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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.

Thanks

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    Over which ring?2012-11-29
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    @student yes, of course, thank you to remind me2012-11-29
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    It is a cyclotomic polynomial.2012-11-29

2 Answers 2

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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 :)2012-11-29
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    @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 $x$.2012-11-29
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    @Rafael: Given $n\in\Bbb Z$, $f(x)$ is reducible if and only if $f(x+n)$ is. Try to prove the $\implies$ direction right now (that is, if $f(x)$ is reducible, then $f(x+n)$ is too), then apply the same implication with $-n$ to the polynomial $g(x):=f(x+n)$ for the opposite direction. Finally, recall that (in propositional logic), `A iff B` is equivalent to `(not A) iff (not B)`, so irreducibility of one implies irreducibility of the other.2012-11-29
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    @student no, I don't know, I'm studying Bhattacharya's abstract algebra, this book doesn't cover this subject.2012-11-29
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    I'm trying to do this right now :)2012-11-29
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    @RafaelChavez In any case, it's not hard to prove that you have a homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[x]$ that fixes the constants and sends $x$ to $x+1$. This has an obvious inverse, that maps $x$ to $x-1$ and leaves constants fixed. Thus, this is an isomorphism, and isomorphisms preserve irreducibility.2012-11-29
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    @student so do you mean $f:\mathbb Q[x]\to \mathbb Q[x]$, defined by $f(x)=x+1$ and $f(c)=c; c\in \mathbb Q$?2012-11-29
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    @RafaelChavez exactly. That already determines what $f$ does on any other element in $\mathbb{Q}[x]$.2012-11-29
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    @student the problem is this function is not an homomorphism.2012-11-29
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    @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?2012-11-29
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    @student yes, of course, thank you for helping me :)2012-11-29