Let $K$ be a field and set $R= K[x]/(x^2)$ so that $R=K \dotplus K\bar{x}$ with $\bar{x}^2=0$. Show that $R \ncong \bar{x}R \oplus \bar{x}R$

Question

Let $K$ be a field and set $R= K[x]/(x^2)$ so that $R=K \dotplus K\bar{x}$ with $\bar{x}^2=0$. In the textbook, there is a sentence :

"$R \ncong \bar{x}R \oplus \bar{x}R$ since $\bar{x}$ annihilates the right-hand module but not the left."

Can anybody explain me why $\bar{x}$ annihilates the right-hand module but not the left implies $R \ncong \bar{x}R \oplus \bar{x}R$?

Note that $\dotplus$ : internal direct sum and $\oplus$ : the external direct sum.

Thanks in advance

Answer 1

We proceed by contradiction. Assume that $f\colon R\to \bar{x}R\oplus \bar{x}R$ is an isomorphism of $R$-modules.

Note that $f(\bar{x}\cdot 1)=\bar{x}\cdot f(1)=0$ as $\bar{x}$ annihilates everything on the right-hand side. Thus $\bar{x}\in\ker(f)$ and thus $f$ is not injective. This contradicts $f$ being an isomorphism.