Are the $k[x]$-modules $M = k[x]/\langle x + 1 \rangle \oplus k[x]/\langle x + 1 \rangle $ and $N = k[x]/\langle (x + 1)^2 \rangle$ isomorphic?

Question

Let $k$ be a field and consider the $k[x]$-modules

$$M = \frac{k[x]}{\langle x + 1 \rangle} \oplus \frac{k[x]}{\langle x + 1 \rangle} \quad \text{ and } \quad N = \frac{k[x]}{\langle (x + 1)^2 \rangle}.$$

How can I show that $ M \not\cong N $? It feels obvious but I cannot prove it.

Answer 1

$M$ is annihilated by $x+1$ but $N$ isn't.

Compare the Abelian groups $\Bbb Z/2\Bbb Z\oplus \Bbb Z/2\Bbb Z$ and $\Bbb Z/4\Bbb Z$.

Answer 2

In $N$, there is a non-zero element whose square is zero, namely $x+1$. Such an element does not exists in $M$, because if we consider a polynomial $P(x)$ such that $x+1$ divides $P(x)^2$, then $x+1$ also divides $P$ ($x+1$ is a prime element of the ring $k[x]$).