Polynomial Rings Over Fields

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May 31st 2013, 02:53 AM
Bernhard

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.
June 1st 2013, 01:36 AM
Gusbob

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.
June 1st 2013, 02:17 AM
Bernhard

Re: Polynomial Rings Over Fields

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

Peter
June 1st 2013, 02:19 AM
Gusbob

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?
June 1st 2013, 02:25 AM
Bernhard

Re: Polynomial Rings Over Fields

Be great to see your proof of both

Peter
June 1st 2013, 03:10 AM
Gusbob

Re: Polynomial Rings Over Fields

Theorem: $I$ is maximal iff $R/I$ is a field

Proof: Let $J$ be a ideal such that $I\subseteq J \subseteq R$ . By the correspondence theorem $\overline{I}=(0)\subseteq\overline{J}\subseteq R/I$ , where $\overline{I},\overline{J}$ denotes the image of $I,J$ under the homomorphism $r\mapsto r+I$ .

If $R/I$ is a field, there are only two possible ideals, so $\overline{J}$ is either $\overline{I}$ or $R/I$ . This means $J=I$ or $J=R$ , so $I$ is maximal.

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

Theorem: $f(x)$ is irreducible iff $(f(x))$ is prime.

Proof: Suppose $(f(x))$ is prime. If $f(x)$ is constant and non-zero, then $(f(x))=F[x]$ , violating the primality of $f(x)$ . So $f(x)$ cannot be constant. If $f(x)=g(x)h(x)$ , then $g(x)h(x) \in (f(x))$ . Since $(f(x))$ is prime by assumption, without loss of generality assume $g(x)\in (f(x))$ (recall that an ideal $I$ is prime if $xy \in I \implies x\in I$ or $y\in I$ ). Then the degree of $g(x)$ is at least that of the degree of $f(x)$ . But $\deg f(x)=\deg g(x)+\deg h(x)$ , so it must be that $h(x)$ is a constant polynomial. Hence $f(x)$ is irreducible.

Now suppose $f(x)$ is irreducible. Since we're in a UFD, it is enough to show that $(f(x))$ is maximal. Let $I \subsetnoteq F[x]$ be an ideal containing $f(x)$ . We'll need to show that $I=(f(x))$ . One has to note that $F[x]$ is a principal ideal domain. So let $I=(g(x))$ . Then $(f(x))\subseteq(g(x))\subsetnoteq F[x]$ . The first inclusion means $f(x)=g(x)h(x)$ for some polynomial $h(x)$ . The irreducibility of $f(x)$ forces $h(x)$ to be a unit (non-zero constant). Then $f(x)$ divides $g(x)$ as well, so $(f(x))=(g(x))=I$ .
June 1st 2013, 04:12 AM
Gusbob

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 $IF[x]$ should read $I \subsetneq F[x]$ and the $(g(x))F[x]$ should read $(g(x))\subsetneq F[x]$ .
June 1st 2013, 04:15 AM
Bernhard

Re: Polynomial Rings Over Fields

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

Thanks again!!! So very helpful!!!

Peter