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I have this ideal in $\mathbb{Z}[X]$: $I=\left \langle X^2+2X-3 \right \rangle$.

Let's say $P$ and $Q$ are polynomials: $P(X)=a_0+a_1X+a_2X^2+...$, $Q(X)=b_0+b_1X+b_2X^2+...$.

$PQ=X^2+2X-3$. Since $PQ=a_0b_0+(a_1b_0+a_0b_1)X+...$, we have system:

$$-3=a_0b_0$$

$$2=a_1b_0+a_0b_1$$

and so on.

I'm not really sure what should be my next step. Can anyone help me?

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    $\begingroup$ $x^2 + 2x - 3 = (x-1)(x+3)$. But neither $(x-1)$ or $(x+3)$ are in $I$. So it is not prime. $\endgroup$ – Good Morning Captain Feb 5 at 22:42
  • $\begingroup$ Given a Ring: R and an Ideal: I We say I is Prime if I = (a), $ a \in R$, a is Prime Since $ \mathbb{Z} [X]$ is a E.D. it's also a P.I.D and we get {p| p is irreducible}$ = I = P = ${p| p is prime} in $ \mathbb{Z} [X]$ Given these, we can say that only the Irreducible polynomials generate Prime ideals. Then our strategy when asking if $p(x) \in \mathbb{Z} [X]$ is irreducible is to look for roots in $\mathbb{Z}$ so that $p(x)$ has divisors in $\mathbb{Z} [X]$ Given this try to factor (x^2+2x-3) and see what you can say about it's ideal. $\endgroup$ – Ross Flaxman Feb 5 at 23:01
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Rational roots theorem:

$X^2+2X-3=(X-1)(X+3)$, so it is not irreducible, hence not prime.

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