Hilbert series of $k[x,y]/(x^2,y^3)$ This is Example 1.7 in Graded Syzygies by Peeva, and I am wondering how to work this out. The author writes

Let $A=k[x,y]$ and let $J=(x^2,y^3)$. Then $A/J$ is graded with basis $\{1\}$ in degree $0$, basis $\{x,y\}$ in degree $1$, basis $\{xy,y^2\}$ in degree $2$, and basis $\{xy^2\}$ in degree $3$. Its Hilbert series is
$$
\text{Hilb}_{A/J}(t)=1+2t+2t^2+t^3.
$$

I am very new to the topic of graded rings and Hilbert series, so I was wondering if someone could help me understand how to compute these bases? I understand how the Hilbert series is formed once we know the sizes of these bases, but I am stuck on actually working them out.
What I have tried:
We know that as a graded ring, $A$ decomposes as $A=\oplus A_i$ where $$
\begin{align}
A_0&=k\\
A_1&=\text{span}_k(x,y)\\
A_2&=\text{span}_k(x^2,xy,y^2)\\
A_3&=\text{span}_k(x^3,xy^2,yx^2,y^3)\\
&\;\;\vdots
\end{align}
$$
and so forth. Furthermore, we know $J$ decomposes as $J=\oplus J_i$, where
$$
\begin{align}
J_0=A_0\cap J&=k\cap (x^2,y^3)=\{0\}\\
J_1=A_1\cap J&=\text{span}_k(x,y)\cap (x^2,y^3)\\
J_2=A_2\cap J&=\text{span}_k(x^2,xy,y^2)\cap (x^2,y^3)\\
J_3=A_3\cap J&=\text{span}_k(x^3,xy^2,yx^2,y^3)\cap (x^2,y^3)\\
&\;\;\vdots
\end{align}
$$
and so on. Finally, $A/J$ decomposes as $A/J=\oplus A_i/J_i$. From here, I see that $A_0/J_0=k/\{0\}\cong k$, so it is obvious that $A/J$ has basis $\{1\}$ in degree $0$. However, how can I understand the bases for $A_1/J_1,A_2/J_2,\ldots$? For example, I'm not sure what to make of $$
A_1/J_1=\text{span}_k(x,y)/(\text{span}_k(x,y)\cap (x^2,y^3)).
$$
The only idea I have off the top of my head right now is to use the Second Isomorphism Theorem to write
$$
A_1/J_1\cong (\text{span}_k(x,y)+(x^2,y^3))/(x^2,y^3),
$$
but I'm not sure if this helps me find a basis. Any feedback or help would be appreciated. Is there a "standard" procedure that one follows in general to compute these sorts of things?
 A: Like you mention, one can decompose $A = k[x, y]$ as a $k$-vector space in the following manner. $$A = k \oplus k \langle x, y \rangle \oplus k \langle x^2, xy, y^2 \rangle \oplus \cdots \oplus k \langle x^{n - i} y^i \,|\, 0 \leq i \leq n \rangle \oplus \cdots$$ Considering that $J = (x^2, y^3)$ is a homogeneous ideal of $A,$ it follows that $A / J$ is graded, as you mention. One obtains the graded pieces of $A / J$ in a fashion analogous to the case of $A.$ $$A / J = k \oplus k \langle \bar x, \bar y \rangle \oplus k \langle \bar x^2, \bar x \bar y, \bar y^2 \rangle \oplus \cdots \oplus k \langle \bar x^{n - i} \bar y^i \,|\, 0 \leq i \leq n \rangle \oplus \cdots$$ Observe that $\bar x^{n - i} = \bar 0$ for each integer $n - i \geq 2$ and $\bar y^i = \bar 0$ for each integer $i \geq 3.$ Consequently, all of the graded pieces beyond $n = 4$ are zero, as the Pigeonhole Principle guarantees in this case that $n - i \geq 2$ or $i \geq 3.$ Further, the first four graded pieces are easily seen to be $k,$ $k \langle \bar x, \bar y \rangle,$ $k \langle \bar x \bar y, \bar y^2 \rangle,$ and $k \langle \bar x \bar y^2 \rangle.$ Considering that the Hilbert series is defined as $H_A(t) = \sum_{n = 0}^\infty \dim_k (A / J)_n t^n,$ we find that $H_A(t) = 1 + 2t + 2t^2 + t^3,$ as desired.
