CR refers to *Methods of Representation Theory* by Charles Curtis and Irving Reiner.


Let $F$ be a finite extension of $\mathbb{Q}\_p$ with valuation ring $\mathcal{O}\_F$.
Let $G$ be a finite group and let $A$ be the group algebra $F[G]$.
Let $\Lambda$ be an $\mathcal{O}\_F$-order in $A$.
Let $K\_0(\Lambda)$ (resp. $K\_0(A)$) denote the Grothendieck group of the category of finitely generated projective left $\Lambda$-modules (resp. $A$-modules). 


Let $SK\_0(\Lambda)=\ker \varphi$ where $\varphi: K\_0(\Lambda) \longrightarrow K\_0(A)$ is the map given by 
$\varphi([M])=[F \otimes\_{\mathcal{O}\_F} M]$ (this definition is taken from CR vol 2, top of page 222.)


Is $SK\_0(\Lambda)$ always trivial? I know that this is true in the following cases:


* $\Lambda=\mathcal{O}\_{F}[G]$ - see CR vol 1, Theorem 32.1
* more generally, when $\Lambda$ satisfies the conditions of CR vol 1, Theorem 32.5
* $\Lambda$ is a maximal order - see CR vol 1, Theorem 26.24iii
* $\Lambda$ is commutative - see CR vol 1, Proposition 35.7


Can anyone provide (a reference to) a proof of the general case or counterexample?



Triviality of SK_0(Lambda) for Lambda an order in a group algebra over a $p$-adic field
No, the kernel of the map $\varphi: K\_0(\Lambda)\longrightarrow K\_0(A)$ is not necessarily trivial for all orders $\Lambda$ in $A$. In fact, it is only trivial for all orders in $A$ if $A$ is a direct sum of fields and skew fields. 


Assume the Wedderburn decomposition of $A$ has the matrix algebra $M\_n(D)$ as a simple component (with $n > 1$ and $D$ a skew field). Let $\Delta$ be the maximal order in $D$ and let $\Pi$ be its maximal ideal. Then define an order
$$
 \Lambda\_1 := \left[ \begin{array}{ccccc} \Delta & \Pi & \Pi& \cdots & \Pi \newline \Delta & \Delta & \Pi & \cdots & \Pi \newline \vdots & \vdots & \ddots & & \vdots \newline \Delta &\Delta &\Delta &\cdots & \Delta \end{array}\right] \subset M\_n(D)
$$ 
Note that $\Lambda\_1$ has $n$ non-isomorphic projective lattices (corresponding to the rows or, if you prefer left modules, the columns ). All of these become isomorphic to $D^{n}$ upon tensoring with the ground field. Thus the map $K\_0(\Lambda\_1)\cong \mathbb Z^n\longrightarrow K\_0(A)\cong \mathbb Z$ has non-trivial kernel. Now if $A$ decomposes as $A \cong M\_n(D)\oplus B$, then just choose an any order $\Lambda\_2 \subset B$ and you get a counterexample $\Lambda=\Lambda\_1\oplus \Lambda\_2 \subset A$.


Note that the choice of $\Lambda\_1$ above is arbitrary. Any order in $M\_n(D)$ with at least two non-isomorphic projective modules would do the trick.


