Polynomial Rings Over Fields

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 reads as follow:

**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.

Re: Polynomial Rings Over Fields

Quote:

Originally Posted by

**Bernhard**

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)

Every field is a UFD, so F[x] is also a UFD. In light of Proposition 7 of Section 8.2, this means every non-zero prime ideal I of F[x] is maximal. I'm not sure if you've encountered this theorem, but it is immensely useful:

Let R be any ring.

An ideal I is maximal iff R/I is a field.

An ideal I is prime iff R/I is an integral domain

If you accept the theorem above (particularly the first one), all you'll need to show is that the irreducible polynomials are precisely the prime elements of F[x]. If you haven't read about the theorem above, or you're having trouble showing irreducible polynomials are prime, I can sketch the proof.

Re: Polynomial Rings Over Fields

Thanks Gusbob, it would be very helpful if you sketched the proof

Peter

Re: Polynomial Rings Over Fields

Quote:

Originally Posted by

**Bernhard** Thanks Gusbob, it would be very helpful if you sketched the proof

Peter

Which one? Or both?

Re: Polynomial Rings Over Fields

Be great to see your proof of both

Peter

Re: Polynomial Rings Over Fields

Theorem: is maximal iff is a field

Proof: Let be a ideal such that . By the correspondence theorem , where denotes the image of under the homomorphism .

If is a field, there are only two possible ideals, so is either or . This means or , so is maximal.

Conversely, if is maximal, any ideal between and is or itself, so only has two ideals, i.e. is a field.

Theorem: is irreducible iff is prime.

Proof: Suppose is prime. If is constant and non-zero, then , violating the primality of . So cannot be constant. If , then . Since is prime by assumption, without loss of generality assume (recall that an ideal is prime if or ). Then the degree of is at least that of the degree of . But , so it must be that is a constant polynomial. Hence is irreducible.

Now suppose is irreducible. Since we're in a UFD, it is enough to show that is maximal. Let be an ideal containing . We'll need to show that . One has to note that is a principal ideal domain. So let . Then . The first inclusion means for some polynomial . The irreducibility of forces to be a unit (non-zero constant). Then divides as well, so .

Re: Polynomial Rings Over Fields

I made a mistake with my latex code, but now I can't edit it. In the last paragraph, the should read and the should read .

Re: Polynomial Rings Over Fields

Thank you so much for your help ... Working through your posts carefully

Thanks again!!! So very helpful!!!

Peter