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

Question

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 $F$

2. 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.

Answer 1

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.

Answer 2

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 $E=F(a_1,\dots,a_n)$.

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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 $F$.