How to show that $K[t]/(t^d)$ is indecomposable as a $K[t]$-module and $\operatorname{End}_{K[t]} (K[t]/(t^d)) \cong K[t]/(t^d)$? How to show that 
(1) $K[t]/(t^d)$ is indecomposable as a $K[t]$-module?
(2)$\operatorname{End}_{K[t]} (K[t]/(t^d)) \cong K[t]/(t^d)$? 
I think that if (2) is true, then $\operatorname{End}_{K[t]} (K[t]/(t^d))$ is local since $K[t]/(t^d)$ has only two idempotents $0, 1$. Therefore $\operatorname{End}_{K[t]} (K[t]/(t^d))$ is local and hence $K[t]/(t^d)$ is indecomposable as a $K[t]$-module. Is this true? How to show (2)? Thank you very much.
 A: In general if $\varphi: R \rightarrow S$ is a surjective ring homomorphism and $M$ is a (left) $S$-module, $M$ is also an $R$-module via $\varphi$ and $\operatorname{End}_R(M) = \operatorname{End}_S(M)$. (Note that we always have $\operatorname{End}_S(M) \subseteq \operatorname{End}_R(M)$ and for the reverse inclusion, actually $\varphi$ being an epimorhism in the category of rings is enough: see these notes) 
Thus if $I$ is a two-sided ideal of a ring $R$, considering $R/I$ naturally as a left $R/I$-module, we have $\operatorname{End}_R(R/I) = \operatorname{End}_{R/I}(R/I) \cong (R/I)^{\text{op}}$, where op denotes the opposite ring - you can ignore it in the commutative case. The last ring isomorphism above comes from a general fact: if $A$ is a ring and we consider $A$ as a left $A$-module and write $_{A}A$ for this module, we have $\operatorname{End}_A(_AA) \cong A^{\text{op}}$ via the maps $f \mapsto f(1)$ and $ \text{"right multiplication by $a$"}\leftarrow a$.
The assignments $R = K[t]$ and $I = (t^d)$ yields (2) , noting that $R$ is commutative here.
