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

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 $$F$$.

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

$$F(\alpha_1,\dots,\alpha_n)$$ 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.

• Thank you very much! – Silent Mar 26 '19 at 6:50

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)

• Thanks. Is there some expression for $F(\alpha_1,\alpha_2,\ldots,\alpha_n)$ similar to $F(\alpha)\cong F[x]/m_{\alpha,F}(x)$? – Silent Mar 26 '19 at 6:39
• 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 $F_i$. – P Vanchinathan Mar 26 '19 at 7:20
• ok. thank you sir – Silent Mar 26 '19 at 7:35

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).