# How can I show that $x^4+6$ is reducible over $\mathbb{R}$?

How can I show that the polynomial $x^4 + 6$ is reducible over $\mathbb{R}$ without explicitly finding factors?

I was trying to find a non-prime ideal that would generate it but I'm kind of lost as to how to proceed. Is there some sort of criterion that will allow me to show that it's reducible in $\mathbb{R}$?

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Since you are new, I want to give some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. –  Zev Chonoles May 15 '13 at 23:05
Sorry about that, it's not homework I'm just studying for a test. –  Danny May 15 '13 at 23:08
Let $a$ be positive. Thn $x^4+a=(x^2-kx+\sqrt{a})(x^2+kx+\sqrt{a})$ where $k=\sqrt{2\sqrt{a}}$. –  André Nicolas May 15 '13 at 23:32

## 1 Answer

Then use the fact that if a complex number $z$ is a root of a real polynomial, then $\bar z$ is also a root.

It follows that in $\Bbb R$ only the polynomials of degree one, and the quadratic polynomials with negative discriminants are the irreducible.

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Wouldn't the fundamental theorem only allow me to show that the polynomial is reducible over $\mathbb{C}$? –  Danny May 15 '13 at 23:07
@Danny yes. But every complex number is a root of a polinomial of degree two. Therefore, you can split it up into two polinomials of degree two. –  CBenni May 15 '13 at 23:09
That makes sense. Is there a theorem that shows that every complex number is a root of a polynomial of degree two? –  Danny May 15 '13 at 23:13
If $x=a+bi$ then $(x-a)^2=-b^2$. –  Berci May 15 '13 at 23:13
@Berci Thank you. –  Danny May 15 '13 at 23:15