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Let $R$ be a ring (commutative, with unit). Show that $A=\{\sum a_iT^i\in R[T] \; : \; a_1=0\}$ is a subring of $R[T]$ and isomorphic to $R[X][Y]/(X^2-Y^3)$.

Of course, I'm trying to find a ring homomorphism of $R[X][Y]$ into $R[T]$ with image $A$ and kernel $(X^2-Y^3)$. Can you give me a hint?

Srivatsan
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Stefan
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    One of the frequent confusions in this sort of problem is the duplicate usage of $X$ in both rings. This prolem would be more obvious if you had expressed $A$ as a subset of $R[Z]$ rather than $R[X]$. – Thomas Andrews May 11 '11 at 17:04
  • @Thomas: Yes, you're right. Should I change it? If I do, it will make lhf's answer incomprehensible. – Stefan May 11 '11 at 17:57
  • I wasn't necessarily recommending changing it, just noting that the reason the solution doesn't seem as obvious as it is is due to this. – Thomas Andrews May 11 '11 at 19:48
  • Related: http://math.stackexchange.com/questions/1130534 – Watson Feb 05 '17 at 13:32

2 Answers2

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How about $X \mapsto T^3$ and $Y \mapsto T^2$?

lhf
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  • Thank you. I see that $A$ is the image, but not yet why every element of the kernel is a multiple of $X^2-Y^3$. – Stefan May 11 '11 at 17:54
  • @Stefan, you need to consider the positive integer solutions of $3i+2j=k$ for each $k$. You go from one solution to another by adding or subtracting $2$ from $i$ while subtracting or adding $3$ to $j$. Then you can collect terms and they'll all be multiples of $X^2-Y^3$. Do a few cases... – lhf May 11 '11 at 18:33
  • @Stefan, in other words, it's all about knowing the shape of the general solution of $3i+2j=k$. – lhf May 11 '11 at 18:49
  • Got it! Thanks again. – Stefan May 12 '11 at 15:45
  • @Stefan, for fun, consider now the general case $R[X,Y]\to R[T]$, with $X \mapsto T^a$ and $Y \mapsto T^b$. What is the image? What is the kernel? – lhf May 12 '11 at 18:12
  • Can $\mathbb{N}a+\mathbb{N}b$ be further simplified? – Stefan May 19 '11 at 21:44
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    @Stefan, it's related to the Frobenius problem. The set $\mathbb{N}a+\mathbb{N}b$ contains all numbers greater than a certain limit and some less than this limit, but I'm not sure the smaller set is easy to describe. – lhf May 19 '11 at 22:28
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If you choose a basis for $R[X][Y]/(X^2-Y^3)$ (as a free $R$-module), you will have a clearer image of what it looks like. For example you can show that $(1,Y,Y^2,Y^3,\ldots,X,XY,XY^2,XY^3,\ldots)$ is one possible basis.

And then look at what is the image of the basis under the isomorphism.

mercio
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