# Module over a commutative ring

#### Siron

##### Active member
Hi,

Let $A$ be a commutative ring, $M$ an $A-$module and $U,V,V'$ submodules of $M$ such that $U \cap V = U \cap V'$ and $U+V=U+V'$. Does it follow that $V=V'$?

The answer is no because the condition that $V \subset V'$ is necessary though I can't find a counterexample.

Does someone has a good counterexample for this wrong statement?

Cheers,
Siron

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
By $U+V$, do you mean $\{u+v\mid u\in U,v\in V\}$?

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Then it's easy to get a counterexample by considering $\mathbb{R}^3$ as a module over $\mathbb{R}$.

#### Deveno

##### Well-known member
MHB Math Scholar
Easy counter-example:

Let our parent $\Bbb R$-module be $\Bbb R^3$ as Evgeny suggests.

Take $U = \text{span}((0,1,0))$

$V = \text{span}((1,0,0))$

$V' = \text{span}((1,1,0))$.

It is clear that:

$U + V = U + V' = \Bbb R^2 \times \{0\}$ and:

$U \cap V = U \cap V' = \{(0,0,0)\}$

but $V \neq V'$.

#### Siron

##### Active member
Easy counter-example:

Let our parent $\Bbb R$-module be $\Bbb R^3$ as Evgeny suggests.

Take $U = \text{span}((0,1,0))$

$V = \text{span}((1,0,0))$

$V' = \text{span}((1,1,0))$.

It is clear that:

$U + V = U + V' = \Bbb R^2 \times \{0\}$ and:

$U \cap V = U \cap V' = \{(0,0,0)\}$

but $V \neq V'$.
Thanks Deveno and Evgeny!