2

I know $(x^2+1)$ is prime ideal of $\Bbb Z[x]$ without being maximal ideal. It can be easily proved as quotient ring isomorphic to integral domain $\Bbb Z[i]$. My question is the generalization of this i. e. if $(x^2+1)$ is replaced by arbitrary irreducible polynomial $f(x)$ over $\Bbb Z$. Can anyone suggest me an outline of this proof ? Thank you.

cqfd
  • 12,219
  • 6
  • 21
  • 51
Pradip
  • 840
  • 2
    How $\langle x^2+1\rangle$ is not maximal in $\mathbb{Z}[x]$? I guess is because $\langle 2, x^2+1\rangle$ contains it right? I just want to remember this – Julian Mejia Jun 02 '19 at 02:32
  • 1
    Yes, true but you can think different way as Z[i] is integral domain without being field. – Pradip Jun 02 '19 at 02:52

3 Answers3

1

Fix $s$ irreducible. Now consider $p,q$ polynomials such that $pq = rs$ for some other polynomial $r$. Now, since $\mathbb{Z}[X]$ is a unique factorization domain, the polynomials $p,q$ and $r$ have a decomposition in irreducibles,

$$ p = f_1^{\alpha_1} \cdots f_n^{\alpha_n}, \quad q = g_1^{\beta_1} \cdots g_m^{\beta_m}, \quad r = h_1^{\gamma_1} \cdots h_k^{\gamma_k}. $$

So the former equality says that

$$ f_1^{\alpha_1} \cdots f_n^{\alpha_n} \cdot g_1^{\beta_1} \cdots g_m^{\beta_m} = h_1^{\gamma_1} \cdots h_k^{\gamma_k} \cdot s. $$

Once again, by the uniqueness of factorization, one of the irreducibles of the left must coincide with $s$ up to a unit. Without loss of generality, suppose that $f_i = \pm s$ for some $1 \leq n \leq n$. Then,

$$ f = \pm s(f_1^{\alpha_1} \cdots \widehat{f_i^{\alpha_i}} \cdots f_n^{\alpha_n}) $$

and so $s$ divides $p$.

This proves that if $s$ divides $pq$ then it divides either $p$ or $q$, which is to say that $(s)$ is prime. In general $R[X]$ is a UFD if $R$ is one, and that's all we have used. So $(s)$ will be prime if $s$ is irreducible in any UFD.

In general, an irreducible element $s$ of an UDF is always prime (by the argument above), which is to say that $(s)$ is prime.

qualcuno
  • 17,121
0

A simple argument could be:

  • If $f(x)$ is irreducible in $\Bbb{Z}[x]$, including that it has no non-unit constant factor, then it is also irreducible in $\Bbb{Q}[x]$ (the usual argument invokes Gauss's lemma at a key step).
  • Therefore $f(x)$ generates a maximal ideal in $\Bbb{Q}[x]$.
  • Therefore $\Bbb{Q}[x]/\langle f(x)\rangle$ is a field.
  • But $\Bbb{Z}[x]/\langle f(x)\rangle$ is a subring of $\Bbb{Q}[x]/\langle f(x)\rangle$, so it is an integral domain.
  • Therefore $\langle f(x)\rangle$ is a prime ideal of $\Bbb{Z}[x]$.
Jyrki Lahtonen
  • 133,153
0

S'pose $g(x)h(x)\in\langle f(x)\rangle $ and that $f(x)$ is irreducible in $\Bbb Z[x]$.

Then $\exists k(x)\in\Bbb Z[x]$ such that $g(x)h(x)=k(x)f(x)$. So, $f(x)\mid g(x)h(x)$. But since $f$ is irreducible, $f(x)\mid g(x)\lor f(x)\mid h(x)$. Then $g(x)\in\langle f(x)\rangle \lor h(x)\in \langle f(x)\rangle $.

(The key is that primes and irreducibles are the same in the UFD $\Bbb Z[x]$.)