Is the ideal $I := (7X+14, X^3+2X^2+1) \subseteq \mathbb Z[X]$ prime? Is it maximal? The question is fully contained in the title.
I tried to prove maximality (if that happens, $I$ is prime as well) in $\mathbb Z[X]$, but I am not able to figure a strategy out for that purpouse. Obviously, if $I$ is not maximal, I am expected to say whether $I$ is a prime ideal, which is a problem too.
How would you solve such exercise?
 A: Let's just play around and see what kinds of polynomials we can find in $I$.
First of all, we can try to cancel the cubic term from the cubic generator, and we see that
$$
7(X^3+2X^2+1)-X^2(7X+14)=7
$$
is an element of the ideal. And since $7X+14=7(X+2)$, we get
$$
I=(7,X^3+2X^2+1)
$$
which is a lot simpler.
Now what? Well, since we have a constant in our ideal, it is apt to use the third isomorphism theorem. It states that we can divide out by one generator at a time (presumably you know all the basic connections between ideals and the corresponding quotients).
So if we're interested in whether $(7,X^3+2X+1)\subseteq \Bbb Z[X]$ is maximal or prime, we might just as well look at $(X^3+2X^2+1)\subseteq \Bbb Z[X]/(7)$. And therefore we can just check whether $X^3+2X+1$ has any roots modulo 7. If it does have roots the ideal isn't prime, and if it doesn't have roots, then the ideal is prime and also maximal (because $\Bbb Z[X]/(7,X^3+2X+1)$ would then be a finite integral domain and therefore a field, or alternatively because the quotient is a degree 3 extension of the field $\Bbb Z_7$).
A: Hint: $I$ contains $\;7(X^3+2X^2+1)-X^2(7X+14)=7$, hence
$$\mathbf Z[X]/I\simeq \mathbf Z/7 \mathbf Z[X]/(I/7 \mathbf Z[X])= \mathbf Z/7 \mathbf Z[X]/(X^3+\bar 2X^2+\bar 1).$$
Can you show that $X^3+\bar 2X^2+\bar 1$ is irreducible in $ \mathbf Z/7 \mathbf Z[X]$?
