Let L be an extension field of K and let $α_1, . . . , α_n$ ∈ L be algebraic over K. Then $K[α_1, . . . , α_n]$ is a field. 
Let $L$ be an extension field of $K$ and let $\alpha_1, \dotsc , \alpha_n \in L$ be algebraic over $K$. Then $K[\alpha_1, \dotsc , \alpha_n]$ is a field. Indeed, $K[\alpha_1, \dotsc , \alpha_n]$ is a finite extension of $K$.
Proof
We argue by induction on $n$, with the case $n = 1$ being given by Proposition 8.2.5.
  Suppose then by induction that $K[\alpha_1, \dotsc , \alpha_{n−1}]$ is a finite extension field of $K$.
  Since $K[\alpha_1, \dotsc , \alpha_n]$ is the smallest $K$-subalgebra of $L$ that contains the elements
  $\alpha_1, \dotsc , \alpha_n$, it must be the case that $K[\alpha_1, \dotsc , \alpha_n] = K[\alpha_1, \dotsc , \alpha_{n−1}][\alpha_n]$.
Now $\alpha_n$ is algebraic over $K$, and hence is a root of a nonzero polynomial $f(X) \in K[X]$.
  But $f$ may be regarded as a polynomial over $K[\alpha_1, \dotsc , \alpha_{n−1}]$, so $\alpha_n$ is algebraic over $K[\alpha_1, \dotsc , \alpha_{n−1}]$ as well. But Proposition 8.2.5 now shows that $K[\alpha_1, \dotsc , \alpha_n]$ is a finite extension of $K[\alpha_1, \dotsc , \alpha_{n−1}]$, and hence also of $K$ by Lemma 8.2.2.

There are two things that confuse me with this proof:


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*When the proof says "$K[\alpha_1, \dotsc , \alpha_n] = K[\alpha_1, \dotsc , \alpha_{n−1}][\alpha_n]$", we know that $K[\alpha_1, \dotsc , \alpha_{n−1}]$ is a field and it's being multiplied/composed with $[\alpha_n]$, right? But what is $[\alpha_n]$? It's just a single element right? So that just means that we are adding the element $\alpha_n$ to the field $K[\alpha_1, \dotsc , \alpha_{n−1}]$, right? I'm just asking to make sure that I understood it correctly.

*"Now $\alpha_n$ is algebraic over $K$, and hence is a root of a nonzero polynomial $f(X) \in K[X]$.
But $f$ may be regarded as a polynomial over $K[\alpha_1, \dotsc , \alpha_{n−1}]$" But what if the function $f(X)$ only has $\alpha_n$ as a root? Then how can we say that one of $\alpha_1, \dotsc , \alpha_{n−1}$ is also a root and hence $f(X)$ is contained in $K[\alpha_1, \dotsc , \alpha_{n−1}]$?
Thanks in advance.

For the second question, $f$ is a polynomial with coefficients in $K$, so, a fortiori, $f$ is a polynomial with coefficients in $K[\alpha_1,\dots,\alpha_{n-1}]$. That's what is meant by "$f$ may be regarded as a polynomial over $K[\alpha_1,\dots,\alpha_{n-1}]$." 
For the first question, $K[\alpha_1,\dots,\alpha_{n-1}][\alpha_n]$ means the smallest field containing $K[\alpha_1,\dots,\alpha_{n-1}]$ and $\alpha_n$. 
