# Is ideal $(X^2+2X-3)$ prime in $\mathbb{Z}[X]$?

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?

• $x^2 + 2x - 3 = (x-1)(x+3)$. But neither $(x-1)$ or $(x+3)$ are in $I$. So it is not prime. – Good Morning Captain Feb 5 at 22:42
• 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. – Ross Flaxman Feb 5 at 23:01

## 1 Answer

Rational roots theorem:

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