9
$\begingroup$

I've just started to work myself into the topic of irreducibility of polynomials and I would like to show the following:

Prove that $x^4+1$ is irreducible over $\mathbb{Q}$.

I've managed to come up with a proof that seems reasonable to me but I'm not sure if it is actually correct. Could someone tell me if I made any beginner's mistakes?

My proof: Clearly $x^4+1$ has no root in $\mathbb{Q}$, so it can only be decomposed into two polynomials of degree $2$. Therefore there must be $a, b \in \mathbb{Q}$ s.t. $x^4+1=(x^2+ax+1)(x^2+bx+1)$. If we multiply this out we get $x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1$. If we now equate the coefficients we get that $a+b = 0$ and therefore $a = -b$ and that $ab+2=0$ and therefore that $a^2=2$, which is not possible in $\mathbb{Q}$.

Any feedback is appreciated!

$\endgroup$
5
  • 2
    $\begingroup$ Why could it not be $(x^2+ax-1)(x^2+bx-1)$? $\endgroup$
    – lulu
    Jan 13 '17 at 19:42
  • $\begingroup$ How do you know that the constant terms of the two quadratic factors have to be $1$? I think you need to invoke Gauß's lemma explicitly and work over $\mathbb Z$ to do that. (And even then, it would seem that @lulu's objection still applies). $\endgroup$ Jan 13 '17 at 19:43
  • 5
    $\begingroup$ For a different approach you can use a substitution $x\mapsto y+1$ to obtain a polynomial to which Eisenstein's criterion can be applied $\endgroup$ Jan 13 '17 at 19:49
  • $\begingroup$ @lulu Could we also use the quadratic formula to obtain $x^2= \pm i$? So, $x^4+1=(x^2-i)(x^2+i)$ which is not a factorization in $\mathbb Q[x]$? $\endgroup$ Mar 11 '20 at 21:51
  • 1
    $\begingroup$ @AlJebr The fact that there are complex factorings does not immediately rule out the possibility of rational factorings. For example, $x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1$ has complex roots (all the ninth roots of $1$ except for $1$ itself) but it also factors as $(x^2 + x + 1) (x^6 + x^3 + 1)$ over $\mathbb Q$. $\endgroup$
    – lulu
    Mar 11 '20 at 22:09
11
$\begingroup$

For this polynomial, it can simplified, since we know its factorisation over $\mathbf R$: $$x^4+1=(x^2+1)^2-2x^2=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1),$$ and this factorisation does not have rational coefficients. So by the uniqueness of factorisation, it cannot have another factorisation over $\mathbf Q$.

$\endgroup$
6
  • 8
    $\begingroup$ Nice idea to show it like this. To avoid potential over-generalization of this method by some reader, it might be worth stressing that this works because we know this factorization over the reals is the factorization into irreducible polynomials (over the reals). Just writing down some decomposition into two factors over the reals that is not one over the rationals does in general not suffice. $\endgroup$
    – quid
    Jan 13 '17 at 22:24
  • $\begingroup$ You're quite right. I just started from the initial observation of the O.P. $\endgroup$
    – Bernard
    Jan 13 '17 at 22:49
  • $\begingroup$ @Bernard, what if we notice that $f$ factorizes over $\mathbb{C}$ as $f = (X^2 - i)(X^2+i) = (X-\sqrt{i})(X+\sqrt{i})(X-i\sqrt{i})(X+i\sqrt{i})$, and then say that none of these factors is in $\mathbb{Q}[X]$, and therefore $f$ is irreducible over $\mathbb{Q}$. Does this proof also work? $\endgroup$
    – Sigurd
    Mar 18 '20 at 16:29
  • $\begingroup$ One point disturbs me: $\sqrt i$ is not well-defined in $\mathbf C$. Other than that, it is correct, but a bit longer, as you would have to specify you would group factors by conjugate pairs.,. $\endgroup$
    – Bernard
    Mar 18 '20 at 16:59
  • $\begingroup$ @Bernard Thanks, I see, what is an alternative for writing $\sqrt{i}$? Or solving this in a similar way without using $\sqrt{i}$ explicitly? And could you explain the second point, about specifying to group factors by conjugate pairs? $\endgroup$
    – Sigurd
    Mar 18 '20 at 18:13
7
$\begingroup$

Your approach in principle works fine you just tried to take a short-cut that is not available, yet is not really needed. Instead:

$$(x^4+1) = (x^2+ax+c)(x^2+bx+d)$$

We get

$$x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd.$$

You get $cd=1$ and thus $c=d^{-1}$. You still get $a=-b$ (from the third power), and then $ad +bc$ becomes $ad-ac$ and further $a(c^{-1}-c)$.

Thus, $a=0$ or $c= c^{-1}$ (that is, $c = \pm 1$).

In the former case, you get $c+ c^{-1}= c+d = 0$ which is impossible. In the latter case, you already dealt with $c=d=1$ and $c= d= -1$ is essentially the same, it yields the impossible $a^2 = -2$.

That said, the Eisenstein approach mentioned in a comment is more elegant, if you have it available.

$\endgroup$
5
$\begingroup$

Your proof is almost fine - you just have to write more generally $$ x^4+1=(x^2+ax+b)(x^2+cx+d), $$ where we obtain of course $bd=1$, which over the integers only leaves the possibilities $b=d=1$, or $b=d=-1$. In both cases, the associated Diophantine equations have no solution, and this is easy to see (we obtain $c=−a$, and $a^2 =b+d=±2$, which is not possible).

Furthermore, the polynomial $x^4+1$ is the $8$-th cyclotomic polynomial, and hence irreducible by the theorem of your algebra lecture. Of course, there are several other methods to prove this directly, e.g., writing $f(x)=x^4+1$, and applying Eisenstein to $$ f(x+1)=x^4+4x^3+6x^2+4x+2. $$

$\endgroup$
2
  • $\begingroup$ I can prove using Eisenstein that $f(x+1)$ is irreducible in $\mathbb{Q}[X]$, what does that say about $f(x)$? $\endgroup$ Apr 2 '19 at 13:48
  • $\begingroup$ @TheCodingWombat It says that $f$ is irreducible, see my answer. $\endgroup$ Apr 2 '19 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.