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Thread: k[x]-module

  1. #1
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    k[x]-module

    Let K be a field, and (x) -> K[x] be an injective map where (x) is the principal ideal generated by x.

    (i) Prove that the quotient K[x]-module, K[x]/(x), is isomorphic to K as an abelian group and also as a K-vector space.

    Hence there is an exact sequence of K[x]-modules :

    0 -> (x) -> K[x] -> K -> 0

    (ii) Is this sequence split as a sequence of K[x]-modules ?

    (iii) Is this sequence split as a sequence of K-modules ?


    (i) is clear. But I can't make sense nor find a solution for (ii) and (iii).
    Which is why I require your help !
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  2. #2
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    Quote Originally Posted by Simply82358 View Post
    Let K be a field, and (x) -> K[x] be an injective map where (x) is the principal ideal generated by x.

    (i) Prove that the quotient K[x]-module, K[x]/(x), is isomorphic to K as an abelian group and also as a K-vector space.

    Hence there is an exact sequence of K[x]-modules :

    0 -> (x) -> K[x] -> K -> 0

    (ii) Is this sequence split as a sequence of K[x]-modules ?

    (iii) Is this sequence split as a sequence of K-modules ?


    (i) is clear. But I can't make sense nor find a solution for (ii) and (iii).
    Which is why I require your help !
    the answer to (ii) is negative and to (iii) is positive. to see this, let $\displaystyle I=\langle x \rangle$ and $\displaystyle \pi: K[x] \longrightarrow K[x]/I$ be the natural homomorphism. let $\displaystyle \text{id}$ be the identity map over $\displaystyle K[x]/I.$

    suppose that $\displaystyle \alpha: K[x]/I \longrightarrow K[x]$ is a $\displaystyle K[x]$ homomorhism such that $\displaystyle \pi \alpha = \text{id}.$ now, for any $\displaystyle p(x) \in K[x]$ we have $\displaystyle xp(x) \in I$ and thus $\displaystyle 0=\alpha(xp(x)+I)=x \alpha(p(x)+I),$

    which means $\displaystyle \alpha(p(x)+I)=0,$ i.e. $\displaystyle \alpha = 0$ and thus $\displaystyle \text{id}=\pi \alpha = 0,$ which is non-sense.

    for (ii) just define $\displaystyle \beta : K[x]/I \longrightarrow K[x]$ by $\displaystyle \beta(p(x)+I)=p(0).$ see that $\displaystyle \beta$ is a well-defined $\displaystyle K$ homomorphism and $\displaystyle \pi \beta = \text{id}.$ (note that $\displaystyle p(0)+I=p(x)+I.$)
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