A field extension L: K is finite if and only if there exists a finite subset ${α_1, α_2, . . . , α_k}$ of L such that $α_i$ is algebraic over K for i = 1, 2, . . . , k and $$L = K(α_1, α_2, . . . , α_k)$$.
I get that If it has n algebraic numbers and $L = K(α_1, α_2, . . . , α_k)$then its degree has to be finite.
I also get if The extension is finite then it is algebraic and then the basis must consists of finite algebraic numbers .WHat i dont get is Why $$L = K(α_1, α_2, . . . , α_k)$$ Maybe i do not understand what the Set $K(α_1, α_2, . . . , α_k)$ is.I understand that is the the Fractions Field of $K[α_1, α_2, . . . , α_k]$ which is the all the polynomials P of arbitary degree $P(a_1,a_2....a_n)$.So my question is L has a basis n with algebraic numbers so L is all the linear combinations of the set ${α_1, α_2, . . . , α_k}$ Why that makes $$L = K(α_1, α_2, . . . , α_k)$$
By definition, $K(\alpha_1,\dots,\alpha_k)$ is a subfield of $L$ which contains every element of $K$ and also contains $\alpha_1,\dots,\alpha_k$. For any $x\in L$, we can write $x=\sum_{i=1}^k c_i\alpha_i$ for some $c_1,\dots,c_k\in K$, since $\{\alpha_1,\dots,\alpha_k\}$ is a basis for $L$ over $K$. Since each $c_i\in K\subseteq K(\alpha_1,\dots,\alpha_k)$, each $\alpha_i\in K(\alpha_1,\dots,\alpha_k)$, and $K(\alpha_1,\dots,\alpha_k)$ is a subfield, it follows that $\sum_{i=1}^k c_i\alpha_i\in K(\alpha_1,\dots,\alpha_k)$. That is, $x\in K(\alpha_1,\dots,\alpha_k)$. Since $x\in L$ was arbitrary, this means $K(\alpha_1,\dots,\alpha_k)=L$.