Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I prove that $x^4+1$ is an irreducible polynomial over $\mathbb Q$? I've already tried the Eisenstein criterion which gives to me any results to solve this question, I need help here.

Thanks

share|improve this question
    
Over which ring? –  student Nov 29 '12 at 6:08
    
@student yes, of course, thank you to remind me –  user42912 Nov 29 '12 at 6:09
1  
It is a cyclotomic polynomial. –  PAD Nov 29 '12 at 12:55

2 Answers 2

up vote 9 down vote accepted

Although the problem may be intended to be solved using a variant of the Eisenstein Criterion, one can also solve it using only elementary facts.

The polynomial $x^4+1$ has no real roots, so if we can reduce over $\mathbb{Q}$, it is as a product of quadratics.

Without loss of generality these quadratics each have lead coefficient $1$. Since $x^4+1$ has no $x^3$ term, the two quadratics must have shape $x^2-ax+b$ and $x^2+ax+c$.

The coefficient of $x$ in the product is $a(b-c)$. But it must be $0$. It is clear that we cannot have $a=0$. So $b=c$. That forces $b=c=1$ or $b=c=-1$.

But the coefficient of $x^2$ in the product is $b+c-a^2$. Thus $a^2=\pm 2$. This is not solvable in rationals.

Remark: The polynomial does have the nice factorization $x^4+1=\left(x^2-\sqrt{2}x+1\right)\left(x^2+\sqrt{2}x+1\right)$. This can be useful in some integration problems. The variant $x^4+4=(x^2-2x+2)(x^2+2x+2)$ has a habit of turning up in contest problems.

share|improve this answer

Try Eisenstein's test on $(x+1)^4+1=x^4+4x^3+6x^2+4x+2$. Can you pick the prime number?

Do convince yourself that (ir)reducibility is preserved by translations in the variable. Nifty trick.

share|improve this answer
    
yes, I saw this argument in the hint of my book, but I didn't understand why the irreducibility is preserved by translations, can you give me a hint? Thank you for your answer :) –  user42912 Nov 29 '12 at 6:15
    
@RafaelChavez do you know the universal property of $\mathbb{Q}[x]$? You can get homomorphisms from $\mathbb{Q}[x]$ into any other ring $R$ by specifying the image of scalars and the image of $x$. –  student Nov 29 '12 at 6:17
1  
@student the problem is this function is not an homomorphism. –  user42912 Nov 29 '12 at 6:36
2  
@RafaelChavez it is. You should be careful with the expression $f(x) = x + 1$; this does not mean that $f(x^2) = x^2 + 1$, but rather that $f(x^2) = f(x)f(x) = (x+1)^2$. Do not think of $x$ as a variable here; $x$ is just an element, and you're extending $f$ as a homomorphism by requiring that it maps the element $x$ to the element $x+1$, and fixes constants. This is not the function $X \mapsto X+1$, where $X$ is anything. Was that the problem? –  student Nov 29 '12 at 6:40
1  
@student yes, of course, thank you for helping me :) –  user42912 Nov 29 '12 at 7:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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