Question on the Irreducibility of Polynomials

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The discussion centers on the irreducibility of polynomials as outlined in Dummit and Foote's "Abstract Algebra," specifically Section 9.4. Proposition 9 states that a polynomial \( p(x) \) over a field \( F \) has a linear factor if and only if it has a root in \( F \). Proposition 10 further clarifies that for polynomials of degree two or three, having a root in \( F \) guarantees reducibility. However, the confusion arises regarding why this does not extend to all polynomial degrees, illustrated by the example of \( p(x) = (x^2+1)(x^2+2) \) in \( \mathbb{R}[x] \), which is reducible but lacks real roots.

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I am reading Dummit and Foote on Polynomial Rings. In particular I am seeking to understand Section 9.4 on Irreducibility Criteria.

Proposition 9 in Section 9.4 reads as follows:

Proposition 9. Let F be a field and let p(x) \in F[x]. Then p(x) has a factor of degree one if and only if p(x) has a root in F i.e. there is an \alpha \in F with p( \alpha ) = 0

Then D&F state that Proposition 9 gives a criterion for irreducibility for polynomials of small degree

D&F then state Proposition 10 as follows:

Proposition 10: A polynomial of degree two or three over a field F is reducible if and only if it has a root in FBUT! Here is my problem - why does not a root in F imply reducibility in polynomials of all degrees? A root in F means, I think, that the polynomial concerned has a linear factor and hence can be factored into a linear factor times a polynomial of degree n-1?

Can anyone clarify this for me?

Peter

[This question has also been posted on MHF]
 
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Peter said:
BUT! Here is my problem - why does not a root in F imply reducibility in polynomials of all degrees? A root in F means, I think, that the polynomial concerned has a linear factor and hence can be factored into a linear factor times a polynomial of degree n-1?

You are right, but take into account that proposition 10 says if and only if. For example consider $F=\mathbb{R}$ and $p(x)=(x^2+1)(x^2+2)$. This polynomial is obviously reducible in $\mathbb{R} [x]$ and has no real roots.
 

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