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Module over a commutative ring

Siron

Active member
Jan 28, 2012
150
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?

Thanks in advance!
Cheers,
Siron
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
By $U+V$, do you mean $\{u+v\mid u\in U,v\in V\}$?
 

Siron

Active member
Jan 28, 2012
150

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
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
Feb 15, 2012
1,967
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
Jan 28, 2012
150
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!