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Let $\mathfrak{R}$ denote the Jacobson radical of a ring.

If $f=\sum_{i=0}^{\infty} a_i x^i \in R[[x]]$, I wish to show that $$f\in\mathfrak{R}(R[[x]])\iff a_0\in\mathfrak{R}(R).$$

I have already proved the forward direction, by defining the maximal ideal $\mathfrak{n}=(\mathfrak{m},x)\subset R[[x]]$ for any maximal ideal $\mathfrak{m}\subset R$, but I'm struggling to adapt this technique to prove the converse. Given a maximal ideal $\mathfrak{n}\subset R[[x]]$, it doesn't necessarily seem like there would exist some $\mathfrak{m}\subset R$ s.t. $\mathfrak{n}=(\mathfrak{m},x)$.

If I could prove that $\mathfrak{m}\cap R$ was maximal, that would probably be sufficient.

user26857
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Dominic Wynter
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1 Answers1

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Just to check whether my solution is correct or not: $\\f \in (A[[x]]) \iff 1-fg \\$ is a non unit for some $g$. In that case, say that the maximal ideal containing $\\1-fg$ is $M$. If $x$ is in $M$ then $f$ is in $M$ which would mean $1$ is in $M$ (contradiction). Else, $x$ is not in $M$, in which case there is some $g$ in $M$ and $h$ in $A[[x]]$ such that $g+hx=1 \implies 1-hx \in M$ which is a contradiction because its a unit.

uzumymw
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