Why principal ideals in Z[x] are not maximal?

Question

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: $Z[x]/(p, f(x)) = (Z/pZ)[x]/(f(x))$?

Answer 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 $\langle f(x),2\rangle$

Answer 2

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.

Answer 3

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 $(p,f(x))$.