Rings exam problem

**Inti** · February 18th 2009, 07:36 AM

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

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

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

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

4) Find explicitly the image of $\phi$

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

My views:
1) No clue. I´m thinking that it has to do with some division but I don´t know what...

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

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

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

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

Thanks, guys!!

**clic-clac** · February 18th 2009, 10:59 AM

Hi

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

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 $p\in K[x,y],$ assume it belongs to $ker\phi ,$ and write it with the identity from 1). Your goal is to prove that $a(x)+<br /> b(x)y=0$

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

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

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