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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.

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    $\begingroup$ Another mistake in Fulton-Lang's book is the assertion that the $j$-th Bott element of a positive element in a $\lambda$-ring $R$ is invertible in $R[1/j]$; see p. 24 (this is only true if $R$ is augmented with a locally nilpotent $\gamma$-filtration). $\endgroup$ Sep 30, 2012 at 7:59
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    $\begingroup$ @Damian: Thank you! Your comment gave me the idea to ask about other known errors in the book. $\endgroup$ Sep 30, 2012 at 14:38
  • $\begingroup$ This comment is very late, but I'm now reading this book and also want to find 'a list of errata' but don't. Do you think that if we modify the sum and product on page 6 so that the term P_k(...)t^k is changed to P_k(...)t^{2k}, and similarly the one with P_{k, j}(...)t^k to P_{k, j}(...)t^kj when we def operations in Λ∘(A)? $\endgroup$
    – Lao-tzu
    Feb 16, 2017 at 14:02
  • $\begingroup$ The link to "see here" has rotted. Do you remember to what it pointed? $\endgroup$
    – LSpice
    Mar 24, 2022 at 18:18
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    $\begingroup$ @LSpice Unfortunately I do not remember what it was! $\endgroup$ Mar 25, 2022 at 19:44

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As others have said, the definition of the Chern ring there is wrong. But if memory serves, the only mistake is that they forgot to introduce the right multiplication law on the sets of power series they consider. The usual one in the theory is given by the universal formulas for exterior powers of tensor products $\Lambda^n(E\otimes F)$, but the one they want is for Chern classes $c_n(E\otimes F)$. When $n=1$ and $E$ and $F$ are line bundles, the first is multiplication and the second is addition. So it's obviously just an oversight, but one that can be confusing if you're seeing these things for the first time. (In the copy at U Chicago, someone mercifully added a warning note in the margin. There are a few obvious suspects.)

If you want a reference where the details are correct, I'd recommend SGA6. Grothendieck's introduction in expose 0 is very clear. Page 28 is where the discussion of the Chern ring starts. If I remember, Berthelot's expose goes into more depth, but I found Grothendieck's easier to read. Berthelot gets to the Chern ring on page 344. Atiyah-Tall is also generally a good reference, but I think they don't cover the Chern ring (although they do introduce the gamma-filtration).

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  • $\begingroup$ Dear James: Are you saying that one can correct the issue by using a different set of universal polynomials to define multiplication in the graded ring that they consider? $\endgroup$ Sep 30, 2012 at 5:26
  • $\begingroup$ Yes, I was saying that and also that the universal polynomials are the ones that describe the Chern classes of tensor products. Note that the polynomials are not completely universal -- they do depend on the ranks of the factors. The usual way around this is to assume both factors have rank zero (which is why you get non-unital rings). $\endgroup$
    – JBorger
    Sep 30, 2012 at 6:09
  • $\begingroup$ What is the definition of universal as in universal polynomials? I mean I know how they are defined, but why are they called universal? Does it simply mean that the polynomials have integral coefficients or is it more to it? $\endgroup$ Oct 7, 2012 at 23:14
  • $\begingroup$ 'Universal' just emphasizes that the polynomial is independent of the ring. For instance, the Leibniz rule $d(xy)=xdy+ydx$ for derivations can be expressed as $d(xy)=P(x,y,dx,dy)$, where $P(a,b,c,d)=ad+bc$, and the polynomial $P$ is independent of the ring on which you have a derivation. You can imagine other kind of operators where $d(xy)$ is a polynomial in $x,y,dx,dy$ but that polynomial can depend on the ring. $\endgroup$
    – JBorger
    Oct 8, 2012 at 1:21
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Hazewinkel in Witt vectors. Part 1 warns about an error on page 15, second paragraph of this book. In fact he advises to "steer clear" of the book!

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  • $\begingroup$ For those who don't have easy access to Fulton-Lang: what is the error? $\endgroup$
    – Todd Trimble
    Sep 30, 2012 at 1:39
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    $\begingroup$ I see! He also mentions the book review by K. R. Coombes in Math. Rev. (88h:14011). I should read the review to see what Coombes wrote about the book. Thanks! $\endgroup$ Sep 30, 2012 at 2:07
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This is not an answer, as I don't exactly know what Fulton and Lang are trying to achieve with the $\lambda$-ring structure on $\Lambda^{\circ}\left(A\right)$ (I must admit that, while I had the quixotic intent to read and rewrite Fulton-Lang's Chapter I in the notes that you cited, I never found the resolve to walk that talk). I can confirm your counterexample.

What I think can be done (don't know if it is of any help) is the following: For every $i\in\mathbb N$, let $\Lambda^{i}_{\circ}\left(A\right)$ be the subset of $\Lambda\left(A\right)$ consisting of all formal power series of the form $1+a_1t+a_2t^2+a_3t^3+...$ with every $k$ satisfying $a_k\in A^{ik}$. Then, each such $\Lambda^{i} _ {\circ}\left(A\right)$ is an additive subgroup of $\Lambda\left(A\right)$, and the direct sum $\bigoplus\limits_{i\in\mathbb N}\Lambda^{i}_{\circ}\left(A\right)$ is well-defined and a sub-$\lambda$-ring of $\Lambda\left(A\right)$. (This is easy to prove by means of the usual grading on the ring of symmetric functions.) This sub-$\lambda$-ring, of course, is graded (and does have a $1$). I have no idea in how far it is what Fulton and Lang wanted.

We could also construct a greater graded sub-$\lambda$-ring of $\Lambda\left(A\right)$ by allowing $i$ rational (with $A^x$ defined as $0$ when $x\not\in\mathbb Z$), but then it will be graded by rationals. This greater graded sub-$\lambda$-ring is actually dense in $\Lambda\left(A\right)$ (in the usual topology on formal power series).

Does it make sense to replace $\Lambda^{\circ}\left(A\right)$ by $\Lambda^{\geq 1}_{\circ}\left(A\right)$ in the definition of a Chern class homomorphism? I don't know. It seems that most notions in Fulton-Lang are motivated by geometry, and without understanding it I am not the one to judge.

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  • $\begingroup$ Dear Darij: Thank you for your suggestion. Are you sure $1+t$ (the $1$ of $\Lambda(A)$) is in $\bigoplus_{i\in\mathbb{N}}\Lambda^i_\circ(A)$? In any case, I think Fulton & Lang wanted a $\lambda$-ring structure on $\Lambda^\circ(A)$ to prove a Splitting Principle for abstract Chern classes (Theorem 3.1), which in the geometric case is evident. But even with your suggestion, I am not sure if it is possible to fix the proof of Theorem 3.1, because I believe the proof has other issues, e.g., I think they mix up the notion of $\lambda$-homomorphism introduced on page 15 with $\lambda$-ring homomo. $\endgroup$ Sep 30, 2012 at 5:22
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    $\begingroup$ $1+t$ lies in $\Lambda^0_{\circ}\left(A\right)$, the $0$-th graded component of $\Lambda_{\circ}\left(A\right)$. Anyway, I guess James Borger understands better what Fulton and Lang were trying to say. $\endgroup$ Sep 30, 2012 at 14:53
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Just to save people some work, here are some problems with of Riemann-Roch Algebra pointed out by K. R. Coombes in his review on MathSciNet:

The beginner, however, may find the going rough at first. Chapters I and III in particular could have been written more carefully. Nowhere is there an unambiguous definition of "special'' λ-ring, even though the term is prominently introduced on p. 6. The reader should perhaps follow the authors' repeated advice to look at a paper by M. F. Atiyah and D. O. Tall [Topology 8 (1969), 253–297; MR0244387] for a "readable account", and also consult one of two papers by A. Grothendieck [Théorie des intersections et théorème de Riemann-Roch (SGA 6), Exposé 0, 1–19, Lecture Notes in Math., 225, Springer, Berlin, 1971; see MR0354655; Bull. Soc. Math. France 86 (1958), 137–154; MR0418782]. It is rarely made clear when the hypothesis "special" is used. It is not in the statement of Theorem I.2.1, for instance, but is in the proof. The statement of the graded splitting principle on p. 49 leaves out one of the main points, namely, that there exists an extension which splits a given element. That these inaccuracies can be removed by reference to other sources should not relieve the authors of this book, which explicitly addresses itself to beginners and claims to be elementary and self-contained, from an obligation to meet standards of exposition higher than those applied to an ordinary advanced monograph.

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I'm not sure what product you are thinking of on $\Lambda^0(A)$, but the one I'm thinking of, and the one that I believe is implicitly used in Fulton-Lang is the usual product on power series. So in particular, $(1+a_1 t+\cdots)\cdot (1+b_1 t+\cdots) = 1 + (a_1 + b_1)t + \cdots$.

There is no problem of grading.

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  • $\begingroup$ This product is not the product, but the sum of $\Lambda\left(A\right)$. $\endgroup$ Sep 29, 2012 at 17:04
  • $\begingroup$ As Darij wrote, in the λ-ring structure the ordinary product that you wrote above is considered the addition operation. So what you did here is you added the elements. Multiplication is given by universal polynomials $\endgroup$ Sep 29, 2012 at 17:07
  • $\begingroup$ Yes, you are right, sorry for this too prompt answer. $\endgroup$ Oct 4, 2012 at 4:51

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