# Polynomial Rings Over Fields

#### Peter

##### Well-known member
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I am reading Dummit and Foote Section 9.5 Polynomial Rings Over Fields II and need some help and guidance with the proof of Proposition 15.

Proposition 15. The maximal ideals in F[x] are the ideals (f(x)) generated by irreducible polynomials. In particular, F[x]/(f(x)) is a field if and only if f(x) is irreducible.

Dummit and Foote give the proof as follows:

Proof: This follows from Proposition 7 of Section 8.2 applied to the Principal Ideal Domain F[x].

My problem
- can someone show me how Proposition 15 above follows from Proposition 7 of Section 8.2 (see below for Proposition 7 of Section 8.2)

I would be grateful for any help or guidance in this matter.

Peter

Dummit and Foote - Section 8.2 - Proposition 7

Proposition 7. Every nonzero prime ideal in a Principal Ideal Domain is a maximal ideal.

[This problem has also been posted on MHF]

##### Active member
I am reading Dummit and Foote Section 9.5 Polynomial Rings Over Fields II and need some help and guidance with the proof of Proposition 15.

Proposition 15. The maximal ideals in F[x] are the ideals (f(x)) generated by irreducible polynomials. In particular, F[x]/(f(x)) is a field if and only if f(x) is irreducible.

Dummit and Foote give the proof as follows:

Proof: This follows from Proposition 7 of Section 8.2 applied to the Principal Ideal Domain F[x].

My problem
- can someone show me how Proposition 15 above follows from Proposition 7 of Section 8.2 (see below for Proposition 7 of Section 8.2)

I would be grateful for any help or guidance in this matter.

Peter

Dummit and Foote - Section 8.2 - Proposition 7

Proposition 7. Every nonzero prime ideal in a Principal Ideal Domain is a maximal ideal.

[This problem has also been posted on MHF]
So in one direction, the conditional is obvious. That is,

Suppose f(x) is reducible:

Then f(x) = g(x)*h(x) for polynomials g and h of degree less than f, which means that
$$(g(x) + \langle f(x) \rangle)\,(h(x) + \langle f(x) \rangle) =(f(x) + \langle f(x) \rangle) =(0 + \langle f(x) \rangle)$$
Which means that $F[x]/\langle f(x) \rangle$ is not a domain, and therefore not a field.
Now for the other direction, we'll need that proposition:

Suppose f(x) is irreducible:

Because f(x) is irreducible, we know that $\langle f(x) \rangle$ is a prime ideal.
By proposition 7, since F[x] is a principal ideal domain, we know that $\langle f(x) \rangle$ is a maximal ideal.
Finally, since $\langle f(x) \rangle$ is a maximal ideal and F[x]≠{0} is a commutative ring, we know that $F[x]/\langle f(x) \rangle$ is a field.

Thus, the proof is complete.