# Finite extension $\Rightarrow$ a finite number of algebraic elements proof

I'm looking at Theorem 21.22 here, and specifically the proof of statement 1 implying statement 2, where:

1. $$E$$ is a finite extension of $$F$$
1. There exists a finite number of algebraic elements $$\alpha_1, \dots, \alpha_n \in E$$, such that $$E=F(\alpha_1, \dots, \alpha_n)$$.

The given proof of this goes (emphasis mine):

Let $$E$$ be a finite algebraic extension of $$F$$. Then $$E$$ is a finite dimensional vector space over $$F$$ and there exists a basis consisting of elements $$\alpha_1, \dots ,\alpha_n$$ in $$E$$ such that $$E=F(\alpha_1, \dots, \alpha_n)$$ . Each $$\alpha_i$$ is algebraic over $$F$$ by [a previous theorem].

My confusion arises in the statement that $$\alpha_1, \dots ,\alpha_n$$, are a basis for $$E$$ over $$F$$. This statement is a bit too quick for me.

Previously, for the case of the algebraic extension $$F(\beta)$$, we see that a basis for $$F(\beta)$$ is spanned by $$\{1, \beta, \beta^2, \dots , \beta^{d-1}\}$$, where $$d = [F(\beta):F]$$.

From this, I would then expect that for $$F(\alpha_1, \dots, \alpha_n)$$, we have a basis spanned by $$\{1, \alpha_1, \dots, \alpha_1^{d_1 -1}, \alpha_2, \dots, \alpha_2^{d_2 -1}, \dots, \alpha_n, \dots, \alpha_n^{d_n -1} \}$$, where $$d_i$$ is the degree of $$\alpha_i$$ over $$F$$. Presumably some of the $$\alpha_i^{p_i}$$ s are linearly dependent, so there are fewer than $$1 + \sum_{i=1}^n (d_i -1)$$ basis vectors.

So am I missing something in the statement that the algebraic elements $$\alpha_1, \dots ,\alpha_n$$ form a basis for $$E$$ over $$F$$ (should it be containing rather than consisting in the highlighted statement?)? Or is this just shorthand for the statement in the previous paragraph.

• Note that $F(\beta)=F(1,\beta,\beta^2,\ldots,\beta^{d-1})$. You can always extend a set that generates $E$ as a field into one that generates $E$ as a vector space, and then pare it down to a basis. Sep 14 at 18:37
• It seems like the following is true. Since $E$ is a finite algebraic extension, it is therefore a field. Since it is also finite vector space over $F$, it is spanned by some set of elements $\alpha_i$. Since $E$ is a field, all powers of $\alpha_i$ must also be in $E$. I think a way to think about this is that $E$ is the same as, say, $E(\beta,\gamma)$, which is the same as $E(\beta,\beta^2, \dots, \gamma, gamma^2, \dots$, and we're just calling the $\alpha_i$ the set of powers of $\beta$ and $\gamma$ that are linearly independent. Sep 14 at 18:52
• OK, so it seems that this is just an issue of the condensed notation hiding the full picture somewhat, which is understandable. Sep 14 at 18:57

Since $$E$$ is a field, it is closed under multiplication. In particular, all powers $$\alpha_j^m$$ of the the basis elements $$\alpha_1,\dots\alpha_n$$ (of the vector space $$E$$ over the field $$F$$) are also in $$E$$. Consider the extension $$F(\alpha_1,\dots,\alpha_n).$$ The field $$E$$ is a subfield of this extension because all elements of $$E$$ can be written as a linear combination of the $$\alpha_i$$. The extension is a subfield of $$E$$ because any finite linear combination in $$\{\alpha_j^k\}$$ is inside $$E$$. Thus, they are equivalent.
Let us observe that if $$a_1,a_2,\dots,a_n$$ are elements of $$E$$ then $$F(a_1,a_2,\dots,a_n) \subseteq E$$. Why?
Because we have $$F\subseteq E$$ and $$a_i\in E$$. Hence any element which is made up of elements of $$F$$ and $$a_i$$ using field operations lies in $$E$$. All these elements are precisely the members of $$F(a_1,\dots,a_n)$$.
Now assume that $$E$$ is finite dimensional extension of $$F$$ with $$[E:F] =n$$. Then we have a basis $$\{a_1,\dots,a_n\}\subseteq E$$ and every element of $$E$$ can be written as a linear combination of $$a_i$$ with coefficients in $$F$$. And since this linear combination lies in $$F(a_1,\dots,a_n)$$ we have $$E\subseteq F(a_1,\dots,a_n)$$ and thus we get $$E=F(a_1,\dots,a_n)$$.
If $$a$$ is algebraic over $$F$$ with degree $$n$$ then it can be proved that $$\{1,a,\dots,a^{n-1}\}$$ is a basis of $$F(a)$$ over $$F$$ but we can also find another set of $$n$$ members $$a_1,\dots,a_n$$ in $$F(a)$$ such that $$a_i$$ is a not a power of $$a_j$$ and yet the $$a_i$$ form a basis for $$F(a)$$ over $$F$$.
For example let $$F=\mathbb {Q},a=\sqrt{2}+\sqrt{3}$$ and $$E=F(a)$$. Then $$\{1,a,a^2,a^3\}$$ as well as $$\{1,\sqrt{2},\sqrt {3},\sqrt{6}\}$$ are bases of $$E$$ over $$F$$.