# $x^4 + 1$ reducible over $\mathbb{R}$… is this possible?

I am seeing this on a homework and am wondering if this is a typo. I am aware that $x^4 + 1$ is irreducible over $\mathbb{Q}$. I know the following:

A polynomial being irreducible over some ring does not imply that it is irreducible over some superring.

$x^4 + 1$ has no linear factors

and if it has no linear factors, then it can only be factored into polynomials of degree 2, and for there to be no term $a_1x^1$, the factor must be of the form $a_2x^2 + a_0$, and this can be done in $\mathbb{C}$ but not in $\mathbb{R}$.

• I don't understand your question, but $x^4+1=\left(x^2+\sqrt2x+1\right)\left(x^2-\sqrt2x+1\right)$. – Git Gud Apr 23 '15 at 23:01
• A further hilarious fact is that $x^4+1$ is reducible modulo every prime, also. – paul garrett Apr 23 '15 at 23:18
• As Steven Stadnicki wrote in his answer below, "polynomials of degree 2" does not mean "polynomials without a linear term". – mweiss Apr 23 '15 at 23:24

Even if you don't know the explicit factorization, you certainly know that a quartic polynomial in $\mathbb{R}[X]$ is reducible.

Indeed, a consequence of the fact that every polynomial of positive degree in $\mathbb{C}[X]$ has a root, we can see that the irreducible polynomials over $\mathbb{R}[X]$ are those of degree $1$ and the quadratic polynomials with negative discriminant.

To wit, let $f(X)\in\mathbb{R}[X]$ be monic of degree $>1$. If $f$ has a real root, then it is reducible. So, assume it has no real root. Since it has a complex root $a$, also $\bar{a}$ is a root, because $$f(\bar{a})=\overline{f(a)}=\bar{0}=0$$ (overlining means conjugation). Thus $f$ is divisible (in $\mathbb{C}$) by $X-a$ and $X-\bar{a}$, so also by their product $$g(X)=(X-a)(X-\bar{a})=X^2-(a+\bar{a})X+a\bar{a}$$ (because $X-a$ and $X-\bar{a}$ are coprime in $\mathbb{C}[X]$). However, $g(X)$ has real coefficients, so the quotient of the division of $f(X)$ by $g(X)$ is again a polynomial in $\mathbb{R}[X]$. So, unless $f$ has degree $2$ (and so $f(X)=g(X)$), $f$ is reducible.

• Thanks. I wondered why the other answers didn't (explicitly) mention this, so I was just about to post something along those very lines. ^_^ – A.P. Apr 23 '15 at 23:42
• @A.P.: well, in my answer I did mention this, although somewhat implicitely and in between the lines: a quadratic polynomial having complex conjugates roots has real coefficients. – AdLibitum Apr 24 '15 at 16:35
• @AdLibitum Indeed I upvoted it, but I preferred to give a more general argument. – egreg Apr 24 '15 at 16:53
• @AdLibitum I know you used this, which is why I upvoted your answer. Still, neither your answer nor mweiss's mentioned that every polynomial (with real coefficients) of degree $>2$ splits over $\Bbb{R}$, although a general proof could be derived from those. – A.P. Apr 24 '15 at 17:09

Remember Sophie Germain: $$x^4+4 = (x^4+4x^2+4)-(4x^2) = (x^2+2)^2-(2x)^2 = (x^2+2x+2)(x^2-2x+2)$$ and adjust the above line by replacing $x$ with $\sqrt{2}\,x$.

The four complex roots of $X^4+1$ are the 4th roots of $-1=e^{\pi i}$, i.e. $$z_1=e^{\frac14\pi i},\quad z_2=e^{\frac34\pi i},\quad z_3=e^{\frac54\pi i},\quad z_4=e^{\frac74\pi i}.$$ Note that $$z_4=\bar z_1\quad\text{and}\quad z_3=\bar z_2.$$ Thus $P(X)=(X-z_1)(X-z_4)$ and $Q(X)=(X-z_2)(X-z_3)$ are polynomials with real coefficients and $$X^4+1=P(X)Q(X).$$

To understand where you went wrong, note that '$x^4+1$ has no linear factors' doesn't imply 'there is no term linear in $x$ in a factor'. Indeed, the restrictions on terms in the factorization can be derived straightforwardly: suppose that $x^4+1$ is a product of two quadratics $a_2x^2+a_1x+a_0$ and $b_2x^2+b_1x+b_0$. It should be obvious that via multiplication by scalars we can force $a_2=b_2=1$; now, the product is $x^4+(a_1+b_1)x^3+(\text{some complicated term in }x^2)+(a_0b_1+b_0a_1)x+a_0b_0$. Comparing terms, we get $a_1=-b_1$ (from the $x^3$ term) and then $a_0=b_0$ (from the $x$ term, using the new knowledge about $a_1$ and $b_1$) and $a_0b_0=1$ (from the constant term). The last two obviously imply that either $a_0=b_0=1$ or $a_0=b_0=-1$, so the product is either $(x^2+a_1x+1)\cdot(x^2-a_1x+1)$ or $(x^2+a_1x-1)\cdot(x^2-a_1x-1)$. Now you can go back to the 'complicated term in $x^2$'; you should find a simple quadratic equation in $a_1$ that you can solve.

We know that $x^4+1=0$ has four solutions in $\mathbb C$, none of which are real. We also know that complex solutions to a polynomial equation with real coefficients come in conjugate pairs. So the solutions must be some set of the form $\{w, \bar{w}, z, \bar{z} \}$, which means that it factors into complex linear factors as: $$x^4 + 1 = (x-w)(x-\bar w)(x-z)(x-\bar z)$$ Now if you multiply these together in pairs, you get two quadratic factors with real coefficients. So even without knowing what the factors are, we know that $x^4+1$ factors into two quadratics with real coefficients.

You can go even further with this if you know what the complex roots are, of course. Let $w=e^{\pi i/4}$, which is certainly one of them. The other three are then found by multiplying by $i$: $z = iw = e^{3 \pi i / 4}$, $\bar z = - w = e^{5 \pi i/r}$, and $\bar w = -iw = e^{- \pi i/4}$. Then you can work out explicitly what the real quadratic factors are.