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.

  • $\begingroup$ It doesn't matter either way, @user26857, since we are really only concerned about the sets $I,J,I\cap J\subset A\subset R[[X]]$. But we can define $A^e$ for any $A\subset R[[X]]$. $\endgroup$ – Thomas Andrews Feb 22 '15 at 0:59
  • $\begingroup$ @user26857 thank you for correction. I edited the question. $\endgroup$ – jgeilberg Feb 22 '15 at 1:33
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    $\begingroup$ At first, I thought this was an easy question, but $I^e$ is not as obvious as I thought. It is clearly a sub-ideal of $I[[x]]$, but it turns out it is not, in general, equal to $I[[x]]$. $\endgroup$ – Thomas Andrews Feb 22 '15 at 1:48
  • $\begingroup$ But if $R$ is a PID, then $I^e=I[[x]]$, so it is easy to show that they are equal then. $\endgroup$ – Thomas Andrews Feb 22 '15 at 1:50
  • $\begingroup$ @ThomasAndrews seems that if $R$ is noetherian, or just if $I$ is finitely generated, then $I^e$ is still $I[[x]]$, so the same argument will work. $\endgroup$ – Kevin Carlson Feb 22 '15 at 2:37

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.

  • $\begingroup$ Drat, I had just worked out the same solution! It's a very nice example and problem. $\endgroup$ – Kevin Carlson Feb 22 '15 at 3:20

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,

  • $\begingroup$ @user26857 : As you can see from my comment in the comments I noticed this a bit too late... $\endgroup$ – Patrick Da Silva Feb 23 '15 at 2:12

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