Nontrivial example of $M$ such that $\text{Ass}(M)=\varnothing$? 
Suppose $R$ is a commutative ring, and $M$ an $R$-module. Is there a nontrivial example of such $M$ where the set of associated primes $\text{Ass}(M)=\varnothing$? 

Taking $M=0$ feels kind of cheap, since any annihilator of $x\in M$ is necessarily all of $R$, and hence not prime.
Is there a better example that is still relatively simple, preferably with some explanation about why $\text{Ass}(M)=\varnothing$? 
 A: Let $k$ be a field and $R=k[X_0,X_1,X_2,..]/ \langle X_0^2,X_1^2,X_2^2,...\rangle=k[x_0,x_1,x_2,..]$. Then $\operatorname{Ass}_R(R)=\emptyset$ .
Edit: a detailed proof
At Lando's request I'll give a complete proof of the above. 
1) Let's show that the only prime ideal of $R$ is the obviously maximal ideal $\mathfrak m=\langle x_0,x_1,x_2,...\rangle$.
Indeed,  if $\mathfrak p \subset R$ is  prime then for all $i\geq 0$ we deduce from $0=x_i^2\in \mathfrak p$ that  $x_i\in \mathfrak p$, so that $\mathfrak m=\langle x_0,x_1,x_2,...\rangle \subset \mathfrak p$ and since $\mathfrak m$ is maximal we have $\mathfrak p=\mathfrak m$.  
2) It remains to prove that $\mathfrak m$ is not an associated ideal and we will have the desired conclusion $\operatorname{Ass}(R)=\emptyset$ ($R$ is seen as a module over itself).
To say that  $\mathfrak m$ is associated means that there exists $0\neq P\in R$ with $\mathfrak m=\operatorname{Ann}(P)$.
Such a $P$ can 
be written as $P(x_0,...,x_N)$, i.e. can involve only finitely many $x_i$'s.
But then we see that it is not true that $\mathfrak m=\operatorname{Ann}(P)$, because $x_{N+1}\notin \operatorname{Ann}(P(x_0,...,x_N))$: any non-zero term $rx_{i_0}...x_{i_s} \neq 0\; (0\leq i_0\lt\cdots\lt i_s\leq N$ ) of  $P(x_0,...,x_N)$ will satisfy $rx_{i_0}...x_{i_s}\cdot x_{N+1} \neq 0$ and so $ P(x_0,...,x_N)\cdot x_{N+1}\neq 0 $.
