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Thread: Rings exam problem

  1. #1
    Junior Member
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    Rings exam problem

    Let $\displaystyle \mathbb{K}$ be a field, consider the ring $\displaystyle \mathbb{K}[x,y]$.

    1) Let $\displaystyle p(x,y) \in \mathbb{K}[x,y]$, prove that there exist $\displaystyle q(x,y) \in \mathbb{K}[x,y]$ and $\displaystyle a(x), b(x) \in \mathbb{K}[x]$ unique such that $\displaystyle p(x,y)=(x^3-y^2)q(x,y)+a(x)+b(x)y$.

    2)Prove that the map $\displaystyle \phi : \mathbb{K}[x,y]\longrightarrow \mathbb{K}[t]$ defined by $\displaystyle \phi(p(x,y)) = p(t^2,t^3)$ is a ring morphism.

    3)Find the kernel of $\displaystyle \phi$ (i.e: find generators of it)

    4) Find explicitly the image of $\displaystyle \phi$

    5)Prove that $\displaystyle x^3-y^2$ is irreducible in $\displaystyle \mathbb{K}[x,y]$ . Deduce that the image of $\displaystyle \phi$ is a domain.



    My views:
    1) No clue. Iīm thinking that it has to do with some division but I donīt know what...

    2) Ugly computation, I used the identity given in part 1) and I think I pulled it off (if thereīs a simpler way of doing it please let me know).

    3) It should be $\displaystyle <x^3-y^2>$ I suppose... Am I right?

    4) $\displaystyle p(t^2,t^3) = a(t^2)+t^3b(t^2)$ means something??

    5) Maybe I can use that $\displaystyle \frac{\mathbb{K}[x,y]}{<x^3-y^2>} \cong Im\phi$ once I proved that $\displaystyle x^3-y^2$ is irreducible, but I donīt know how (Eisenstein doesnīt work, I think).


    Thanks, guys!!
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  2. #2
    Senior Member
    Joined
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    Paris
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    Hi

    1) Division, you said it! Use the fact that $\displaystyle K[x][y]=K[x,y]$.
    $\displaystyle -1$ is invertible in $\displaystyle K[x],$ so $\displaystyle \forall p\in K[x][y],\ \exists !q,r\in K[x][y]$ such that:
    $\displaystyle p(x,y)=(-y^2+x^3)q(x,y)+r(x,y)\ $, and $\displaystyle deg(r)<1,$ i.e. $\displaystyle r(x,y)=b(x)y+a(x)$ for two unique $\displaystyle a,b\in K[x]$. (can be written bette, when I write the identity $\displaystyle p(x,y)$ refers to a $\displaystyle P(y)$ whose coefficients are in $\displaystyle K[x]$ so it's equal to a unique polynomial in $\displaystyle K[x,y]$ and there is no problem, same thing fo $\displaystyle q$ and $\displaystyle r$)

    2) is quite easy! I don't think you really need 1)

    3) Here 1) can be useful. You've guessed it, but how to prove that? Take a $\displaystyle p\in K[x,y],$ assume it belongs to $\displaystyle ker\phi ,$ and write it with the identity from 1). Your goal is to prove that $\displaystyle a(x)+
    b(x)y=0$

    4)Take a $\displaystyle n\in\mathbb{N}$. Is $\displaystyle t^n$ in $\displaystyle Im\phi$ ?

    5)Suppose $\displaystyle -y^2+x^3\in K[x][y]$ is reducible, i.e.$\displaystyle \exists\alpha ,\beta\in K[x]\ \text{s.t.}\ -y^2+x^3=-(uy+\alpha(x))(u^{-1}y+\beta(x))$.
    Then $\displaystyle u^{-1}\alpha(x)+u\beta(x)=0$ and $\displaystyle \alpha(x)\beta(x)=x^3$ since $\displaystyle x$ is irreducible in $\displaystyle K[x]$ which is factorial we have $\displaystyle \alpha(x)=vx^i$ and $\displaystyle \beta(x)=v^{-1}x^{3-i}$ for an invertible $\displaystyle v$ and a $\displaystyle i\in\{0,1,2,3\}$. This makes $\displaystyle u^{-1}\alpha(x)+u\beta(x)=0$ impossible, contradiction.
    Therefore $\displaystyle x^3-y^2$ is irreducible in $\displaystyle K[x][y]=K[x,y],$ and you go on as you said.
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