1
$\begingroup$

Answer:

No principal ideal (f(x)) is maximal. If f(x) is an integer $ n\ne ±1$, then (n, x) is a bigger ideal that is not the whole ring. If f(x) has positive degree, then take any prime number p that does not divide the leading coefficient of f(x). (p, f(x)) is a bigger ideal and it's not the whole ring, since Z[x]/(p, f(x)) = Z/pZ[x]/(f(x)) is not the zero ring.

I have a few questions about the answer. If someone can please clarify, that would be great.

Firstly, (n,x) is the ideal generated by a linear combination of n and x, is that correct? So that is the set of polynomials having n as coefficients?

For the second part, can someone clarify why (p,f(x)) is a bigger ideal than (f(x))? Namely, this part: Z[x]/(p, f(x)) = Z/pZ[x]/(f(x))?

  • 0
    the maximal ideals are $(p,f(x))$ where $f(x)$ is irreducible in $\mathbb{F}_p[x]$ – 2017-01-29

3 Answers 3

1

For your first question, yes, $(n,x)=\{nf+xg\mid f,g\in\Bbb{Z}[x]\}$.

For your second question, $p$ is not in $(f(x))$ because $f$ has positive degree and $\Bbb{Z}[x]$ is an integral domain, so there can't ever be a $g\in\Bbb{Z}[x]$ with $p=f(x)g(x)$. Remember that $\deg(fg)=\deg(f)+\deg(g)$.

For the isomorphism, note there is a natural pair of maps

$$\Bbb{Z}[x]\to(\Bbb{Z}/p\Bbb{Z})[x]\to(\Bbb Z/p\Bbb Z)[x]/(f(x))$$

and you can check that the kernel of this composition is $(p,f(x))$.

1

It might be helpful to think about this in terms of homomorphisms $\phi, \gamma$. If we have $\phi\colon \mathbb{Z}[x]/(x)\to \mathbb{Z}$ then we are basically just evaluating our polynomial at $0$ (i.e. only considering the constant term). This is, however, not a field and therefore $(x)$ is not a maximal ideal. Note, however, that $\mathbb{Z}$ is an integer domain so $(x)$ is a prime ideal.

If we, however, reduce our integer domain, $\mathbb{Z}$, further (say by a prime $p$) with $\gamma\colon \mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}$ then we can see that this now maps to a field and if we thus take $\phi\circ\gamma$ as our surjective homomorphism we can show that the kernel of this composition is exactly $(p,x)$ and is thus maximal.

  • 0
    @Lana Fields have _everything_ to do with our proplem, in some sense. Given a ring $R$, and an ideal $I\subseteq R$, then $I$ is maximal in $R$ iff $R/I$ is a field. – 2017-01-29
  • 0
    Yup, sorry! just revised the definition! – 2017-01-29
1

As for your first question: The ideal $\langle n \rangle$ consists of the polynomials in which all of the terms are multiples of $n$, the ideal $\langle n,x\rangle$ consists of the polynomials in which the constant coefficient is a multiple of $n$.

For the second part, in the ideal $\langle f(x)\rangle$ all of the polynomials have constant coefficients which are multiples of $p$, while the new one does not. This ideal is not the whole ring because all of the polynomials have positive degree or are integers that are multiples of $p$.

Notice that one case is still missing: The case in which $f(x)$ is a polynomial with positive degree and constant coefficient equal to $0$. in that case we can consider $\langle f(x),2\rangle$

  • 0
    By that third case, are you referring to the constant polynomials? – 2017-01-29
  • 0
    No, I had made a mistake. I meant the polynomials of the form $a_nx^n+\dots + a_1x$. The ones with constant coefficient $0$. – 2017-01-29
  • 0
    Yes, but these are just the numbers aren't they? they have no variable x – 2017-01-29
  • 0
    No, the number are the ones that have no x, I am talking about the ones in which everything has x – 2017-01-29