Sum of ideals $ I = (X^2+3X+2)$ and $ J = (X^2-1)$ in $\mathbb{Z}[X]$.

Asked Jan 27 '24 at 14:59

Active Jan 27 '24 at 22:25

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I am stuck finding the ideal $I+J$ for $ I = (X^2+3X+2)$ and $ J = (X^2-1)$ in $\mathbb{Z}[X]$.

So far I factored the two polynomials $ X^2+3X+2 = (X+1)(X+2)$ and $ X^2-1 = (X+1)(X-1)$. Therefore, the gcd of the polynomials is $ X+1$. I have also shown that $ X^2+3X+2 \in (X+1)$ and $ X^2-1 \in (X+1)$. However, I am unable to show that $ X+1 \in I+J $, yet I showed that $ 3X+3 \in I+J$, which is not enough since $3$ is not invertible in $\mathbb{Z}$.

Any tips on how to find $I+J$ would be very much appreciated.

edited Jan 27 '24 at 22:25

user26857

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asked Jan 27 '24 at 14:59

MathMaestro

1

You have shown that $3X+3\in I+J$. You cannot expect $X+1$ to be in it. For some details about ideals in $\Bbb Z[x]$ see for example this post. – Dietrich Burde Jan 27 '24 at 16:03
$I+J$ is the product of ideals $(X+1)$ and $(3,X-1)$. If $X+1\in I+J$ as you wanted, then $1\in(3,X-1)$, which is false. – user26857 Jan 27 '24 at 22:28
@user26857 Thank you for your answer. Is there a clever way to find out that the Ideal $I+J$ is the product of these two Ideals? – MathMaestro Jan 28 '24 at 08:41
If $I=(ab)$ and $J=(ac)$, then $I+J=(ab,ac)=(a)(b,c)$. Nothing clever in this. – user26857 Jan 28 '24 at 11:15

Sum of ideals $ I = (X^2+3X+2)$ and $ J = (X^2-1)$ in $\mathbb{Z}[X]$.

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