0
$\begingroup$

How to prove that if A, B, C are submodules of M module then $$ A \cap (B+C) = (A\cap B)+C \iff C \subseteq A $$

Any help would be appreciated!

  • 2
    What is $B$? Do you want the left hand side to hold for all $B$? If $B$ is given, $\Rightarrow$ needn't hold, does it? Consider $B= M$ ...2012-04-30
  • 1
    Yes, it looks like you are missing a quantifier for this theorem.2012-04-30
  • 1
    Also, if $A=\{0\}$, then $A\cup(B+C) = B+C = (A\cup B) + C$, for any $B,C$, which would negate this statement with the most obvious quantifier added to the formula.2012-04-30
  • 2
    The $\cup$ should be $\cap$.2012-04-30
  • 1
    At least the $\impliedby$ statement is true as written, but I also suspect the question was other than intended. I am also baffled by the mention of associativity mentioned in the title, when this looks more like distributivity.2012-04-30
  • 0
    I am very sorry. It is not my day, I did two mistakes in one line. Shame! Thank you very much that you warned me.2012-04-30
  • 1
    How is $A\cap (B+C) = (A\cap B)+C$ an instance of "distributivity"? Distributivity of what over what?2012-04-30

1 Answers 1

2

Suppose that $A\cap(B+C)=(A\cap B)+C$; since $C\subseteq (A\cap B)+C=A\cap(B+C)\subseteq A$, it follows that $C\subseteq A$.

Conversely, suppose that $C\subseteq A$. If $x\in A\cap (B+C)$, then $x\in A$ and there exist $b\in B$, $c\in C$ such that $x=b+c$. Then $b\in B$, and $b=x-c\in A+C = A$ (since $C\subseteq A$), so $b\in (A\cap B)$. Therefore, $x = b+c\in (A\cap B)+C$. Thus, $A\cap(B+C)\subseteq (A\cap B)+C$.

Conversely, if $x\in (A\cap B)+C$, then there exists $y\in A\cap B$ and $c\in C$ such that $x=y+c$. Then $x\in A$, since $y\in A$, $c\in C\subseteq A$, so $y+c\in A$. And $y\in B$, $c\in C$, so $x=y+c\in B+C$. Therefore, $x\in A\cap(B+C)$. This proves $(A\cap B)+C\subseteq A\cap(B+C)$, proving the equality.