Proposition

Prove

$\Ann(M + N) = \Ann(M) \cap \Ann(N)$.
$(N:P) = \Ann((N + P) / N)$.

Solution

1

$\Ann(M + N) = (0:M + N)$, and $\Ann(M) \cap \Ann(N) = (0:M) \cap (0:N)$. We showed that $(0:M + N) = (0:M) \cap (0:N)$ in a previous post.

2

\[\begin{align*} \Ann((N + P) / N) &= (0:(N + P) / N) \\ &= \{ a \in A \mid a((N + P) / N) = 0 \} \\ &= \{ a \in A \mid a(N + P) \subset N \} \\ &= (N:N + P) \\ &= (N:N) \cap (N:P) \\ &= A \cap (N:P) \\ &= (N:P). \end{align*}\]