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.

Quote:

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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)

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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.

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

Peter

Quote:

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Originally Posted by Bernhard

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

Peter

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Which one? Or both?

Be great to see your proof of both

Peter

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

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

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

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

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

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

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

I made a mistake with my latex code, but now I can't edit it. In the last paragraph, the $\displaystyle IF[x]$ should read $\displaystyle I \subsetneq F[x]$ and the $\displaystyle (g(x))F[x]$ should read $\displaystyle (g(x))\subsetneq F[x]$.

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

Thanks again!!! So very helpful!!!

Peter