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This question is a variation on another one : related question

Let $f(x)$ and $g(x)$ be polynomials with integer coefficients and discriminant of $f(x)$ > discriminant of $g(x)$ , which are irreducible over $\mathbb{Z}$ and have degree $>1$.

It appears that $f(g(x))$ and/or $g(f(x))$ always factors over an extension of degree $p$ where $p$ is the degree of $f(x)$ and/or $g(x)$ , and that extension does not belong to the extension(s) needed for the zero's of $f(g(x))$ resp $g(f(x))$ (*) or their extensions.

(* if you factor $f(g(x))$ its $f(g(x))$ and likewise for the other)

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I'm confused as to how factoring $f(g(x))$ over an extension "does not belong to the extension(s) needed for the zero's of $f(g(x))$". Don't you mean factoring into linear factors, and thus having the required zeros in any such extension? Maybe a simple example with two quadratic polynomials would clarify your meaning? – hardmath Sep 19 '12 at 12:51
See the related question : same comment about confusion and a comment by me to examplify. If examplify is a word. – mick Sep 19 '12 at 12:57
@mick I think it's an interesting tack you've taken here by asking what the roots of polynomials have to do with the roots of their compositions, and what this means in terms of Galois theory. However, I'd like you to consider that this connection does not seem to be very strong at all. For example, the roots of $f=x+1$ and $g=x-1$ have little to do with the roots of $f\circ g(x)=g\circ f(x)=x$. A better example is $f(x)=x^2$ and $g(x)=x+1$. Here $f$ and $g$ even have real roots, while $g\circ f(x)=x^2+1$ does not. – rschwieb Sep 19 '12 at 13:16
Please don't ask people to go to another question to read what should be in this one. If there is confusion here, clear it up here. – Gerry Myerson Sep 19 '12 at 13:23
@rschwieb: Thanks for comments. However. Two remarks. 1) I did not mention your strong connection so I feel you are commenting on your own thoughts rather than mine. 2) Your first example violates my conditions since of degree 1 and your second example is trivial and unrelated to my ideas about ..well.. anything. In fact i did not even mention Galois theory in this OP here actually. I did not speak out about its ( Galois theory )importance in neither direction here. With respect but you only argumented $x^2 + 1 = 0$ has no real solutions , im not that new to math :) – mick Sep 19 '12 at 13:26

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