Irreducible polynomials over the reals

[R.P.F.]

Homework Statement

How to prove that the only irreducible polys over the reals are the linear ones and the quadratic ones no real roots?

What about the ones with higher degree? I feel that I'm missing something that's really obvious.

Homework Equations

The Attempt at a Solution

 

[micromass]

I wouldn't immediately say that it's obvious, but it's certainly something that you can prove:

Let P(X) be a real polynomial. The fundamental theorem of algebra states that there exists complex number a1,...an such that P(X)=(X-a1)...(X-an). Thus we have factorized the polynomial P over the field C. However, we must factorize it over R.

For this: notice the following fact (try to prove this!): if z is a complex solution of P(X) (thus is P(z)=0), then the complex conjugate $$\overline{z}$$ is a solution of P(X).

Thus, in our factorization P(X)=(X-a1)...(X-an), if ai is non-real, then one of the a1,...,an must be $$\overline{a_i}$$. And if we multiply $$(X-a_i)(X-\overline{a_i})$$, then we get a REAL quadratic polynomial!
Thus P(X) can always be written as a product of linear terms and quadratic terms. This implies that the only possible irreducible polynomials are linear or quadratic.

 

[R.P.F.]

I wouldn't immediately say that it's obvious, but it's certainly something that you can prove:

Let P(X) be a real polynomial. The fundamental theorem of algebra states that there exists complex number a1,...an such that P(X)=(X-a1)...(X-an). Thus we have factorized the polynomial P over the field C. However, we must factorize it over R.

For this: notice the following fact (try to prove this!): if z is a complex solution of P(X) (thus is P(z)=0), then the complex conjugate $$\overline{z}$$ is a solution of P(X).

Thus, in our factorization P(X)=(X-a1)...(X-an), if ai is non-real, then one of the a1,...,an must be $$\overline{a_i}$$. And if we multiply $$(X-a_i)(X-\overline{a_i})$$, then we get a REAL quadratic polynomial!
Thus P(X) can always be written as a product of linear terms and quadratic terms. This implies that the only possible irreducible polynomials are linear or quadratic.

:) The proof is so cute. Thanks a lot for helping!