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I need to show that $\{\sum a_i X^i\in R[x]\mid a_1=0\}\cong R[X,Y]/(X^2-Y^3)$, where $R$ is a commutative ring with identity.

Now I've made a hommomorphism $\psi:f\mapsto f(x^3,x^2)$. Now the only thing left, after showing that $\psi$ is surjective, I have to show is that if $f\in \ker(\psi)$ then $f\in(X^2-Y^3)$. Now I've made a polynomial division, $g=X^2-Y^3$, so $f=gq+r$ as usual (division in $(R[Y])[X]$). So I get $\deg_X r\le1$, so as I know $r=0$, not possible as $r(x^3,x^2)=0$, or $\deg(r)=-\infty$, then nothing to show, or $\deg_Xr=1$. In this case $r=(\sum a_i Y^i)X+(\sum b_i Y^i)$, with $m=\deg(\sum a_i Y^i)$ (here of course in $R[Y]$) so $r(x^3,x^2)=a_mX^{3+2m}+\cdots\neq0$.

(1) Now I was wondering if there is a shorter/simpler/more elegant proof?

Other question:

$K$ a field, $R=K[X,Y]/(XY^2)$, $\bar X=X+(XY^2)$. Now I have to prove that $\bar X$ and $\bar X+\bar{XY}$ are not associated and that they produce the same principal ideal.

What I've got: (I let away the bars for the eq. classes)

$X(1+Y)=(X+XY)$, first inclusion $(X+XY)(1-Y)=X-XY+XY-XY^2=X$, second

(2) Is there a general method to get those divisors? I tried for a little while.

Now for "not associated" there's a tip: look at the ideal which contains elements $f\in R$ with $fx=0$. Can't go from here with the tip (the ideal should be $(Y^2)$)

What I did: Assume its true so $X=XY+X$ so we get $0=XY+gXY^2$ (here $X,Y$ are the monoids, not the classes, $g\in K[X,Y]$.) But now if $g\neq 0$ then $\deg(g(XY^2))=\deg(g)+3\geq3$ esp. all monoids "in" $gXY^2$ have higher rank than $3$, so $XY$ can't disappear (as a field is integral domain, so its polynomial domain), so there's no unit $u$ with $uX=XY+X$.

(3) How can I show this more smoothly?

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  • I honestly like the proof in (1) quite a bit. I also think Artin gives roughly this proof in Algebra -- if something significantly better were out there my guess is he would have found it. [Appeal to authority, whoops.] – Hoot Jan 10 '15 at 20:10
  • what is the meaning of $\text{rank}(r)$? do you mean $\text{deg}(r)$? and one word of caution: you have to be careful when you are saying "usual division in $(R[Y])[X]$" because it's not a Euclidean domain (it may not even a domain). – Krish Jan 10 '15 at 20:13
  • Yes I mean deg! The highest coefficient of g is 1, a unit in R[Y], so the division is allowed. I missed to mention it! Thanks, changed it – user206998 Jan 10 '15 at 20:19
  • At the end, I don't see why you start off by writing $X=XY+X$. – Hoot Jan 10 '15 at 20:20
  • I was thinking $X=(1+gXY^2)(X+XY)$ – user206998 Jan 10 '15 at 20:32
  • But I don't think that's what associate means. $1 - xy$ is a unit in $R$, for example. – Hoot Jan 10 '15 at 20:51
  • Yes, true, have to rethink my point then! – user206998 Jan 10 '15 at 20:55
  • I don't quite understand the hint, though — it seems to me that the annihilators of $x$ and $x(1 + y)$ are the same. – Hoot Jan 10 '15 at 21:01
  • $gXY^2=Xu - XY - X = \sum \limits{i\geq0} a_i X^{i+2} Y + \sum b_i X^{i+2} + (b_1-1)X + (a_1-1)XY$ with $u=\sum a_i X^{i} Y + \sum b_i X^{i}$ as with $Y^2$ the terms disappear. Impossible as at the right side $Y^2$ doesnt divide any term (i.e. polynom division through $Y^2$) – user206998 Jan 10 '15 at 21:20

1 Answers1

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(1) This is the usual way to prove that $R[X,Y]/(X^2-Y^3)\simeq R[T^2,T^3]$.

(2) It's already solved here.

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