Timeline for Finding $q(x)$ such that $p(q(x))$ is reducible over $\mathbb{Q}[x]$

9 events

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Apr 19, 2020 at 7:44	comment	added	Gerry Myerson		Hmm. I was following the argument in the answer. If $\beta=\alpha^2$, then ${\bf Q}(\beta)={\bf Q}(\alpha)$, and $\{\,1,\alpha,\beta\,\}$ is linearly independent over ${\bf Q}$, so $\beta$ is a root of $p(x^2)$, so $p(x^2)$ must have a factor of degree equal to the degree of $p$ while it has degree twice that. What goes wrong? Ah! found it! It's $\sqrt{\alpha}$ that's a zero of $p(x^2)$. Sorry.
Apr 19, 2020 at 7:36	comment	added	Jeremy Rickard		@GerryMyerson For example, if $p(x)=x^3-2$, then $p(x^2)$ is irreducible.
Apr 19, 2020 at 7:25	comment	added	R.P.		I don't think so, how would it follow?
Apr 19, 2020 at 0:10	comment	added	Gerry Myerson		Does it not follow that, if $p$ is irreducible and of odd degree, then $q(x)=x^2$ will do?
Apr 18, 2020 at 19:11	history	edited	R.P.	CC BY-SA 4.0	Fixed error
Apr 18, 2020 at 16:19	history	edited	R.P.	CC BY-SA 4.0	added 250 characters in body
Apr 18, 2020 at 16:13	comment	added	R.P.		Yes, I guess that's what I should have written... Thank you. Also I noticed after writing this that the question asked for $q$ to have integral coefficients, I don't know whether my method can easily give that.
Apr 18, 2020 at 16:11	comment	added	Jeremy Rickard		When you write "by subtracting the appropriate multiple of $p$", do you mean "by subtracting the appropriate multiple of the minimal polynomial of $\beta$"?
Apr 18, 2020 at 15:47	history	answered	R.P.	CC BY-SA 4.0