generators of associated graded ring

Question

Let $k$ a field, $A = k[x,y]/(y^2-x^3)$ and $\mathfrak{a} = (x,y)$ and let $G_{\mathfrak{a}}(A) := \bigoplus_{n \geq 0} \mathfrak{a}^n/\mathfrak{a}^{n+1}$ be the associated graded ring of $A$ with respect to $\mathfrak{a}$.

For each $n$, $\mathfrak{a}^n$ is generated by the monomials $x^n, x^{n-1}y, \dots, y^n$ as a $k$-module, thus generated by $x$ and $y$ as a $k$-algebra. Thus can we then say that $G_{\mathfrak{a}}(A)$ is generated by $x$ and $y$ as a $k$-algebra?

I want to define a graded $k$-algebra morphism $\varphi : G_{\mathfrak{a}}(A) \to k[s,t]/(t^2)$ by sending $x \mapsto s$ and $y \mapsto t$ and then extending the map polynomially from the generators to the whole $k$-algebra, but I am wondering if I am forgetting to check something or made a mistake somewhere.

Answer 1

$\DeclareMathOperator{\G}{G} \newcommand{\fraka}{\mathfrak{a}}$To your first question, yes, the associated graded $\G_\fraka(A)$ is always generated in degree 1 over $A/\fraka$ since by definition, the elements of $\G_\fraka(A)_n=\fraka^n/\fraka^{n+1}$ are products of elements of $\G_\fraka(A)_1=\fraka/\fraka^2$.

For your second question, we can make the isomorphism $\varphi:\G_\fraka(A)\to k[x,y]/(y^2)$ explicit (I will abuse notation and use the same variables $x$ and $y$). This follows Milne's AG notes. Let $I\subset k[x,y]$ be any ideal and $I^*\subset k[x,y]$ the ideal generated by the initial forms of $I$, which in this setting just means the lowest-degree homogeneous term. Set $\fraka=(x,y)$. Then $$\begin{align} \G_\fraka(k[x,y]/I) &= \bigoplus_{n\geq 0} \frac{(\fraka+I)^n}{(\fraka+I)^{n+1}} \\ &\cong \bigoplus_{n\geq 0} \frac{\fraka^n+I}{\fraka^{n+1}+I} \\ &\cong \bigoplus_{n\geq 0} \frac{\fraka^n}{\fraka^n\cap (\fraka^{n+1}+I)} & \text{(2nd isomorphism theorem)} \\ &= \bigoplus_{n\geq 0} \frac{\fraka^n}{\fraka^{n+1}+(\fraka^n\cap I)} & \text{(modular law)} \\ &\cong \bigoplus_{n\geq 0} \frac{\fraka^n}{\fraka^{n+1}+I_n} \end{align}$$ where $I_n$ is the degree $n$ homogeneous part of $I^*$. This last isomorphism follows since any element of $\fraka^n\cap I$ of degree $\geq n+1$ is already being killed by $\fraka^{n+1}$, and everything else belongs to $I_n$. But now we can identify $\fraka^n/(\fraka^{n+1}+I_n)$ with the degree $n$ homogeneous part of $k[x,y]/I^*$ and we're done. Note the same proof works for $k[x_1,\dots,x_r]$ and $\fraka=(x_1,\dots,x_r)$.

For a principal ideal $I=(f)$ in a polynomial ring, we have $I^*=(f^*)$. Thus $\G_\fraka(k[x,y]/(y^2-x^3))\cong k[x,y]/(y^2)$ under the above isomorphism.