Graded Rings , divided polynomial algebra

Question

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.

Answer 1

This sounds like it is in a cohomology ring, where the degree can determine the commutativity of a product. For a fully commutative ring you would want generators to be in even degrees.

For a ring like a polynomial ring, the degree of the generator is generally seen to be arbitrary. Outside a cohomology ring, you could make the degree any positive integer and get the same behavior. You'd just be multiplying the degree function by that integer but up to that multiple using the same one.

There doesn't need to be anything in the degrees in between.