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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 $F$.

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    $\begingroup$ “I am stuck at…” : are you dubious that $K\subset F(\alpha_a,\cdots,\alpha_n)$ or the reverse inclusion? $\endgroup$ – Lubin Mar 26 '19 at 4:16
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    $\begingroup$ @Lubin Yes, the direction $K\subset F(\alpha_1,\alpha_2,\ldots,\alpha_n)$ dubious $\endgroup$ – Silent Mar 26 '19 at 4:21
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$F(\alpha_1,\dots,\alpha_n)$ 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.

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  • $\begingroup$ Thank you very much! $\endgroup$ – Silent Mar 26 '19 at 6:50
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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)

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  • $\begingroup$ Thanks. Is there some expression for $F(\alpha_1,\alpha_2,\ldots,\alpha_n)$ similar to $F(\alpha)\cong F[x]/m_{\alpha,F}(x)$? $\endgroup$ – Silent Mar 26 '19 at 6:39
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    $\begingroup$ No single expression is guaranteed. $F_1=F(\alpha_1), F_2= F_1(\alpha_2),\ldots, K=F_n=F_{n-1}(\alpha_n)$, is what we can say definitely possible. With $F_{i+1}=F_i[x]/m_i(x)$ where $m_i(x) $ is the minimal polynomial of $\alpha_{i+1}$ over the field $F_i$. $\endgroup$ – P Vanchinathan Mar 26 '19 at 7:20
  • $\begingroup$ ok. thank you sir $\endgroup$ – Silent Mar 26 '19 at 7:35
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This answer is almost same as the answer P. Vanchinathan gave. I hope this slightly different view helps.


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).

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