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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.


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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
It is a cyclotomic polynomial. – PAD Nov 29 '12 at 12:55
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.

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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.

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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
@student the problem is this function is not an homomorphism. – user42912 Nov 29 '12 at 6:36
@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
@student yes, of course, thank you for helping me :) – user42912 Nov 29 '12 at 7:05

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