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Let $\mathcal{O}_{k^2,(0,0)}$ be the local ring at origin, i.e. $k[x,y]_{(x,y)}$.

I want to show that $\dim_k \mathcal{O}_{k^2,(0,0)}/(y-x^2,x^3)=3$.

My rough argument is the following, but I feel that this is somewhat not rigorous:

Suppose $\alpha=\dfrac {f(x,y)}{g(x,y)},\ g(0,0)\neq 0$ is an element of $\mathcal{O}_{k^2,(0,0)}/(y-x^2,x^3)$. Then substituting $y=x^2, > x^3=0$, we can make $\alpha = \dfrac{f_0+f_1x+f_2x^2}{g_0+g_1x+g_2x^2}$. Also by multiplying an appropriate polynomial $1+ax+bx^2$ to denominator and numerator, we can make the denominator to be constant so that $\alpha= A+Bx^2+Cx^2$. So the dimension is 3.

Can I make it mathematically rigorous? or is it the best?

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1 Answer 1

up vote 4 down vote accepted

Lets compute $k[x,y]/(y-x^2,x^3)$ first and localize afterwards. In the quotient ring, we have $y=x^2$, and we divide out by $x^3$, so the quotient is (isomorphic to) $k[x]/x^3$. This ring has a $k$-basis consisting of $1,x$ and $x^2$, so has dimension $3$ over $k$.

Now, localizing corresponds to bealing allowed to divide by everything outside $(x)$. But $k[x]/x^3$ is a local ring, so everything not in $(x)$ are units. This means that localizing wont change anything (being allowed to divide by something invertible doesn't give any new elements). So in fact $k[x,y]_{(x,y)}/(y-x^2,x^3)=k[x,y]/(y-x^2,x^3)=k[x]/x^3$. So by the above, the dimension is 3.

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Isn't there something outside $(x)$? For example, $1+x$ is in $k[x]/(x^3)$, but it is not in $(x)$. So localization contains such elements, how can I deal this? –  Gobi Dec 1 '12 at 1:58
1  
@Gobi: Thank you! I've edited my answer. The key is that $k[x]/(x^3)$ is a local ring. –  Fredrik Meyer Dec 1 '12 at 11:10

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