5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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,

$\endgroup$
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.