Let $K$ be a field and let $f\in K[x]$ be an irreducible polynomial, of degree $n>0$.

Say that $f(x)$ has atleast two roots, $\alpha,\beta$, where both are not in $K$.

We take $K[x]/\langle f\rangle$ and looking at $\alpha = x+\langle f\rangle$ has:

$$f(\alpha) = a_n\alpha^n+a_{n-1}\alpha^{n-1}+\cdots+a_0$$ $$= a_n(x+\langle f \rangle)^n+a_{n-1}(x+\langle f \rangle)^{n-1}+\cdots + (1+\langle f\rangle)a_0$$ $$=a_n x^n+a_{n-1}x^{n-1}+\cdots+a_0 + \langle f\rangle = 0$$

So then $\alpha$ is a root of $f(x)\in K[x]/\langle f\rangle = L$.

Now, say that this factors $f(x)=(x-\alpha)g(x)$ and $g(x)$ is irreducible. We can now take:

$$K(\alpha)[y]/\langle g(y)\rangle$$

and look at $\beta = y+\langle g\rangle$.

$$g(\beta) =\cdots$$

So then $\beta$ is a root of $g(x)\in K(\alpha)(\beta)$

An arbitary element of $K(\alpha)$ looks like $K(\alpha)=\{a_{n-1}\alpha^{n-1}+\cdots + a_0:a_i\in K\}$

  • One what does an arbitrary element of $K(\alpha)(\beta)$ look like.
  • How do I reconcile the $f$ being of degree $n$, and the proof of $L= K[x]/\langle f \rangle$ utilises the degree $n$?
  • $\begingroup$ I think it will help if you instead of $L[x]$ use $L[y]$. Then you will see that a general element in $L(\beta)$ can be represented by an element of $K[x,y]$. $\endgroup$
    – Arthur
    Dec 5 '15 at 10:01

If you know what $L(\alpha)$ looks like, for any field $L$ and algebraic element $\alpha$ (over $L$), then you can easily figure out your first question. You have already noted that $$K(\alpha)=\{a_{n-1}\alpha^{n-1}+\cdots + a_0:a_i\in K\}$$

So an arbitrary element of $K(\alpha)$ can be written slightly more concisely as $$\sum_{i = 0}^{n-1} a_i\alpha^i,\quad a_i\in K$$

Now just apply the same reasoning to $K(\alpha,\beta) = K(\alpha)(\beta)$. An arbitrary element of this field will be:

$$\sum_{j = 0}^{m} b_j\beta^j,\quad b_j\in K(\alpha)$$

for an appropriate $m$. Now just expand each of the $b_j$:

$$\sum_{j = 0}^{m} \left(\sum_{i = 0}^{n-1} a_{ij}\alpha^i\right)\beta^j,\quad a_i\in K$$

which simplifies to

$$\sum_{j = 0}^{m} \sum_{i = 0}^{n-1} a_{ij}\alpha^i \beta^j,\quad a_i\in K$$

In other words, they are just bivariate polynomials in $\alpha,\beta$, of limited degree in each.

I'm not quite sure what your second bullet means, could you clarify?

  • $\begingroup$ Thanks, I am reading your answer to my first bullet point, it makes sense to me. For the second bullet point, I meant that my proof for $K(\alpha)$ relies on the degree $n$ of $f$, where, when I get to $K(\alpha)[y]/\langle g\rangle$ I am not sure what degree I should be using for $g(y)$. Just $n-1$ if I assume that $\alpha$ is not a repeated root? $\endgroup$ Dec 5 '15 at 12:43
  • $\begingroup$ Oh, I see now. That's actually why I only said "for appropriate $m$". I may be wrong (I can't think of an example right now), but I don't think that $g$ can be assumed to be irreducible. The reason I think this is because $\alpha$ could be a multiple root in $L$. Though, upon searching a little bit, it seems this only happens in characteristic $p > 0$: math.uconn.edu/~kconrad/blurbs/galoistheory/separable1.pdf (starting at theorem 2.8 is what I refer to). However, in characteristic $0$, $g(x)$ is irreducible and of degree $n-1$, by which in my answer the appropriate $m$ is $n-2$. $\endgroup$
    – GPerez
    Dec 5 '15 at 13:05

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