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, 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, 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.

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 Algebra, 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, 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.

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 Algebra, p. 11, Fulton and Lang define $\Lambda^{\circ}(A):=\{1+a_1t+a_2t^2\ldots\mid a_i\in A_i\}$. 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 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 $1+a_1t+a_2t^2\ldots$ and $1+b_1t+b_2t^2\ldots$ are elements in $\Lambda^\circ(A)$, then their product using the product of $\Lambda(A)$ is given by $1+P_1(a_1;b_1)t+P_2(a_1,a_2;b_1,b_2)t^2+\ldots$, where $P_1,P_2,\ldots$ are certain universal polynomials. But $P_1(a_1;b_1)$ turns out to be $a_1b_1$ (see here, 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 it here.

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, 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, 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.

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 $A=\bigoplus_{i=o}^\infty A_i$. In Riemann-Roch Algebra, p. 11, Fulton and Lang define $\Lambda^{\circ}(A):=\{1+a_1t+a_2t^2\ldots\mid a_i\in A_i\}$. 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 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 $1+a_1t+a_2t^2\ldots$ and $1+b_1t+b_2t^2\ldots$ are elements in $\Lambda^\circ(A)$, then their product using the product of $\Lambda(A)$ is given by $1+P_1(a_1;b_1)t+P_2(a_1,a_2;b_1,b_2)t^2+\ldots$, where $P_1,P_2,\ldots$ are certain universal polynomials. But $P_1(a_1;b_1)$ turns out to be $a_1b_1$ (see here, 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 it here.

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 Algebra, p. 11, Fulton and Lang define $\Lambda^{\circ}(A):=\{1+a_1t+a_2t^2\ldots\mid a_i\in A_i\}$. 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 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 $1+a_1t+a_2t^2\ldots$ and $1+b_1t+b_2t^2\ldots$ are elements in $\Lambda^\circ(A)$, then their product using the product of $\Lambda(A)$ is given by $1+P_1(a_1;b_1)t+P_2(a_1,a_2;b_1,b_2)t^2+\ldots$, where $P_1,P_2,\ldots$ are certain universal polynomials. But $P_1(a_1;b_1)$ turns out to be $a_1b_1$ (see here, 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 it here.