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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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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
up vote 8 down vote accepted

Hint: Fundamental Theorem of Algebra.

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

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