I'm working through Kassel's Quantum Groups book and I'm stuck on what looks like a pretty simple problem (ch3, ex 2 in the book):
Let $k$ be a field and $C= k[t]$ be a coalgebra with coproduct: $ \Delta (t^n) = \sum_{p+q = n} t^p \otimes t^q$
and product
$ t^n \cdot t^m = \binom{n+m}{n} t^{n+m} $
The book asks me to prove that this, together with the co/unit, form a bialgebra. However, if I try to calculate $\Delta(x)\Delta(y)$ with that product I get:
$ \Delta(t^n)\Delta(t^m) = \Large( \sum_{i+j=n} t^i\otimes t^j \Large) \Large( \sum_{k+l=m} t^k\otimes t^l \Large)$ $ = \sum_{i+j+k+l=n+m} \binom{i+k}{i}\binom{j+l}{j} t^{i+k} \otimes t^{j+l}$
Which does equal $\Delta(t^n \cdot t^m) = \binom{n+m}{n}\sum_{p+q=n+m} t^p\otimes t^q$
However, if I ignore Kassel's product with the binom coeff, then I have bialgebra without issue:
$ \Delta(t^n)\Delta(t^m) = \Large( \sum_{i+j=n} t^i\otimes t^j \Large) \Large( \sum_{k+l=m} t^k\otimes t^l \Large)$ $ = \sum_{i+j+k+l=n+m} t^{i+k} \otimes t^{j+l} =\Delta(t^{n+m})$
Clearly the book contains an error or I don't understand something. Probably the latter. Can you help me figure out what?