# 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:

1. 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.

2. "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 $K[\alpha_1, \dotsc , \alpha_{n−1}]$?

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 $\alpha_n$.
• Well, technically, it means "the smallest ring containing ...", or more technically, "the smallest subring of $L$ containing ...". – Hurkyl Apr 15 '13 at 18:37
• @Hurkyl is correct. In this problem, the $\alpha_i$ were given as algebraic over $K$ so, fortunately, I am also correct. – Gerry Myerson Apr 16 '13 at 0:00