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$x^4+1$ is irreducible over $\Bbb Z$ (hence $\Bbb Q$). The proof I saw is to try transforming it into $(x+1)^4+1=x^4+4x^3+6x^2+4x+2$ and use the Eisenstein criterion with prime $2$. I can understand the proof. However, how do we come up with this idea? How do we know we should try to transfer $x$ into $x+1$, rather than, say $3x-5$? And by the way, I also wonder is there other way to proof the irreducibility?

Eric
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    You can learn about ramification; there is a theory behind the use of Eisenstein’s criterion, which is generalized by what are called Newton polygons. – Qiaochu Yuan Jan 15 '18 at 02:08

2 Answers2

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This is a trick that one sees over and over again, and perhaps in a year or two, you’ll see it as perfectly natural.

However, let me explain how I see it, from a $p$-adic (in this case, $2$-adic) viewpoint. The roots of $f(X)=X^4+1$ are all close to $1$ in the $2$-adic sense, namely for a root $\rho$ of $f$, $\vert1-\rho\vert_2<1$, while they are not close to zero: $\vert\rho-0\vert_2=\vert\rho\vert_2=1$. The Eisenstein criterion deals exclusively with small roots of a polynomial, so if we have any hope of using it, we need to look not at the numbers $\rho$, but at the numbers $\rho-1$. It’s $f(X+1)$ that has these for its roots.

Now, I’m guessing that at your stage of mathematical experience, you haven’t seen the $p$-adic numbers or the $p$-adic absolute value on the rationals. Maybe as you get more experienced, you’ll see another justification for the trick, and maybe you’ll see things through a lens similar to mine. Throughout mathematics, there are many ways to look at a phenomenon.

Lubin
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  • Yes I haven't heard about $p-$adic. I wonder is it custom that when textbooks or teachers taught this exercise, they didn't explicitly give the reason why they try to do so? – Eric Jan 15 '18 at 02:20
  • The subject of the $p$-adic numbers is not usually considered part of elementary number theory. For this particular question, maybe some adventurous individual hit on the trick. But it’s always fair, when worrying about the irreducibility of $f(X)$, to look at $f(X+c)$ for some constant $c$ in the ground field ($\Bbb Q$ in this case). If you play around with polynomials, you’ll find various simple tricks on your own. – Lubin Jan 15 '18 at 02:30
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Another way: the polynomial $x^4+1$ doesn't have an integral root, so $x^4+1 \ne (x-a)(x^3+bx^2+cx+d)$.

we assume that $x^4+1=(x^2+ax+b)(x^2+cx+d)$, then $a+c=0, ac+b+d=0, ad+bc=0, bd=1$, and so $x^4+1=(x^2-\sqrt2 x+1)(x^2+\sqrt2 x+1)$. i.e. $x^4+1$ is irreducible over $\mathbb Z$.

Mustafa
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  • I like this kind of argument, too, and have been using it more and more lately. I think I read somewhere that Leibniz did not know how to factor the related polynomial $X^4+4$, but your factorization leads to $X^4+4=(X^2+2X+2)(X^2-2X+2)$ – Lubin Jan 16 '18 at 00:07