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For $I,J$ ideals, show that $IJ\subseteq (I\cap J)(I+J)$. If it helps $I$ and $J$ are ideals in a Dedekind domain, but as far as I can tell the proof given in the book only uses the fact that we are in a commutative ring, either way I can't follow the proof:

If $ab\in IJ$ for $a\in I$ and $b\in J$, then $ab\in I,J$ and thus $ab\in I\cap J$. Furthermore we can see that $ab\in I+J$. Therefore $ab\in (I\cap J)(I+J)$.

Everything is rather straightforward up to the 'therefore' at the end. Just because something is in two ideals doesn't mean it's in their product, so we know that $(ab)^2\in (I\cap J)(I+J)$, but I can't see how it follows that $ab\in (I\cap J)(I+J)$. Thanks

This is Lemma 1.3 from page 30 of Mollin's Algebraic Number Theory.

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Note that this question is closely related. – Matt E May 17 '13 at 1:54
up vote 4 down vote accepted

It is indeed a mistake for the reasons you cite. For example, if $I=J=4\mathbb Z$ then $2\cdot 2\in I\cap J$ and $2\cdot 2\in I+J$ but $2\cdot 2\notin (I\cap J)(I+J)$. The proof should read:

If $ab\in IJ$ for $a\in I$ and $b\in J$, then $a,b\in I+J$. Thus $ab\in I(I+J)$ and $ab=ba\in J(I+J)$, so $ab\in I(I+J)\cap J(I+J)=(I\cap J)(I+J)$.

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It's that final step: $I(I+J)\cap J(I+J)=(I\cap J)(I+J)$ that requires you be in a Dedekind domain correct? What about being in a Dedekind domain allows that? – Thoth Dec 21 '12 at 4:35
Yes. As Matt E comments, intersection behaves like lcm in a Dedekind domain, and obeys the same distribution rules. – Alex Becker Dec 21 '12 at 4:56

Actually in a general ring the inclusion is the other way around:

$(I\cap J) I \subset JI = IJ$, and $(I \cap J) J \subset IJ,$ therefore $(I\cap J) (I + J) = (I\cap J)I + (I\cap J) J \subset IJ.$

Of course $IJ \subset I \cap J$, and so altogether we get the sequence of inclusions $$(I\cap J) (I + J) \subset IJ \subset (I \cap J).$$

(This is commonly used, for example, to show that $IJ = I \cap J$ if $I$ and $J$ are coprime, i.e. if $I + J = A$, the entire ring.)

But in a Dedekind domain we in fact have equality $$(I\cap J) (I + J) = IJ.$$ For this, note that we can write $I = I' (I+J)$ and $J = J'(I+J)$ for some $I'$, $J'$. (This is where we use the assumption that $A$ is a Dedekind domain.) Then $I' + J' = A,$ so that by the preceding parenthetical remark, $I'\cap J' = I'J',$ giving $$(I \cap J) (I+J) = (I'\cap J')(I+J)^2 = I'J' (I+J)^2 = I J.$$

In a more general ring, the inclusion $(I \cap J)(I+J) \subset IJ$ can be strict. E.g. if $A = k[x,y]$ (for a field $k$), and $I = (x),$ $J = (y)$, then $IJ = I\cap J = (xy)$, while $I + J = (x,y)$, and the inclusion $(xy)(x,y) \subset (xy)$ is strict. In fact, if we take instead $I = (x^2)$ and $J=(xy)$ then the inclusions $(I\cap J)(I+J) \subset IJ \subset I\cap J$ become $$(x^2y)(x^2,xy) = (x^3y)(x,y) \subset (x^3y) \subset (x^2 y),$$ and all of them are strict inclusions.

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Could you talk a bit more about how being in a Dedekind domain allows you to do what you did in the paragraph that begins "For this, note that..." Thanks. – Thoth Dec 21 '12 at 4:21
Dear ya brah u jelly, In a Dedekind domain, non-zero ideals have unique factorization, and the g.c.d of $I$ and $J$ is $I+J$. I just divided each of $I$ and $J$ by their g.c.d. In Alex Becker's answer you see the same thing proved in a slightly different way: namely he writes $I(I+J) \cap J(I+J) = (I\cap J)(I+J)$, which you can prove by noting that $\cap$ gives the l.c.m. of ideals in a Dedekind domain. (Of course, there are other ways to prove it to, e.g. by working locally and using that ideals in a Dedekind domain are locally principal.) Regards, – Matt E Dec 21 '12 at 4:50

You're right, the statement is false for general rings. As is remarked in Atiyah-Macdonald (page 6),

enter image description here

An example of a ring where the statement is false is $\mathbb{Z}[x]$, e.g. with $\mathfrak{a}=(2)$ and $\mathfrak{b}=(x)$. We have $$(\mathfrak{a}+\mathfrak{b})(\mathfrak{a}\cap\mathfrak{b})=(2,x)(2x)=(4x,2x^2)\subsetneq (2x)=\mathfrak{a}\mathfrak{b}.$$

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