Using equational Jacobson condition to prove element lies in radical of ideal

Question

Recall the Jacobson radical of a ring consists of elements $f\in A$ such that $1-gf\in A^\times$ for every $g\in A$. Say an ideal $I\vartriangleleft A$ is Jacobson if in the quotient $A/I$ the Jacobson radical coincides with the nilradical.

To prove an ideal is Jacobson it suffices to prove $\mathrm J(A/I)\subset \mathrm{nil}(A/I)$. Let us unpack what this means. Fix $f\in A$ and suppose every $g\in A$ admits $h\in A$ such that $h(1-gf)\in 1+I$. This means $f+I\in \mathrm J(A/I)$. The above inclusion asserts that $f+I\in \mathrm{nil}(A/I)$, i.e $f\in \sqrt I$.

It is possible to prove a ring is Jacobson iff all of its ideals are Jacobson, but I just want to explicitly see how this works for nice ideals of nice Jacobson rings like $\mathbb Z[x],\mathbb Q[x]$.

Fix an ideal $I\vartriangleleft\mathbb Z[x]$ and a polynomial $f\in \mathbb Z[x]$. Suppose every polynomial $g\in \mathbb Z[x]$ admits $h\in \mathbb Z[x]$ such that $h(1-gf)\in 1+I$. Which polynomials $g\in \mathbb Z[x]$ should we choose to show that $f\in \sqrt I$?

I am stuck with even the simple cases $I_n=\langle x^n\rangle \vartriangleleft \mathbb Z[x]$. Any kind of insight into this type of problem would be appreciated.

Answer 1

Fix $I_2=\langle x^2\rangle \vartriangleleft \mathbb Z[x]$. Let $f=x$. For each polynomial $g\in \mathbb{Z}[x]$ we have $h(1-gf)\in 1+I$ when taking $h=1+gf$.

Your question: "Which polynomials $g\in \mathbb{Z}[x]$ should we choose to show that $f\in \sqrt{I}$?" doesn't seem to make sense. The symbol "$g$" already has a meaning (as a universally quantified variable).

To show that $f\in \sqrt{I}$ you need to show that some power of $f$ lies in $I$. That power changes as $f$ changes (or as $I$ changes).