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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.

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    $\begingroup$ You can have $\deg y$ any positive integer, as long as then $\deg y^{(i)}=i\deg y$. $\endgroup$ – Angina Seng May 24 at 19:07
  • $\begingroup$ 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 $y^{(n)}$. $\endgroup$ – Lizard King May 24 at 19:08
  • $\begingroup$ For some reason, your author wants $y^{(n)}$ to generate $A_{2n}$. $\endgroup$ – Angina Seng May 24 at 19:37
  • $\begingroup$ Hm ok so if $y$ is off degree $2$, what is in degree $1$? @AnginaSeng $\endgroup$ – Lizard King May 24 at 19:45
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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.

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  • $\begingroup$ Some topologically trained authors follow this "commuting generators should live in even degrees" ideology even when they are not considering cohomology rings. $\endgroup$ – darij grinberg May 24 at 20:58

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