Showing $x^4 + 1$ is irreducible in $\mathbb{Q}[x]$.

Question

Clearly, none of the roots are in $\mathbb{Q}$ so $f(x) = x^4 + 1$ does not have any linear factors. Thus, the only thing left to check is to show that $f(x)$ cannot reduce to two quadratic factors.

My proposed solution was to state that $f(x) = x^4 + 1 = (x^2 + i)(x^2 - i)$ but $\pm i \not\in \mathbb{Q}$ so $f(x)$ is irreducible.

However, I stumbled across this post $x^4 + 1$ reducible over $\mathbb{R}$... is this possible? with a comment suggesting that $x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$ which turns out to be a case that I did not fully consider. It made me realize that $\mathbb{Q}[x]$ being a UFD only guarantees a unique factorization of irreducible elements in $\mathbb{Q}[x]$ (which $x^2 \pm i$ nor $x^2 \pm \sqrt{2} x + 1$ aren't in $\mathbb{Q}[x]$) so checking a single combination of quadratic products is not sufficient.

Therefore, what is the ideal method for checking that $x^4 + 1$ cannot be reduced to a product of two quadratic polynomials in $\mathbb{Q}[x]$? Am I forced to just brute force check solutions of $x^4 + 1 = (x^2 + ax + b)(x^2 + cx + d)$ don't have rational solutions $(a,b,c,d) \in \mathbb{Q}^4$?

Answer 1

I'd say it's easiest to use Eisenstein's criterion after a shift, although this might be a sledgehammer. $$(x+k)^4 + 1 = x^4 + 4k x^3+6k^2 x^2 + 4k^3 x + k^4 +1$$ So we want a prime $p$ dividing each of $\{4k, 6k^2, 4k^3, k^4 + 1\}$ but $p^2$ not dividing $k^4 + 1$. That looks easy enough: let $p = 2$ and $k = 1$.

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(If you want to use a lot more theory and make the test a little simpler, note that the discriminant of $x^4+1$ is $256$, whose only prime factor is $2$, so in fact $p=2$ is the only prime that could possibly work. I used to understand why this was true, but I no longer do; it's something to do with theorem 1.3 in https://kconrad.math.uconn.edu/blurbs/gradnumthy/disc.pdf . I believe it may be possible to use this theory to show that $p=2$ does work without finding $k$, but that's far beyond my pay grade.)

Answer 2

Alternative approach:

My proposed solution was to state that $f(x) = x^4 + 1 = (x^2 + i)(x^2 - i)$ but $\pm i \not\in \mathbb{Q}$ so $f(x)$ is irreducible.

This isn't quite valid. Even when neither of the two 2nd degree polynomials given by
$[(x - r_1) \times (x - r_2)], ~[(x - r_3) \times (x - r_4)]$ are polynomials in $~\mathbb{Q}[x],~$
it is still possible that the 2nd degree polynomials given by
$~[(x - r_1) \times (x - r_3)]~$ and $~[(x - r_2) \times (x - r_4)]~$ are polynomials in $~\mathbb{Q}[x].$

To me, it is unclear what the intent of the problem composer is. If he intended that brute force be avoided, then my response (below) is not what the composer intended.

Alternatively, if the composer was merely testing your understanding of the basic idea that a 4th degree polynomial in $~\Bbb{Q}[x]~$ would be irreducible, and assuming that you had not been exposed to deeper theory (re the answer of Patrick Stevens), the brute force approach is not that bad.

Using the variable $~z,~$ instead of $~x,~$ to denote a complex root, note that the $4$ roots of $z^4 = 1$ are given by the set of values $\{1,i,-1,-i\} = \{e^{i(0)}, e^{i\pi/2}, e^{i\pi}, e^{3i\pi/2}\}.$

Further, one of the roots of $z^4 = -1$ is given by $z = e^{i\pi/4}$.

Therefore, the $4$ roots of $z^4 = -1$ are given by
$\{e^{i\pi/4}, e^{3i\pi/4}, e^{5i\pi/4}, e^{7i\pi/4} \}.$

In order for the polynomial $[(z-r_1) \times (z - r_2)]$ to be an element in $\Bbb{Q}[x]$, you need both of the following:

• $[r_1 + r_2]$ must be an element in $\Bbb{Q}$.

• $[r_1 \times r_2]$ must be an element in $\Bbb{Q}$.

Consider (for example) attempting to combine $e^{i\pi/4}$ with any of the other three roots.

• $\displaystyle \left[e^{i\pi/4} + e^{3i\pi/4}\right] = [2i\sin\pi/4] \not\in \Bbb{Q}$.

• $\displaystyle \left[e^{i\pi/4} \times e^{5i\pi/4}\right] = [e^{3i\pi/2}] \not\in \Bbb{Q}$.

• $\displaystyle \left[e^{i\pi/4} + e^{7i\pi/4}\right] = [2\cos\pi/4] \not\in \Bbb{Q}$.

This makes it game over.

Answer 3

There is no such thing as an ideal method. Abandon the dream that there should be some kind of technique that "always works" unless you just want to find a general algorithm that works all the time, like the Berlekamp–Zassenhaus algorithm for $\mathbf Z[T]$.

In practice, for monics in $\mathbf Z[T]$, reduction mod $p$ is the simplest technique to use (with a computer), since if it works at all then it works for infinitely many $p$. Of course just one $p$ is enough, if that method works, but knowing there are either no such $p$ or a lot of them makes this technique feel more robust.

There are reasons involving algebraic number theory that reduction mod $p$ need not work at all for some polynomials (it's related to the structure of the Galois group of the polynomial). See here for an example of a quartic in $\mathbf Z[T]$ that is irreducible over $\mathbf Q$ but it is reducible mod $p$ for all $p$ and it has no Eisenstein translate: $T^4 - 10T^2 + 1$. The method used there to prove irreducibility over $\mathbf Q$ is (i) show there's no linear factor by the rational roots test and (ii) show there is no quadratic factor by finding all monic quadratic factorizations in the larger ring $\mathbf R[T]$ and showing none live in $\mathbf Q[T]$: if there were a quadratic irreducible factorization in $\mathbf Q[T]$ then it would have to be a quadratic factorization in $\mathbf R[T]$.

A final observation: if a monic in $\mathbf Z[T]$ is reducible in $\mathbf Q[T]$ with two factors of degree $d$ and $d'$, then it is a product of monics in $\mathbf Z[T]$ with degrees $d$ and $d'$. Therefore the "brute force check" that you mention at the end of your post has a serious omission: you only need to consider $a, b, c, d \in \mathbf Z$.

Answer 4

You can indeed try a factorization of the form

$x^4+1=(x^2+ax+b)(x^2+cx+d).$

Expanding the right side and matching terms with like powers gives

$x^3$ terms: $a+c=0,c=-a$

$x^2$ terms: $ac+b+d=0,b+d=a^2$

$x^1$ terms: $ad+bc=a(d-b)=0,a=0$ or $d=b$

The case $a=0$ leads to $(x^2+i)(x^2-i)$ which fails to lie in $\mathbb Q[x]$. So we try $d=b$, which then means $d=b=a^2/2$ from the matching of $x^2$ terms. Then matching the $x^0$ terms gives:

$x^0$ terms: $bd=b^2=1.$

Then $b=a^2/2=\pm1$ and neither choice of the $\pm$ sign allows a rational value for $a$. In fact only $b=+1,a=\pm\sqrt2$ admits a quadratic-quadratic factorization even over the reals.

Answer 5

A general method: assume that $f\in \Bbb{Q}[x]$ monic of degree $1$ or $2$ divides $x^4+1$. The roots of $x^4+1$ have absolute value $\le 1$ whence so do the roots of $f$.

Gauss lemma: $f\in \Bbb{Z}[x]$

• If $\deg(f)=1$ then it must be that $f\in \{x,x+1,x-1\}$. Well no it doesn't divide $x^4+1$

• If $\deg(f)=2$ then it must be that $f=(x-a)(x-b)=x^2+cx+d$ with $|d|\le 1, |c|\le 2$. Only $15$ polynomials to try, you can check that none of them divide $x^4+1$.

Therefore $x^4+1$ is irreducible.

Answer 6

The only non-trivial decomposition over $\mathbb{R}$ of $x^4+1$ is $x^4 + 1 = (x^2 + \sqrt{2} x + 1)(x^2 - \sqrt{2} x + 1)$. Since $\mathbb{Q} \subset \mathbb{R}$, if there is a non-trivial decomposition over $\mathbb{Q}$, it's gotta be the above. However, $\sqrt{2} \not\in \mathbb{Q}$.

Answer 7

Clearly, $x+a$, $a\in\Bbb{Z}$ can not be a factor since then $a^4=-1$. On the other hand, $x^2+ax+b$, $a,b\in\Bbb{Z}$ can not be a factor since its roots are $\frac{a\pm\sqrt{a^2-4b}}{2}$ and these roots can not be roots of unity wheras the roots of $x^4+1=0$ are root of unity. Because taking modulus, we see that $b=1$. But, then also considering the argument of the roots $\frac{\sqrt{4-a^2}}{a}=\pm1$ gives $a=\sqrt2$.