Intersection of ideals in the ring of formal power series

Question

Let $R$ be a commutative ring and $I,J$ ideals in $R$. Denote by $R[[X]]$ the ring of formal power series with coefficients in $R$. If $A\subseteq R$, denote by $A^e$ the ideal in $R[[X]]$ generated by the set $A$.

It is easy to prove that $(I\cap J)^e\subseteq I^e\cap J^e$ but does the converse $I^e\cap J^e\subseteq(I\cap J)^e$ hold? I ask for a sketch of a proof or a counterexample. Thank you.

Answer 1

Note that in general $\mathfrak aR[[T]]$ can be strictly contained in $\mathfrak a[[T]]$, where $\mathfrak a\subset R$ is an ideal. (Here $\mathfrak aR[[T]]$ denotes the extension of $\mathfrak a$ to $R[[X]]$, and $\mathfrak a[[T]]$ is the ideal of formal power series having all coefficients in $\mathfrak a$.) This happens, for instance, when $\mathfrak a$ is countably (generated) but not finitely generated. (If $\mathfrak a=(a_0,a_1,\dots,a_n,\dots)$, then the series $f=\sum_{n\ge0}a_nT^n$ belongs to $\mathfrak a[[T]]$, but it's not in $\mathfrak aR[[T]]$, otherwise $\mathfrak a$ would be finitely generated.)

Moreover, $(aR\cap bR)R[[T]]\subseteq aR[[T]]\cap bR[[T]]=(aR\cap bR)[[T]]$ for any $a,b\in R$.

Now use the same ring and ideals from this answer.

Answer 2

The answer from user26857 uses non-finitely generated ideals, so I guess it is an interesting question to restrict ourselves to noetherian rings, which takes a big part of commutative algebra and algebraic geometry. I will assume $R$ is a commutative noetherian ring (without necessarily having a $1$) and $\mathfrak a, \mathfrak b$ are ideals of $R$ (I don't want to think about the non-commutative case and speak of left-right ideals although it might also work...)

If we want to show that $(\mathfrak a \cap \mathfrak b) R[[t]] = \mathfrak a R[[t]] \cap \mathfrak b R[[t]]$, it suffices to show that $\mathfrak a R[[t]]= \mathfrak a[[t]]$ since it is quite clear that $(\mathfrak a \cap \mathfrak b)[[t]] = \mathfrak a[[t]] \cap \mathfrak b[[t]]$.

The inclusion ($\subseteq$) is obvious for any such $R$. Conversely, assume $f = \sum_{n \ge 0} c_n t^n \in \mathfrak a[[t]]$ and write $\mathfrak a = (a_1,\cdots,a_m)$ (since $R$ is noetherian, any ideal is finitely generated). In particular, $$ c_n = \sum_{j=0}^m r_{n,j} a_j \quad \Longrightarrow \quad \sum_{n \ge 0} c_n t^n = \sum_{j=0}^m a_j \left( \sum_{n \ge 0} r_{n,j} t^n \right) \in \mathfrak aR[[t]]. $$ Note that the issue with this proof in the non-noetherian case is precisely the finiteness of the sum, which breaks down when the ideal is not finitely generated.

Hope that helps,