Given a commutative ring $A$ with unity, Grothendieck used universal polynomials to define a *special* $\lambda$-ring structure on $\Lambda(A):=1+t\:A[[t]]$. Suppose $A$ is graded, say $A=\bigoplus_{i=0}^\infty A_i$. In [**Riemann–Roch Algebras**](https://doi.org/10.1007/978-1-4757-1858-4), p. 11, Fulton and Lang define $\Lambda^{\circ}(A):=\{1+a_1t+a_2t^2\dotsb\mid a_i\in A_i\}$. Then on page 15 they state that since the product and $\lambda$ operations of $\Lambda(A)$ take $\Lambda^\circ(A)$ to itself, $\Lambda^\circ(A)$ becomes a $\lambda$-ring (without unit). They use this $\lambda$-ring structure of $\Lambda^\circ(A)$ in the proof of Theorem 3.1 on p. 16.

However, a straightforward computation shows that the product in $\Lambda(A)$ does *not* take $\Lambda^\circ(A)$ to itself. For example, if $1+a_1t+a_2t^2\dotsb$ and $1+b_1t+b_2t^2\dotsb$ are elements in $\Lambda^\circ(A)$, then their product using the product of $\Lambda(A)$ is given by $1+P_1(a_1;b_1)t+P_2(a_1,a_2;b_1,b_2)t^2+\dotsb$, where $P_1,P_2,\dotsc$ are certain universal polynomials. But $P_1(a_1;b_1)$ turns out to be $a_1b_1$ ([see here][1], p. 22) and $a_1b_1$ is not in $A_1$, which shows the product is not in $\Lambda^\circ(A)$.

**Question.** Is there an error in the book? If yes, can it be fixed? 

**Edit.** If you know other errors in this book that one should be aware of, please share them here.


  [1]: https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxkYXJpamdyaW5iZXJnfGd4OjEwNzljZThlNDcwMGE5YmU