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Let $a \in \mathbb{Z^{+}}$. Show that $x^4+a$ is reducible over $\mathbb{Q}$ if and only if $a=4b^4$ for some integer $b$.

My idea for one implication was to assume reducibility and write $x^4+a = (x^2+\alpha_1x + \beta_1)(x^2+\alpha_2x + \beta_2)$, leading to the system of equations: $\alpha_1 + \alpha_2 = 0$, $\beta_1+\beta_2+\alpha_1 \alpha_2 = 0$, $\alpha_2 \beta_1 + \alpha_1 \beta_2 = 0$, $\beta_1 \beta_2 = a$. But I wasn't able to get much farther than this. This is a practice problem for an exam, any help is appreciated.

Bill Dubuque
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frnkcl99
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    Is there any missing information about $a$ (for example, is $a$ assumed to be $>0$)? Here is a counterexample: $a=-1$ makes $x^4+a$ reducible, but $a\ne 4b^4$. Also $a=0$ is a trivial counterexample. – Batominovski Apr 05 '20 at 01:33
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    In general, if $a$ is a rational number, then $x^4+a$ is reducible over $\Bbb Q$ if and only if $a=-t^2$ or $a=4t^4$ for some rational number $t$. – Batominovski Apr 05 '20 at 01:41
  • Yes, my bad, $a$ is assumed to be positive. I've edited the question to include this. Thanks. – frnkcl99 Apr 05 '20 at 01:41
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    Here is a general theorem on irreducibility of $,x^n - a,$ over any field (use $1.4(ii)\Rightarrow 1.5(ii)$, quite simple). – Bill Dubuque Apr 05 '20 at 04:19

2 Answers2

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You don't solve the whole system at once. Since the cubic coefficient must be zero, we must have $$ (x^2 + cx + d)(x^2 - cx + e) $$ The linear coefficient is $ce - cd,$ and this must be zero. So, $c(d-e) = 0.$ When $c=0,$ in order to get the quadratic coefficient zero, we need $d+e = 0.$ So the case $c=0$ gives $$ x^4 -d^2, $$ which is ruled out

The other case is $d=e.$ We continue with $$ (x^2 + cx + d)(x^2 - cx + d) = x^4 + (2d-c^2)x^2 + d^2 $$ Since $c,d$ are integers, we have $c = 2 \gamma$ and $d = 2 \gamma^2,$ finally $$ (x^2 + 2 \gamma x + 2 \gamma^2) (x^2 - 2 \gamma x + 2 \gamma^2) $$ which becomes..........

Will Jagy
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Another solution is to first factor out into linear factors over $\mathbb{C}$, and multiply conjugate factors to obtain the real quadratic factors:

$$X^4 + a = \left(X^2 + ia^{1/2}\right)\left(X^2 - ia^{1/2}\right) = \color{red}{\left(X + \frac{1-i}{\sqrt{2}}a^{1/4}\right)}\color{blue}{\left(X - \frac{1-i}{\sqrt{2}}a^{1/4}\right)}\color{red}{\left(X + \frac{1+i}{\sqrt{2}}a^{1/4}\right)}\color{blue}{\left(X - \frac{1+i}{\sqrt{2}}a^{1/4}\right)}$$ $$ = \color{red}{\left(X^2 + \sqrt{2}a^{1/4}X + a^{1/2}\right)}\color{blue}{\left(X^2 - \sqrt{2}a^{1/4}X + a^{1/2}\right)}$$

where we used $\sqrt{i} = \pm \frac{1+i}{\sqrt{2}}$ and $\sqrt{-i} = \pm\frac{1-i}{\sqrt{2}}$.

(you can also check the factorization directly without using complex numbers).

From this expression it is clear that we need $\sqrt{2}a^{1/4} = 2r$ for some $r \in \mathbb{Q}$ which implies $a = 4r^4$.

Tob Ernack
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