$I \cap (J+K) =I \cap J + I\cap K $

Question

Let I, J, K be three ideals in a commutative ring R with unity.

If R is ring of integers then above equation holds.

I know the equation do not hold for arbitrary ring. Can you give me an example of a Ring R where the equation does not hold?

Note that There is famous modular law. I am not saying that. In modular law there is a condition I contains J or K. Here I omit that condition

Thanks....

score 3 · Accepted Answer · answered Jul 01 '19 at 05:43

3

One example, in $R=F[x,y]$, the polynomial ring in two variables over a field $F$: take $$ I = \langle x+y \rangle, \quad J = \langle x \rangle, \quad K = \langle y \rangle. $$ Then $x+y\in I\cap(J+K)$ (indeed $I\cap(J+K)=I$); but $I\cap J = \langle x(x+y) \rangle$ and $I\cap K = \langle y(x+y) \rangle$, and so there are no polynomials of total degree $1$, including $x+y$, in $I\cap J + I\cap K$.

answered Jul 01 '19 at 05:43

Greg Martin

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@Gerg Martin Very good... I understand... It's ok... But can u give another example other than more than one variable ( if possible) – Pradip Jul 01 '19 at 06:14
@Siraj But can u give another example other than more than one variable ( if possible) What is that supposed to mean? You mean "don't use more than one variable"? – rschwieb Jul 01 '19 at 13:28
@rschwieb yes, I mean that – Pradip Jul 01 '19 at 14:14
@Siraj I addressed that idea at the end of my solution. – rschwieb Jul 01 '19 at 15:35

rschwieb · Answer 2 · 2019-07-01T14:21:24.747

A ring with this property on its left ideals is called a left distributive ring, since this a lattice having this property is called a distributive lattice.

(You mentioned you are not talking about the modular condition: of course we should not, because the lattices of left/right/two-sided ideals of a ring are all automatically modular. No such counterexample would exist anyway.)

Here is the DaRT query for rings that aren't left distributive. At the time of posting, it has 20+ examples. (I know you asked for commutative examples, but I wanted to demonstrate that there are also many noncommutative examples which are interesting.)

There are at least two which are commutative, which I will mention explicitly in case the links go out:

$F_2[x,y]/(x,y)^2$ (the one given above.)

$S=k[x_1, x_2, x_3,\ldots ]$ modulo the ideal $I$ generated by $\{x_i^2\mid i\in \mathbb N\}\cup\{x_ix_j\mid i, j\in \mathbb N, j\geq 2i\}$ (for much the same reason as the previous example.)

But can u give another example other than more than one variable ( if possible)

A quotient of a univariate polynomial ring over a field won't work, because that would be a principal ideal ring, and distributivity works for the lattice of ideals of a commutative principal ideal ring.

$I \cap (J+K) =I \cap J + I\cap K $

2 Answers2