## Grothendieck group for projective space over the dual numbers

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Fix a field $k$. For a singular variety $X$, I understand that the Grothendieck group $K^0(X)$ of vector bundles on $X$ is not necessarily isomorphic to the Grothendieck group $K_0(X)$ of coherent sheaves on $X$.

I am curious to learn what is known about these two groups in one family of examples: $\mathbb P^n_{D}$, where $D$ is the dual numbers $D=k[\epsilon]/(\epsilon^2)$.

References would be especially appreciated, as I know very little about K-theory.

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## 2 Answers

If $X$ is a noetherian separated scheme and $X_{red}$ its reduction , we have $K_0(X)=K_o(X_{red})$: in other words $K_o$ doesn't see nilpotents .
Much more generally and profoundly, Quillen has proved that for all his $K$-theory groups, $K_i(X)=K_i(X_{red})$.
In your particular case you thus have (in the following $T$ is an indeterminate) $$K_0(\mathbb P^n_D)=K_0(\mathbb P^n_k)=\mathbb Z[T]/(T^{n+1})$$

As for $K^0$, a special case of a theorem of Berthelot (SGA 6, Exposé VI, Théorème 1.1, page 365) states that, for any commutative ring $A$, we have $K^0(\mathbb P^n_A)=K^o(A)[T]/(T^{n+1})$.
If $A=D=k[\epsilon]$, we have $K^0(D)=\mathbb Z$, since projective modules over local rings (like $D$) are free.
So here too $$K^0(\mathbb P^n_D)=\mathbb Z[T]/(T^{n+1})$$

Bibliography
Srinivas has written this nice book on $K$-theory.

And as an homage to the recently sadly departed Daniel Quillen, let me refer to his groundbreaking paper
"Higher algebraic $K$-theory I", published in Springer's Lecture Notes LNM 341.

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Thank you for the thoughtful answer and the helpful references. – Daniel Erman Oct 4 2011 at 17:07

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The identity map from $k$ to itself factors through $D$. Thus, if $K$ represents either $K^0$ or $K_0$, $K(\mathbb P^n_{k})$ is a direct summand of $K(\mathbb P^n_{D})$.

For $M$ a coherent $\mathbb P^n_{D}$-module, we have an exact sequence

$$0\rightarrow \epsilon M \rightarrow M \rightarrow M/\epsilon M\rightarrow 0$$

The classes of the modules on the left and right, and therefore the class of the module in the middle, all come from the $K_0(\mathbb P^n_{k})$ direct summand. This shows that $K_0(\mathbb P^n_{D})=K_0(\mathbb P^n_{k})$.

For $K^0$, you can apply Nakayama's Lemma.

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Thank you for this excellent answer. – Daniel Erman Oct 4 2011 at 17:07
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