# Minimal polynomial of an algebraic number expressed in terms of another algebraic number

I am working on the following exercise: Let $\alpha \in \mathbb{C}$ be a root of the polynomial $f(X) = X^4 - 3X - 5$.

1. Prove that $f$ is irreducible in $\mathbb{Q}[X]$.
2. Find the minimal polynomial of $2\alpha - 3$ over $\mathbb{Q}$.
3. Find the minimal polynomial of $\alpha^2$ over $\mathbb{Q}$.

Here are my thoughts:

I am okay with question one ($f$ is irreducible of $\mathbb{Z}_2$ and hence over $\mathbb{Q}$) but struggling with the rest of the exercise. I have found this relevant question but fail to apply Gerry’s answer to this example. Could someone give me a hint?

• for 2) look at $f(aX+b)$ – reuns Jan 7 '17 at 23:19
• for 3) Set $\beta:=\alpha^2$, and then $\beta^2=3\alpha+5$ gives $\alpha=(\beta^2-5)/3$. This implies that $f((x^2-5)/3)$ vanishes at $\beta$. Expanding the latter, you will get a polynomial of degree $8$, and one of its factors (perhaps of degree 4) will be the minimal polynomial of $\beta$. – math Jan 7 '17 at 23:47
• See this question for part 3. – Jyrki Lahtonen Jan 8 '17 at 7:31

2) Let $\beta = 2 \alpha - 3$. As user1952009 points out, for suitably chosen $a,b$, you can show that $f(aX+b)$ has $\beta$ as a root. As a further hint, note that $\alpha = \frac{\beta+3}{2}$.

3) Here's a solution in a similar vein to Gerry Myerson's linked answer. As a vector space over $\mathbb{Q}$, $\mathbb{Q}(\alpha)$ has $1, \alpha, \alpha^2, \alpha^3$ as a basis. We compute the matrix of the multiplication map \begin{align*} \lambda_{\alpha^2}: \mathbb{Q}(\alpha) &\to \mathbb{Q}(\alpha)\\ x &\mapsto \alpha^2 x \end{align*} with respect to this basis, which is $$A = \left(\begin{array}{rrrr} 0 & 0 & 5 & 0 \\ 0 & 0 & 3 & 5 \\ 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 0 \end{array}\right) \, .$$ (The fancy name for this is the regular representation.) One can show that $\alpha^2$ is a root of the characteristic polynomial of $A$. (This is a general fact and there is nothing special about considering $\alpha^2$; see Dummit and Foote, $\S13.2$, Exercise $20$, p. $531$.) You'll still have to show that this polynomial is irreducible, but actually the same trick you used for 1) should work.

• Thank you very much. I am accepting this answer because it is the most complete and general one. – bbrot Jan 8 '17 at 17:43

3. $(\alpha^2)^2-5=3\alpha$ implies that $((\alpha^2)^2-5)^2=9\alpha^2$.$\alpha^2$ is a root of $(X^2-5)^2-9X=X^4-10X^2-9X+25$ this implies that $g(\alpha^2)=0$ where $g(X)=X^4-10X^2-9X+25$.

Remark that $\alpha={{(\alpha^2)^2-5}\over 3}$ we deduce that $\alpha\in Q(\alpha^2)$ and $Q(\alpha^2)=Q(\alpha)$. This implies that the degree of $[Q(\alpha^2):Q]=4$ and $g$ is the minimal polynomial of $\alpha^2$.

• Thank you, so that’s the clever answer to question three :) – bbrot Jan 8 '17 at 17:46
1. put $x=2\alpha-3 \Rightarrow x+3=2\alpha\Rightarrow (x+3)^4=16\alpha^4 \Rightarrow \frac {(x+3)^4}{16}=3\alpha+5; \alpha^4=3\alpha+5 \Rightarrow \frac {(x+3)^4}{16}=3(\frac{x+3}{2}) +5 \Rightarrow \frac {(x+3)^4}{16}-3(\frac{x+3}{2}) -5=0$ then $f(x)= \frac {x^4}{16}+\frac{3x^3}{4}+\frac{27x^2}{8}+\frac{21x}{4}-\frac{71}{16}=0$ is minimal polynomial of $2\alpha-3$
• Thank you, this is indeed nicer then evaluating $f(\frac{\beta + 3}{2})$ directly... – bbrot Jan 8 '17 at 17:49