How to show $\langle 7\rangle$ is a prime ideal of $\Bbb Z[x]$

Question

I know that since $\Bbb Z$ is commutative and unital $\langle 7\rangle$ is just the polynomials whose coefficients are all in $7\Bbb Z$.

I know that since $\Bbb Z[x]$ is commutative that in order to show $\langle 7\rangle$ is a prime ideal show: $ab \in \langle 7\rangle$ implies $a \in \langle 7\rangle$ or $b \in \langle 7\rangle$ for all a,b in $\Bbb Z[x]$.

I have calculated what form ab is using Cauchy product. I now have stated each coefficient is in $7\Bbb Z$ but Im finding it hard to show how this requires either a or b lies in $\langle 7\rangle$

I suggest finding a homomorphism from $\Bbb Z[X]$ to an integral domain whose kernel is $\left<7\right>$. — Angina Seng, Nov 16 '19 at 18:47
If you want to do things using the cauchy product a nice approach is as follows: show that if $a$ is not in $\langle 7 \rangle$, then $b$ must be in $\langle 7 \rangle$. — Ben Grossmann, Nov 16 '19 at 19:03
Possible duplicate of If $p$ is a prime ideal then $p[X]$ is a prime ideal. — Dietrich Burde, Nov 16 '19 at 19:10

score 2 · Answer 1 · answered Nov 17 '19 at 00:03

2

$\mathbb{Z}[x]/(7) = \frac{\mathbb{Z}}{(7)} [x]$

the latter is clearly an integral domain, therefore (7) is prime in $\mathbb{Z}[X]$.

answered Nov 17 '19 at 00:03

Alessandro

1,334

How to show $\langle 7\rangle$ is a prime ideal of $\Bbb Z[x]$

1 Answers1