# Graded Rings , divided polynomial algebra

I just read about the notion of a divided polynomial algebra, which is defined as follows: Consider the elements $$y^{(i)}=y^i/i!, i\geq 0$$ in the polynomial ring $$\mathbb{Q}[y]$$. They satisfy $$y^{(i)}y^{(j)}=(i,j)y^{(i+j)}$$, so the $$\mathbb{Z}$$-submodule of $$\mathbb{Q}[y]$$ generated by $$y^{(i)}$$ is a subring, this ring is denoted $$\tau(y)$$. Now the author states that we can see $$\tau(y)$$ as a graded ring with deg $$y=2$$, and this is the part I don understand, shouldnt $$y$$ have degree $$1$$? I am new to the idea of graded rings and such so there must something I am mixing up and would appreciate some help. Thanks in advance.

• You can have $\deg y$ any positive integer, as long as then $\deg y^{(i)}=i\deg y$. – Angina Seng May 24 at 19:07
• I am confused I though the way this would be a graded ring was that $\tau(y)=\bigoplus_{n=0}^{\infty}A_n$ where $A_n$ would be generated by $y^{(n)}$. – Lizard King May 24 at 19:08
• For some reason, your author wants $y^{(n)}$ to generate $A_{2n}$. – Angina Seng May 24 at 19:37
• Hm ok so if $y$ is off degree $2$, what is in degree $1$? @AnginaSeng – Lizard King May 24 at 19:45