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I thought that $X^4+1$ was irreducible, but in fact, $$X^4+1=(X^2+\sqrt 2X+1)(X^2-\sqrt 2X+1).$$

In general, how can I have the intuition of such a factorisation if I don't know it ?

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    $\begingroup$ Over what field are you working? $\endgroup$ – lulu Feb 11 '17 at 16:41
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    $\begingroup$ Every polynomial with complex coefficients factors into linear and quadratic terms over $\mathbb R$. It is easy to see that there are no real roots, so it must factor as two quadratics. $\endgroup$ – lulu Feb 11 '17 at 17:01
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    $\begingroup$ @lulu Typo in your comment: "Every polynomial with complex coefficients" - should be REAL coefficients. $\endgroup$ – Dustan Levenstein Feb 11 '17 at 17:04
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    $\begingroup$ This question is one of those that could be adduced as evidence that the "algebra" tag should not be deprecated. $\endgroup$ – Michael Hardy Feb 11 '17 at 17:05
  • $\begingroup$ @DustanLevenstein Of course you are correct. $\endgroup$ – lulu Feb 11 '17 at 17:05

11 Answers 11

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There's a trick here, that is useful in other circumstances. I will do it over the real numbers $$ x^{4} + 1 = x^{4} + 2 x^{2} + 1 - 2 x^{2} = (x^{2} + 1)^{2} - (\sqrt{2} x)^{2} = (x^{2} + 1 - \sqrt{2} x) (x^{2} + 1 + \sqrt{2} x). $$ So it's just completing the square.

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  • $\begingroup$ :-) I had to wonder what took you so long as to undelete this. $\endgroup$ – Simply Beautiful Art Feb 12 '17 at 1:06
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Hint

You can easily solve $X^4+1=0$ in $\mathbb C$ and identify which product of two monic are in $\mathbb R[X]$.

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    $\begingroup$ why the downvote ? $\endgroup$ – Surb Feb 11 '17 at 16:54
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    $\begingroup$ Now there are two downvotes...@Surb. Perhaps it's because your answer is too short? I don't know. People like downvoting this answer for some reason. I upvoted. $\endgroup$ – S.C.B. Feb 11 '17 at 17:10
  • $\begingroup$ @BogdanSimeonov, But the solution given is for this specific polynomial, not for an arbitrary case. $\endgroup$ – Alex Silva Feb 11 '17 at 17:17
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    $\begingroup$ @BogdanSimeonov, If it is a reason to downvote the answer, then you should downvote all other answers since they do not tackle the general case either. $\endgroup$ – Alex Silva Feb 11 '17 at 17:23
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    $\begingroup$ @BogdanSimeonov: Like every one knows, there is no general result to know if a polynomial is reducible or not. Therefore, you can't give any general answer for such a question. $\endgroup$ – Surb Feb 11 '17 at 17:23
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There's a sort of completion of the square that goes like this: \begin{align} x^4+1 & = \underbrace{(x^4+2x^2 + 1)}_\text{This is a square.} - \underbrace{(2x^2)}_\text{So is this.} \\[10pt] & = \left( x^2+ 1 \right)^2 - (\sqrt 2\ x)^2 \\[10pt] & = (x^2 + 1 - \sqrt 2\ x)(x^2 + 1 + \sqrt2\ x). \end{align}

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In fact $$(x^2 - x+1)(x^2 + x+1)$$ $$ = (x^2+1 -x)(x^2+1 +x)$$ $$= (x^2+1)^2 -x^2$$ (remember the $\alpha^2-\beta^2 =(\alpha - \beta)(\alpha + \beta)$ formula? here $\alpha = x^2+1$ and $\beta = -x $) $$ = x^4 + x^2 +1$$

Well, if you really want to factor $x^4+1$, see here. Hope it helps.

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This is because the splitting field of $x^4+1$ is $K=Q(\zeta_8)$ and a quadratic subfield of that is $Q(\sqrt2)$.This may sound complicated if you are not versed in the terminology, but if you are interested,you can go through a Galois theory textbook, or more generally an abstract algebra textbook.

Basically, the quadratic subfields of K are $Q(\sqrt2),Q(\sqrt2i),Q(i), $ each corresponding to its Galois group (by the so called correspondence theorem).This theorem is what allows us to find these fields easily and what assures their finitude.Here are some factorisations of our polynomial: $x^4+1=(x^2+i)(x^2-i)=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)=(x^2+\sqrt2ix-1)(x^2-\sqrt2ix-1).$

All three of these ways of factoring come from the aforementioned quadratic subfields of K.

EDIT: On an elementary note, $\zeta_8$, being the 8th root of unity, is actually equal to $\frac{i+1}{\sqrt2}$.I urge you to try to write the roots of 2,-2 and -1 as algebraic expressions of $\zeta_8$. This should directly give you that they are quadratic subfields of K.

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  • $\begingroup$ I am wondering if the OP has really understood your answer. $\endgroup$ – Alex Silva Feb 11 '17 at 17:41
  • $\begingroup$ @AlexSilva I'm sure if he put effort into researching Galois theory, he would eventually figure everything out. I find this to be a lot more fruitful for the OP's mathematical interests. $\endgroup$ – Bogdan Simeonov Feb 11 '17 at 17:52
  • $\begingroup$ Kudos, Bogdan ! $\endgroup$ – Namaste Feb 12 '17 at 2:02
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We know that $$a^4+4=a^4+4a^2+4-4a^2=(a^2+2)^2-(2a)^2=(a^2+2a+2)(a^2-2a+2)$$

This is an well known identity, most easily identifiable from the difference between two squares. It is called the Sophie Germain Identity=

Putting in $x=\sqrt{2} a$, we have that $$x^2+1=(x^2+\sqrt{2} x+1)(x^2-\sqrt{2}x+1)$$ Though the first step is unnecessary, I added it as it is a generally useful formula.

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    $\begingroup$ @S.C.B.: lulu wasn't claiming anything, merely asking. $\endgroup$ – Henning Makholm Feb 11 '17 at 16:50
  • $\begingroup$ Why the downvote? $\endgroup$ – S.C.B. Feb 12 '17 at 14:38
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If you're looking for real coefficients, every polynomial factors into a product of linear and quadratic terms. Writing down the coefficients, on the other hand, can be nearly impossible. Therefore, for your given polynomial, you know that it can be factored into quadratic terms.

There are deeper reasons behind the factorization in Galois theory over the rationals, but I'll go for the elementary approach - although we will need a detour through complex numbers.

To factor $x^4+1$, let's start by setting $y=x^2$. Then, $x^4+1$ becomes $y^2+1$. A quadratic can easily be factored, in this case we use the quadratic formula to get that the roots are $y=\pm i$. Hence, this factors as $$ y^2+1=(y-i)(y+i). $$ Since $y=x^2$, we now know that $$ x^4+1=(x^2-i)(x^2+i). $$ Each of these are quadratics and can be factored with the quadratic formula. In particular, $x^2-i$ has solutions $x=\pm\sqrt{i}=\pm\frac{\sqrt{2}}{2}\pm\frac{\sqrt{2}}{2}i$. Therefore, $$ x^2-i=\left(x-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)\left(x+\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right). $$ Similarly, $$ x^2-i=\left(x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)\left(x+\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right). $$ Since we're looking for quadratics with real coefficients, we can try to pair these factors to get real coefficients after multiplying. We can test out a few cases to find that the product $$ \left(x-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)\left(x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)=x^2-\sqrt{2}x+1. $$ Similarly, the other pair results in $x^2+\sqrt{2}x+1$.

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For me, the most intuitive way to see this is by factoring over the complex numbers. Mainly notice that

$$x^4+1=0\implies x^4=-1\implies x=\operatorname{cis}(\pi(1+2k)/4),\quad k=0,1,2,3$$

where

$$\operatorname{cis}(\theta)=\cos(\theta)+i\sin(\theta)$$

Geometrically, it looks like this:

enter image description here

It is then easy to multiply these points back together to get

$$x^4+1=(x^2+\sqrt2x+1)(x^2+\sqrt2x-1)$$

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    $\begingroup$ When you write \text{cis} instead of \operatorname{cis} then you see $a\text{cis} b$ and $a\text{cis}(b)$ instead of $a\operatorname{cis}b$ and $a\operatorname{cis}(b).$ I edited accordingly. $\endgroup$ – Michael Hardy Feb 11 '17 at 16:52
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we have a factoring rule: $a^4+b^4=(a^2-\sqrt2ab+b^2)(a^2+\sqrt2ab+b^2)$, then for question: $x^4+1=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)$

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Iirc, Sophie Germaine in Bending of Isotropic plates has these four roots generated (from the governing biharmonic equation in Theory of Plates she had at first set up ) in complex plane for $z^4+1= 0$ with the four roots $ (\pm \dfrac{1}{\sqrt2}, \pm \dfrac{i}{\sqrt2})$ given also here by Simply Beautiful Art.

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There are many good answers already. Here's a very elementary approach:

Let $f(x) = x^4 + 1$.

Over the real numbers, every polynomial factors into a product of linear and quadratic polynomials. We know that $f(x) = 0$ has no solutions in the real numbers. That means $f(x)$ doesn't have any linear factors.

So, it must have only quadratic factors. Now since $f(x)$ is a polynomial of degree $4$, it must be the product of exactly $2$ quadratic polynomials:

$$ x^4 + 1 = (a_1 x^2 + b_1 x + c_1) (a_2 x^2 + b_2 x + c_2) $$

From above we know that both factors are quadratic; so their $x^2$ terms are nonzero. We might as well assume $a_1=1$ (because if not, we can divide by $a_1$ in the first factor, and multiply the second factor by $a_1$). But then $a_2$ will be $1$ as well, because the $x^4$ term has coefficient $1$.

Similarly, $c_1c_2$ must equal $1$, so $c_2=1/c_1$ (and $c_1\neq 0$).

So let's rewrite the equation, using some new symbols for the coefficients:

$$ x^4 + 1 = (x^2 + ax + b)(x^2 + cx + 1/b) $$

We do some rearranging: $$ \begin{eqnarray} x^4 + 1 & = &(x^2 + ax + b)(x^2 + cx + 1/b)\\ &=& x^4 + ax^3 + bx^2 + cx^3 +acx^2 + bcx + (1/b)x^2 + (a/b)x + 1\\ &=& x^4 + (a+c)(x^3) + (b + ac + 1/b)(x^2) + (bc + a/b)(x) + 1\\ \end{eqnarray} $$

Since the coefficients on both sides have to be equal, we get three new equations:

$$ \begin{eqnarray} 0& =& a + c\\ 0 &=& b + ac + 1/b\\ 0& =& bc + a/b \end{eqnarray} $$

The first equation says $c = -a$, so let's substitute that into the other two equations:

$$ \begin{eqnarray} 0& =& b - a^2 + 1/b\\ 0& = &-ab + a/b\\ \end{eqnarray} $$

Or, multiplying both sides by $b$ in both equations:

$$ \begin{eqnarray} a^2b &=& b^2 + 1\\ 0& =& -ab^2 + a = a(-b^2 + 1) = a(1+b)(1-b)\\ \end{eqnarray} $$

The first equation guarantees that $ a\neq 0$, so the second equation now says that $b=\pm 1$. At this point the first equation says that $\pm a^2 = 2$, so in fact $a^2 = 2$ and $a = \pm \sqrt{2}$. Now the first equation says $2b=2$, so $b=1$.

Case I: $a=\sqrt{2}$. Then we get

$$ x^4+1 = (x^2 + x\sqrt{2} + 1)(x^2 - x\sqrt{2} + 1) $$

Case II: $a=-\sqrt{2}$. Then we get

$$ x^4 + 1 = (x^2 - x\sqrt{2} + 1)(x^2 + x\sqrt{2} + 1) $$

One case or the other has to be correct, but either way we get the same factorization (just written in a different order). Therefore that factorization must be correct, and we're done.

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