# Let $X,Y,Z$ be subspaces of $V$ so that $X$ is a subspace of $Y$. Prove that $Y\cap (X+Z)=X+(Y\cap Z)$

Let $X,Y,Z$ be subspaces of $V$ so that $X$ is a subspace of $Y$. Prove that $Y\cap (X+Z)=X+(Y\cap Z)$

I know that I need to prove that $Y\cap (X+Z)\subseteq X+(Y\cap Z)$ and $X+(Y\cap Z)\subseteq Y\cap (X+Z)$ but I´m having a hard time.

Can you lend me a hand please? I would really appreciate it

• so what precisely have you tried? Sep 4, 2014 at 12:49
• I give a $v\in Y\cap (X+Z)$ so $v\in Y$ and $v\in (X+Z)$ and the latter one implies by definition that $v=v_1+v_2$ where $v_1\in X$ and $v_2\in Z$ but I don´t know what to do from here Sep 4, 2014 at 12:54
• okay I have added an answer to help you out. I just realized that you already mentioned the general hint, but anyways... Sep 4, 2014 at 13:02

This is called a modular law. It holds for subgroups of any abelian (additive) group, thus it holds also for submodules of a module, and in particular for subspaces of a vector space. I prefer to write it in the form that is easy to remember: $\newcommand{\inters}{\cap}$

Let $A$, $X$, $B$ be subgroups of an abelian group. If $A\subseteq B$, then $A+(X\inters B)=(A+X)\inters B$.

You see -- you just move the pair of parentheses.

Proof. $~$Since $A\subseteq A+X$ and $A\subseteq B$, we have $A\subseteq (A+X)\inters B\,$; since $X\subseteq A+X$, we have also $X\inters B\subseteq(A+X)\inters B\,$; it follows that $A+(X\inters B)\subseteq (A+X)\inters B\,$. In order to prove the reverse inclusion, we consider any $y\in(A+X)\inters B\,$, that is, $y=a+x=b$ for some $a\in A$, some $x\in X$, and some $b\in B$. Then $x=b-a\in B$ because $b\in B$ and $a\in A\subseteq B$, thus $x\in X\inters B$, whence $y=a+x\in A+(X\inters B)\,$.$~$ Done.

The proof is essentially the same as the one given by Deuteu, just cleaner and clearer. $\newcommand{\coll}{\mathcal}$ $\newcommand{\Inters}{\bigcap}$

Let $\coll{L}$ be the set of all subgroups of an abelian group (or of all submodules of a module, or of all subspaces of a vector space). When $\coll{L}$ is ordered by inclusion, it becomes a lattice: any $X,Y\!\in\coll{L}$ have the greatest lower bound $X\inters Y$ and the least upper bound $X+Y\!$. (Actually every indexed family of members $X_i$ of $\coll{L}$ has the greatest lower bound $\Inters_iX_i$ and the least upper bound $\sum_iX_i$
-- the lattice $\coll{L}$ is complete). The lattice $\coll{L}$ is a modular lattice because (you guessed it)
it obeys the modular law. See the article "Modular lattice" in Wikipedia (for starters).

As noted, to prove an equality of sets one must prove each side is a subset of the other one.

First, we prove the $\subseteq$ side, i.e. $Y \cap (Z+X) \subseteq X + (Y \cap Z)$:

Take any $a \in Y \cap (Z+X)$. That means $a$ is in both $Y$ and $(Z+X)$. The question is: is $a \in X$?

• if $a \in X$, that means that it certainly is in $X+(Y\cap Z)$.
• if $a\notin X$, then it must be in $Z$ (since it is in $Z+X$). Thus, $a$ is in both $Y$ and $Z$, which means it is in $Y\cap Z$ and therefore it is also in $X+(Y\cap Z)$.

Second, prove that $X+(Y\cap Z) \subseteq Y\cap (Z+X)$:

Take any $a \in X+(Y\cap Z)$. Then, $a$ is either in $X$ or in the intersection $Y\cap Z$. Since $X$ is a subset of $Y$, we have that $a$ must be in $Y$. Now, the same question: is $a\in X$?

• if $a\in X$, it is also in $Z+X$ and we are done.
• if $a \notin X$, it must be in $Y\cap Z$, which means it certainly is in $Y\cap (Z+X)$. $\Box$

A collection of hints:

Hint: $X + Y = Y$

General hint: if you try to show $A \subset B$ then take a generic element $a \in A$ and show $a \in B$.

Hint about elements: a generic element of the LHS is of the form: $Y \ni y=x+z \in X+Z$ (it lies in the intersection). When does this happen to be in the RHS? In which subspace(s) lives $z = y-x$? (hint: it is not only $Z$)

Another hint about elements: if $x + r$ is a generic element on the RHS, then in particular it is an element in $X+Z$. So it suffices to show that it is also in $Y$. Why is that?

If you use those hints you should be able to solve the exercise yourself - hope it helps.

As you said we can do it by double inclusion there is :

If $w \in X+(Y\cap Z)$, we have $w=x+\mu$ with $x\in X$ and $\mu \in (Y\cap Z)$
- $X \subseteq Y$ so $x \in Y$ thus $x,\mu \in Y$ so $w \in Y$
- $\mu \in Z$ so $w \in (X+Z)$
In conclusion $w \in Y\cap (X+Z)$

If $v \in Y\cap (X+Z)$, $v=\alpha \in Y$ and $v=\beta + \gamma$ with $\beta \in X$ and $\gamma \in Z$
$\gamma = \alpha - \beta$ and $\beta \in X \subseteq Y$ so $\gamma \in Y$ thus $\gamma \in Y \cap Z$
In conclusion $v=\beta_{\in X} + \gamma_{\in Y\cap Z}$ so $v\in X+(Y\cap Z)$