I read "Undergraduate Commutative Algebra By Miles Reid".
I don't understand this description、
"Whether an exact sequence is split or not on what ring it is considered over.
For example,$k $ is field.
$0 \rightarrow (x) \rightarrow k[x] \rightarrow k \rightarrow 0 $ is split over $k$ but not over $k[x]$ ".
I guess if over $k$,$k[x]=(x) \oplus k$ and its bases of vector space is $x $ and $1$.
But if over $k[x]$, I can't construct counterexample morphism.
Please show me a not split example.
On
Some easy example of an exact sequence that does not split... Consider the following abelian groups as $\mathbb{Z}$-modules... Then the followin sequence is exact, $$ 0\rightarrow\mathbb{Z}_2\rightarrow\mathbb{Z}_4\rightarrow\mathbb{Z_2}\rightarrow0 $$ where the morphism $\mathbb{Z}_2\rightarrow\mathbb{Z}_4$ is just inclusion and the next arrow, $\mathbb{Z}_4\rightarrow\mathbb{Z}_2$ is reduction modulo $4$. Clearly, this sequence doesn't split! If it splits, then $\mathbb{Z}_4\cong\mathbb{Z}_2\oplus\mathbb{Z}_2$ but the right-hand side is not a cyclic group.
(By "is just inclusion" I mean of course $0\mapsto0$ and $1\mapsto2$.)
That's not quite right. Over $k$, the sequence splits, but a basis for $k$ is $1$ while a basis for $(x)$ is $\{x,x^2,x^3, \dots\}$.
There are many ways to see why the sequence does not split over $k[x]$. For instance, If the sequence were to split, then we would have $k[x] \cong (x) \oplus k$ as $R$-modules. In particular, this would mean $k[x]$ contains a nonzero element annihilated by $x$, which is not the case.