# Associated primes of a non-zero module always exist if corresponding ring is Noetherian

Let $R$ be a commutative Noetherian ring. Let $M$ be a non-zero $R$-module (not necessary finitely generated). Prove that set of associated primes of the module is not empty, $\mathrm{Ass}_R(M) \neq \emptyset$.

For brevity I will use name $A = \{ \mathrm{Ann}_R(m) | m \in M \bullet m \neq 0 \}$ . It's already was proved that maximal elements of $A$ will be primes in $R$ and hence belong to $\mathrm{Ass}_R(M)$. However I'm struggling showing that $A$ actually has maximal elements. Its possible to establish that for any $m \in M$ such that $m \neq 0$

$$\{0\} \subset\mathrm{Ann}(M) \subset \mathrm{Ann}(m)$$ So $A \neq \emptyset$ and it seems that we can apply Zorn's lemma. However, now we need to show that any ascending chain $C$ in $A$ indexed by strictly ordered set $I$ has an upper bound in $A$. As our ring is Noetherian we can use representation $C_i = (c_i)$ for some indexed family $c_i$ of $n_i$ elements in $R$. We know that for each $i < j$ from $I$ there exists $r \in R^{n_j}$ such that $c_i = r \cdot c_j$ (where $(\cdot)$ represents pointwise multiplication). And I don't know how to apply Noetherian nature of $R$ any further to attain actual upper bound for chain $C$. (I don't know how to prove that $\bigcup_{i \in I}C_i \in A$ ). And I don't know how to use decomposition to irreducable elements as I don't know how to bound growth of $n_i$ .

Maybe I forgot some important properties of Noetherian rings?

Maybe Zorn's lemma is a wrong way to go?

Thanks!

p. s.

It turned out that I forgot about ACC for ideals of Noetherian rings.

• Every non-empty set $A$ of ideals in a noetherian ring has a maximal element: Take $I_0 \in A$. If it's not maximal there is $I_1$ in $A$ such that $I_0 \subsetneqq I_1$. If $I_1$ is not a max. element in $A$, there is $I_2$ with $I_0 \subsetneqq I_1\subsetneqq I_2$. Continuing this way you get a strictly ascending chain of ideals which must stop since $R$ is noetherian. By construction the last ideal in the chain is maximal in $A$. – Todd Leason Mar 31 '16 at 12:22
• It turned out that I forgot about Ascending chain condition for Noetherian rings. Do you think I should delete the question? – Nik Pronko Mar 31 '16 at 12:29
• No need to delete it. Other people might have the same question and will find your post helpful. – Todd Leason Mar 31 '16 at 12:31

It suffices just to use Ascending chain condition on Ideals: For Noetherian rings we know that for any infinite chain $C$ $$\exists N \in I \; : \; \forall i > N \; . \; C_N = C_i$$ As all elements of $C$ were drown from $A$ ACC also resolves this situation.