Why is $K=F(\alpha_1,\alpha_2,\ldots,\alpha_n)$ here?

Question

This is from Dummit and Foote:

Theorem: If the extension $K/F$ finite, then $K$ is generated by a finite number of algebraic elements over $F$.

Proof: If $K/F$ is finite of degree $n$, let $\alpha_1, \alpha_2,..., \alpha_n$ be a basis for $K$ as a vector space over $F$. So, $[F(\alpha_i):F]$ divides $[K:F]=n$ for $i=1,\ldots,n$, so that, each $\alpha_i$, is algebraic over $F$. Since $K$ is obviously generated over $F$ by $\alpha_1,\alpha_2,\ldots,\alpha_n$, we see that $K$ is generated by a finite number of algebraic elements over $F$.

I am stuck at that 'obviously': Why is $K=F(\alpha_1,\alpha_2,\ldots,\alpha_n)$?

Also, I will be obliged if you provide something similar as $F(\alpha)\cong F[x]/m_{\alpha,F}(x)$ (where $\alpha$ algebraic over $F$ and $m$ is minimal polynomial) for $F(\alpha_1,\alpha_2,\ldots,\alpha_n)$ where each of $\alpha_i$ is algebraic over $F$.

Answer 1

The hypothesis $K/F$ is finite means, $K$, regarded as a vector space over the field $F$ is of finite dimension. The set $\{\alpha_1,\alpha_2,\ldots, \alpha_n\}$ is a basis for this vector space: SO every element of $K$ is obtained as a linear combination of elements $\alpha_i$'s with coefficients from $F$.

So definitely the same elements suffice to generate $K$ as an extension field over $F$. Note that such an extension may not be simple (i.e generated as a field by a single element) unless additional hypothesis on $F$ is available (characteristic zero, or separable)

Answer 2

$F(\alpha_1,\dots,\alpha_n)$ is by definition the smallest subfield of $K$ containing $F$ and containing $\alpha_1,\dots,\alpha_n$. So $K \supseteq F(\alpha_1,\dots,\alpha_n)$ follows from what it means to be "the smallest subfield containing..."

On the other hand, $\alpha_1,\dots,\alpha_n$ form a basis of $K$ over $F$. This means that every element of $K$ can be written as a linear combination

$$x_1\alpha_1 + \dots + x_n \alpha_n \tag{1}$$

for some $x_1,\dots,x_n \in F$.

Now think about the following: every subfield of $K$ that contains $F$ and contains $\alpha_1,\dots,\alpha_n$ must contain all linear combinations that look like $(1)$. In particular, such a field must be equal to $K$ since $K$ is the set of these linear combinations.

Answer 3

This answer is almost same as the answer P. Vanchinathan gave. I hope this slightly different view helps.

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Suppose $[K:F]=2$. Let $\alpha,\beta$ be a basis for $K$ as a vector space over $F$. So every element of $K$ can be written as $a_1\alpha+a_2\beta,\,$ where $a_1,a_2\in F$. Now every element of $F(\alpha,\beta)$ can be written as $$\sum_{i,j}a_{ij}\alpha^i\beta^j,\,a_{ij}\in F$$ with suitable limits(see Dummit and Foote). Comparing the elements, we see that $K\subseteq F(\alpha,\beta)$. You can generalise the same argument for the case $[K:F]=n$ (but, writing out the elements will be a nasty job).