Let $R$ be a ring (commutative, unital) and $M$ an $R$-module. Let $I \subset R$ be an ideal. We make the following definitions:

$$ A := \{ am \ | \ a \in I,\ m \in M \} $$ $$ B := \left\{ \sum_{\text{finite}} a_i m_i \ | \ a \in I,\ m \in M \right\} $$

Clearly $A \subset B$, and if $I$ is a principal ideal then $B \subset A$ too. I came up with the following counterexample to $B \subset A$ for when $I$ is not principal:

Let $R = k[x_1, x_2, \dotsc]$ be the polynomial ring in infinitely many variables, $M$ be the $R$-submodule of $R$ consisting of those polynomials with zero constant term (and $0$), and $I = (x_1,x_2) \subset R$.

Then $p = x_1 x_3 + x_2 x_4 \in B$, but it is not in $A$. (For if there are $f \in I, g \in M$ such that $fg = p$, then $\deg f = \deg g = 1$ and we can check that there are in fact no such $f$ and $g$.)

Here I haven't actually used the non-Noetherianness of $k[x_1,x_2,\dotsc]$; $R = k[x_1, \dotsc, x_4]$ would have done just as nicely. This counterexample also generalizes to $R = k[x_1, \dotsc, x_{2n}],\ I = (x_1,\dotsc,x_n)$. But it's all pretty ad-hoc to me at the moment, are there any simpler/more general counterexamples?

Incidentally, am I right in assuming that the notation $IM$, for $I$ and $M$ as in the first sentence is to be read in the sense of $B$?

  • 1
    $\begingroup$ If $M$ is flat, then $B$ is isomorphic to $I\otimes M$ and you're asking when $I\otimes M$ (does not) consist of simple tensors. $\endgroup$ Dec 20, 2013 at 9:14
  • $\begingroup$ As I can't find a good answer to your main question, I will just say in a comment that indeed $IM$ is the submodule $B$. $\endgroup$ Dec 20, 2013 at 10:23
  • $\begingroup$ @KarlKronenfeld Rarely does it happen that $I \otimes M$ consists of only elementary tensors. $\endgroup$
    – user38268
    Dec 20, 2013 at 13:58
  • $\begingroup$ $IM$... is to be read in the sense of $B$? That's right. The thing is that we want $IM$ to be closed under addition, and $A$ doesn't guarantee that. $\endgroup$
    – rschwieb
    Dec 20, 2013 at 14:14
  • $\begingroup$ @Benja In fact, that's the reason why I mentioned it in a comment. It suggests that there really should be more general counterexamples. $\endgroup$ Dec 21, 2013 at 2:54

2 Answers 2


For another (in my opinion simpler, but certainly not more general) example:

What about $R= \mathbb{Z}[X]$, $M = (Y) \subset \mathbb{Z}[X,Y]$.

Now take $I = (2,X)$. Then $2Y^2 + XY \in B$, $2Y^2 + XY \notin A$

Even simpler: take $R$, $I$ as above and $M=I$.

$4+X^2 \in B = I^2$, but $4+X^2 \notin A$.


My examples are basically the same as Louis's and the original poster, but maybe the pattern is helpful to see.

If either $I$ or $M$ is cyclic, then $A=B$. Hence we need to choose both $I$ and $M$ non-cyclic. The easiest choice usually works:

Let $P$ be a non-principal ideal of $R$, and take $I=P$ and $M=P$ so that $B=P^2$. Usually $B \neq A$.

Example 1: If $R=\mathbb{Z}[\sqrt{-17}]$ and $I=M=(3,1+\sqrt{-17})$, then $1+\sqrt{-17} \in B \setminus A$.

Proof: Check that $I=\{ (3a+b) + b\sqrt{-17} : a,b \in \mathbb{Z} \}$, or, in other words, that I have given an integral basis. Check that $N:R\to \mathbb{Z}:a+b\sqrt{-17}\mapsto a^2+17b^2$ is multiplicative and zero iff $a+b\sqrt{-17}=0$. Check that $N( (3a+b) + b\sqrt{-17} ) = 3(a+b)^2 + 6a^2 + 15b^2$. Notice that this is which is minimized (with value $9$) when $a=1$, $b=0$ (other than $a=b=0$), so that $N(im) = N(i) N(m) \geq 9 \cdot 9 = 81$. Of course $N(1+\sqrt{-17}) = 18$, so $1+\sqrt{-17} \notin A$ (see also this question). On the other hand $1+\sqrt{-17} = ((1+\sqrt{-17})-6)(3) - (1+\sqrt{-17})^2 \in B$. $\square$

Example 2: If $R=\mathbb{Q}[x,y]$ and $I=(x,y)$, then $x^2 + y^2 \in B \setminus A$.

Proof: Clearly $x^2 + y^2 \in B=(x^2,xy,y^2)$. If $x^2+y^2 = fg$ we can WLOG assume the coefficients on $x$ are both $1$, so $f=x+ay+i$ where $i \in I^2$ and $g=x+by+j$ where $j \in I^2$, so that $fg = (x+ay+i)(x+by+j) = x^2 + (a+b)xy + aby^2 + k$ where $k \in I^3$. Hence we seek a solution for $a+b=0$ and $ab=1$, but this is equivalent to $a=-b$ and $a^2=-1$, which has no solutions in $\mathbb{Q}$ (nor in lots of other fields). Here I did use that $\left(\mathbb{Q}xy \oplus \mathbb{Q}y^2\right) \cap I^3 = 0$. $\square$

Example 3: If $R=\mathbb{Z}[x]$ and $I=(2,x)$, then $4+x^2 \in B \setminus A$.

Proof: Clearly $4+x^2 \in B=(4,2x,x^2)$. We describe $I$ as the polynomials with even value at $0$. Again $4+x^2 = (2+ax+i)(2+bx+j) = 4+(a+b)x+abx^2 +k$ and again $a+b=0$, $ab=1$ has no solution in integers. $\square$

This next example is neat because $B$ is cyclic, even though $I$ and $M$ are not. We still get $A\neq B$.

Example 4: If $R=\mathbb{Z}[\sqrt{-5}]$ and $I=(3,1+\sqrt{-5})$, then $-2+\sqrt{-5} \in B \setminus A$.

Proof: Check that $I=\{ (3a+b) + b\sqrt{-5} : a,b \in \mathbb{Z} \}$, that $N(a+b\sqrt{-5})=a^2 +5b^2$ is a norm, that $N((3a+b)+b\sqrt{-5}) = 3(a+b)^2 + 6a^2 + 3b^2$ takes its minimum on $I \setminus \{0\}$ at $a=0$, $b=1$ with value $6$. Thus $A\setminus \{0\}$ has minimum norm $36$, but $-2+\sqrt{-5}$ has norm $9$, so $-2+\sqrt{-5} \notin A$. Of course, $-2+\sqrt{-5} = -3^2 -(1+\sqrt{-5})^2 + 3(1+\sqrt{-5}) \in B$. $\square$

Nonexample: Now just because $I=M$ is non-principal does not mean $A \neq B$, since $I^2$ might itself be too simple. For example $R=\mathbb{Q}[x,y]/(x^2,xy,y^2)$ with $I=(x,y)$, then $A=B=0$.

I'm not aware of any simple necessary or sufficient criteria not already mentioned.


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