Show that $K(a_1, \dots , a_n)=K[a_1, \dots , a_n]$ Let $L/K$ be a field extension and $a_1, \dots a_n\in L$, such that $a_1$ is algebraic over $K$, $a_2$ is algebraic over $K(a_1)$ and in general, $a_i$ is algebraic over $K(a_1, \dots , a_{i-1})$ for $i=2, \dots , n$. I want to show that $K(a_1, \dots , a_n)=K[a_1, \dots , a_n]$. 
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I have done the following: 
We have that $K[a]=\{f(a), f(x)\in K[x]\}$ and $K(a)=\{\frac{f(a)}{g(a)}, f(x), g(x)\in K[x] \text{ with } g(a)\neq 0\}=\{f(a)g(a)^{-1}, f(x), g(x)\in K[x] \text{ with } g(a)\neq 0\}$. 
So, $K[a]\subseteq K(a)$. 
Therefore, $$K[a_1, \dots , a_n]\subseteq K(a_1, \dots , a_n)$$ or not? 
We have to show that $K[a_1, \dots , a_n]$ is a field, since then every element will be invertible. 
Let $0\neq c\in K[a_1, \dots , a_n]$, then $c=f(a_1, \dots , a_n), f\in K[a_1, \dots , a_{i-1}, x]$. 
Let $p(x)=\text{Irr}(a_n, K[a_1, \dots , a_{n-1}])$. 
We have that $p(x)\not\mid f(a_1, \dots , a_{n-1})(x)$ since $p(a_n)=0$ but $f(a_1, \dots , a_n)\neq 0$. 
So, $(p, f(a_1, \dots , a_{n-1}))=1\Rightarrow \exists h,g\in K[a_1, \dots , a_{n-1}, x]$ with $h(a_1, \dots , a_{n-1}, x)p(x)+g(a_1, \dots , a_{n-1}, x)f(a_1, \dots , a_n)=1\overset {x=a_n}{\Longrightarrow} g(a_1, \dots , a_{n-1}, a_n)f(a_1, \dots , a_n)=1$. 
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Is this correct? Could I improve something? 
 A: $K[a_1, \dots , a_n]$ is a finite dimensional vector space over $K$ because only a finite number of powers of $a_1, \dots , a_n$ suffice to generate it.
Take $a \in K[a_1, \dots , a_n]$, $a\ne0$. Then the map $x \mapsto ax$ is an injective $K$-linear transformation of $K[a_1, \dots , a_n]$, and so must be surjective because of the rank–nullity theorem.
In particular, there is $b \in K[a_1, \dots , a_n]$ such that $ab=1$.
Therefore, every nonzero element in $K[a_1, \dots , a_n]$ has an inverse and so $K[a_1, \dots , a_n]$ is a field.
Since $K(a_1, \dots , a_n)$ is the smallest field containing $K[a_1, \dots , a_n]$, we must have $K(a_1, \dots , a_n)=K[a_1, \dots , a_n]$.
A: The first set inclusion is good, since we can set $g(a_1, ..., a_n) = 1$ to show that $K[a_1, ..., a_n] \subseteq K(a_1, ..., a_n)$ for any $n \in \mathbb{N}$.
The other set inclusion is good as well, but it would've been even faster using induction.  Since $K[a_1, ..., a_{n-1}][a_n] = K[a_1, ..., a_n]$, the induction step is automatically proven by the base case.  Thus, all you really needed to show was that $K(a) = K[a]$ for any field $K$ and any $a \in \overline{K}$. The fundamental idea, however, is the same: every polynomial without $a$ as a root will correspond to a nonzero element in $F[a]$, and further, every such polynomial will be relatively prime the minimal polynomial of $a$.  Therefore, one can use the Euclidean algorithm to find inverses as you suggest.
