Intersection of two primary ideals in $\mathbb{Z}[x]$.

Question

Consider $\mathbb{Z}[x]$, and define $I = (x(x^{2}-2),(x^{2}-2)(x^{2}+2))$, $J = (x^{2}-2)\cap(x^{3},2)$. I want to show that $I = J$.

Notice that $I\subset J$ is clear since the generators of $I$ are clearly in $J$. The other direction is less clear to me. It comes down to showing that $J/I = 0$. So you want to show that for $f\in J$ we have $f + I = 0 + I$. We can write $f = \alpha (x^{2}-2)$ for some $\alpha\in\mathbb{Z}[x]$. Since we also want $f\in (x^{3},2)$ we see that $f = \beta x^{3} + 2\gamma$ for some $\beta,\gamma\in\mathbb{Z}[x]$. So we clearly see that $\alpha(0) = -\gamma(0)$. From here I am stuck.

Answer 1

To show the inclusion $J \subset I$ , we proceed as follows. Let $f(x) \in J$ such that $f(x) = \alpha (x^2-2)$, and $f(x) = \beta x^3 + 2 \gamma$, for $\alpha, \beta, \gamma \in \mathbb{Z}[x]$. As you suggested, we want to show that $f(x) \equiv 0 + I$. Observe that $\beta x^3 +2 \gamma + I = 2x \beta + 2 \gamma + I = 2(x\beta + \gamma) + I$ because $x^{3} \equiv 2x \pmod{x(x^2-2)}$. But $f(x) + I = 2(x\beta + \gamma) + I = \alpha(x^2-2) + I$ implies that $2 \mid \alpha(x)$. So $f(x) + I = k(2(x^2-2)) + I$, for some $k \in \mathbb{Z}[x]$. But $2(x^2-2) \in I$ because $-x(x(x^2-2)) + (x^2 + 2)(x^2 - 2) = -x^4 + 2x^2 + x^4 - 4 = 2x^2 - 4 \in I$, which implies $f(x) + I = k(2(x^2-2)) + I = 0 + I$. Altogether, $$ f(x) + I = \alpha (x^2 - 2) + I = \beta x^3 + 2\gamma + I \implies $$ $$ f(x)+I = \alpha (x^2-2) + I = 2(x\beta + \gamma) + I \implies $$ $$ 2 \mid \alpha(x), 2(x^2 - 2) \in I \implies $$ $$ f(x) + I = k\cdot 2(x^2-2) + I = 0 + I \implies f(x) \in I $$ Please let me know if you have any questions.