha_1)(\alpha_2)\subset\mathbb{k}(\alpha_1,\alpha_2)$$.

Conversely, $$\mathbb{k}(\alpha_1)(\alpha_2)$$ contains $$\mathbb{k},\alpha_1,\alpha_2$$ then it immediately contains $$\mathbb{k}(\alpha_1,\alpha_2)$$. Thus we proved the equality. Is the proof correct?

Question 2. If $$\alpha$$ algebraic over $$\mathbb{k}$$, and $$\mathbb{k}\subset F$$ and $$\mathbb{k}[\alpha],F\subset L$$ then $$\alpha$$ is algebraic over $$F$$.

This proposition seems to me quite weird because the condition $$\mathbb{k}[\alpha],F\subset L$$ is extra.

Indeed, if $$\alpha$$ is algebraic over $$k$$ then there exists $$a_0,\dots,a_n\in k, n\geq 1$$ (not all of them are zero) such that $$a_0+a_1\alpha+\dots+a_n\alpha^n=0$$, but since $$\mathbb{k}\subset F$$ then it follows that $$\alpha$$ is algebraic over $$F$$.

Am I right? Maybe I am misubderstanding something?

Question 3: If $$E=\mathbb{k}(\alpha_1,\dots,\alpha_n)$$, and $$F$$ is an extension of $$\mathbb{k}$$, both $$F,E$$ contained in $$L$$, then $$EF=F(\alpha_1,\dots,\alpha_n).$$

Can anyone show how prove this equality? And what does it mean?

Would be very grateful for detailed answers!

EDIT: How it follows from question 1 that $$k(\alpha_1,\dots,\alpha_{n-1})(\alpha_n)=k(\alpha_1,\dots,\alpha_{n-1},\alpha_n)$$? I don't know how to prove it correctly and rigorously?

For your first two questions, you are correct. (The extra conditions in Question 2 are probably to avoid technical concerns about the existence of a common field containing both $$\mathbb k[\alpha]$$ and $$F$$. E.g. there is not common field containing both $$\mathbb Q$$ and $$\mathbb F_2$$.) For your follow-up to question 1, apply exactly the same logic: $$\mathbb k(\alpha_1,\ldots, \alpha_{n-1})(\alpha_n)$$ contains $$\mathbb k$$, $$\alpha_1,\ldots, \alpha_n$$ and so contains $$\mathbb k (\alpha_1,\ldots, \alpha_n)$$, and the converse is similar.

For 3, essentially the claim is that the compositum $$EF$$ is the same as the extension $$F(\alpha_1,\ldots, \alpha_n)$$ (again, the field $$L$$ is just to avoid technical concerns). In this case, the compositum $$EF$$ is the smallest subfield of $$L$$ containing $$E\cup F$$.

Since $$EF$$ contains $$E$$, it contains $$\alpha_1,\ldots, \alpha_n$$. As $$EF$$ also contains $$F$$, by definition $$F(\alpha_1,\ldots, \alpha_n)\subseteq EF$$.

Conversely, note that $$F(\alpha_1,\ldots, \alpha_n)$$ contains $$\mathbb k$$ (as $$F$$ is an extension) as well as $$\alpha_1,\ldots, \alpha_n$$, hence $$\mathbb k(\alpha_1,\ldots, \alpha_n)=E$$. Obviously $$F\subseteq F(\alpha_1,\ldots, \alpha_n)$$, thus by definition $$EF\subseteq F(\alpha_1,\ldots, \alpha_n)$$. Thus we conclude $$EF= F(\alpha_1,\ldots, \alpha_n)$$.

EDIT: To see why the conditions in 2 are needed, let's break it down. We are given a field $$\renewcommand\k{\mathbb k}$$ $$\k$$ and an element $$\alpha$$ which is algebraic over $$\k$$. This is ambiguous in that algebraic elements are only defined in the context of a field extension (or if we are a bit less literal, it makes sense in any ring containing $$\k$$ and $$\alpha$$). This is because the statement that $$p(\alpha)=0$$ for some $$p(x)\in\k[x]$$ only makes sense if we can parse how to multiply powers of $$\alpha$$ with elements of $$\k$$ and add those things together.

This is usually glossed over in the context of a single field, or when all fields are subfields of an algebraically closed superfield containing everything (like $$\overline{\mathbb Q}$$ or $$\mathbb C$$). This is because in these contexts,there is an obvious context in which we can put $$\alpha$$; in the case of a single field, we can just say $$\alpha=x+q(x)\in \k[x]/[q(x)]$$ for some irreducible $$q$$; or in the case of a superfield $$L$$, we can just say $$\alpha\in L$$ and be done with it.

However, if there are two fields $$\k$$ and $$F$$, there is not necessarily a common field containing both. The obvious example of this is when the characteristics differ, such as $$\mathbb F_2$$ and $$\mathbb Q$$, but even for fields of the same characteristic it is not obvious how to combine arbitrary fields. (Simply saying 'the smallest field containing both fields' is not helpful, since it is not obvious the set of fields containing both subfields is non-empty, and as far as I recall there is not a universal construction of a compositum of arbitrary fields of the same characteristic.)

To avoid such technical details, and since it is most oten the case anyhow, it is often simplest to assume that all fields under examination are subfields of some field, so that the various extensions can be examined in a common context.

That's what is going on here: to avoid ambiguity, we assume $$\k, \alpha,$$ and $$F$$ are all within a common field so that we can parse what it means to be algebraic.

• But in question 2 we do not need $\mathbb{k}[\alpha]$ at all. So why do we make some assumptions on $\mathbb{k}[\alpha]$? It seems pointless – ZFR Feb 5 '19 at 21:52
• I will edit my answer with a response. – Sean Clark Feb 5 '19 at 22:24
• Would be grateful for that! – ZFR Feb 5 '19 at 22:28