I'm looking at Theorem 21.22 here, and specifically the proof of statement 1 implying statement 2, where:
- $E$ is a finite extension of $F$
- 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.
Best Answer
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.