Irreducibility of $X^4 + a^2$ with odd $a$

Question

For my abstract algebra course I have to decide whether $$(*)\qquad X^4 + a^2, \; a \in \mathbb{Z}\;\text{odd}$$ is irreducible over $\mathbb{Q}[X]$ and $\mathbb{Z}[X]$.

Since the degree of the polynomial is 4 and thus even I should be able to split it into two polynomials of degree two $if$ the coefficients of those polynomials lie in $\mathbb{Z}[X]$ and/or $\mathbb{Q}[X]$ respectively. What I know is that, if I could show that it is irreducible over $\mathbb{Z}[X]$ then I would know by Gauss' Lemma it is so also over $\mathbb{Q}[X]$. On the other hand, I suspect that it might be reducible over $\mathbb{Q}[X]$ which, in turn, would imply it is reducible over $\mathbb{Z}[X]$. What I tried is to set up some general product of two second-degree polynomials

$$(x^2 + bx + c)(x^2 + dx + e)$$ and get some conditions on which I could determine a set of parameters $\{b, c, d, e\}$ so that the polynomial in $(*)$ ensues. After coefficient comparison I get several conditions, i.e.

$$b + d = 0 \\ e + bd + c = 0 \\ be + cd = 0 \\ ce = a^2 = (2k + 1)^2 $$

for some $k \in \mathbb{Z}$.Here I did not know how to proceed. I saw that $b = -d$ and then tried to resolve the other equations but I did not make progress in that I found factors of (*). I obviously miss something here and appreciate a nudge in the right direction.

Answer 1

Hint: the roots of $x^4 + a^2$ are

$$ \pm\sqrt{a}\frac{1 \pm i}{\sqrt{2}}. $$

Separate these into conjugate pairs to find the factorization of $x^4 + a^2$ over $\mathbf{R}[x]$.

If you want to save time, or perhaps check your work, the answer is

$$ x^4 + a^2 = (x^2 - \sqrt{2a} \cdot x + a)(x^2 + \sqrt{2a} \cdot x + a). $$

Answer 2

After coefficient comparison I get several conditions

Hint:  it follows that $d=-b\,$, then either $b=0$ (which can be easily discarded) or $e=c=\pm a$ and $b^2 = 2c$. But the square of an integer cannot be twice an odd number, so the latter cannot hold.

Answer 3

First note it can't have a linear factor, as it would imply $X^4+a^2$ has a rational, hence real, root. Second it can be factored in $\mathbf R[X]$: $$X^4+a^2=(X^2+a)^2-2aX^2=(X^2-\sqrt{2a}X+a)(X^2+\sqrt{2a}X+a),$$ so all we have to prove is $\sqrt{2a}$ is irrational, which is equiavalent to $2a$ being a square in $\mathbf Z$. This is impossible since $a$ is odd.

Answer 4

Why did you not finish the computations? You already knew that $c+e=d^2$, that $d(c-e)=0$, and that $ce=a^2$. From the second equality, you can deduce that $d=0$ or $c=e$. But $d=0$ is not really a possibility, because then $c+e=0$, that is $c=-e$. This is not compatible with $ce=a^2$.

So, we know that $c+e=d^2$, that $c=e$ and that $ce=a^2$. Therefore, $c=e=a$ or $c=e=-a$. Since $c+e=d^2$, we can't have $c=e=-a$. So, $c=e=a$, $d=\sqrt{2a}$, and $b=-\sqrt{2a}$. This is impossible, because, since $a$ is an odd integer, $\sqrt{2a}$ cannot be an integer.

Answer 5

Since $d=-b$ you have $e+c=b^2$ and $ec=a^2$, with $a$ odd (hence $b$ even). Then $e,c$ are roots of $x^2-b^2x+a^2=0$, i.e., $$ (2x-b^2)^2=b^4-4a^2. $$ But $b^4-4a^2$ cannot be a square (of a even integer $2c$). Indeed, in the opposite case, setting $b=2B$ you would have $$ (2B)^4-4a^2=(2c)^2 \implies a^2+c^2=4B^4 \equiv 0\pmod{4}, $$ and this is impossible because $c^2 \not\equiv 3\pmod{4}$.

Answer 6

Since $a$ is odd, $a^2+1 \equiv 2 \pmod{4}$. So consider the polynomial $(x+1)^4+a^2$. Irreducibility of this polynomial can be ascertained by Eisenstein's criterion. This will show that your actual polynomial (which is a shift of this one) is also irreducible.