Doesn't completion of a representation ring preserve its indecomposables? For $G = PSU(3)$, it is known that $\dim I(G;\mathbb Q) / I(G;\mathbb Q)^2 = 3$, while $H^{**}(BG;\mathbb Q)$ is obviously a power series ring in two indeterminates since $G$ has rank 2. This would would be fine except that there is an elementary argument the natural map should induce a bijection on spaces of indecomposable elements and $3\neq2$.
What goes wrong?
Here's the argument. Let $T$ be compact torus, $R(T)$ its representation ring, and $\widehat R(T)$ the completion with respect to the augmentation ideal $I(T)$. One has maps
$$R(T) \longrightarrow \widehat R(T) \overset\sim\longrightarrow K^*(BT) \overset{\textrm{ch}}\longrightarrow H^{**}(BT;\mathbb Q),$$
where the double-star insists we view the cohomology ring as the direct product of the $H^n$. If we tensor $R(T)$ with $\mathbb Q$ beforehand, 
we get
$$R(T;\mathbb Q) \longrightarrow \widehat R(T;\mathbb Q) \longrightarrow H^{**}(BT;\mathbb Q).$$
The last map sends a one-dimensional representation $t \in R(T)$
to $e^u$ for $u = c_1(ET \times_T \mathbb C_t)$,
and hence is an isomorphism, as both rings are power series rings on $\dim T$ indeterminates and $\log t$ is a power series in $t-1$.
Since $t-1 \mapsto e^u - 1 = u + u^2/2 + \cdots$, 
the map also sends $\hat I(T;\mathbb Q)$ to the augmentation ideal 
$H^{\geq 1}(BT;\mathbb Q)$ of $H^{**}(BG;\mathbb Q)$.
Now consider a compact, connected Lie group $G$ with maximal torus $T$. If we make the identification $BT = EG / T$, then $BT$ admits a right action of the Weyl group $W$ of $G$. The maps are equivariant with respect to the action of $w \in W$ since $t w$ is sent to $\exp c_1(ET \times_T \mathbb C_{t w})$ and $ET \times_T \mathbb C_{t w}$ is the pullback of $ET \times_T \mathbb C_{t}$ under $w\colon BT \to BT$. This equivariance and the standard isomorphisms 
$$R(T)^W = R(G),$$ 
$$\widehat R(T)^W = \widehat R(G),$$
$$H^{**}(BT;\mathbb Q)^W = H^{**}(BG;\mathbb Q)$$ then show
$\widehat R(G;\mathbb Q) \to H^{**}(BG;\mathbb Q)$
is also an isomorphism preserving the augmentation ideal.
Then, since $R(G)$ is Noetherian, the maps
$$R(G;\mathbb Q) \longrightarrow \widehat R(G;\mathbb Q) \overset\sim\longrightarrow H^{**}(BG;\mathbb Q)$$ 
induces isomorphisms 
$$\frac{I(G;\mathbb Q)\phantom{2}}{I(G;\mathbb Q)^2} \overset\sim\longrightarrow 
\frac{\hat I(G;\mathbb Q)\phantom{2}}{\hat I(G;\mathbb Q)^2} \overset\sim\longrightarrow
\frac{H^{\geq 1}(BG;\mathbb Q)\phantom{2}}{H^{\geq 1}(BG;\mathbb Q)^2}$$ 
of modules of indecomposables.
 A: In fact $\text{dim }I(G)/I(G)^2=2$. Let $x=[V_{3L_1}]$, $y=[V_{2L_1+L_2}]$ and $z=[V_{3L_1+3L_2}]$. Then 
\begin{eqnarray}R(PSU(3))\cong\mathbb{Z}[x, y, z]/(y^3-y^2-xz-2y(x+z)-x-y-z).\end{eqnarray}
Note that $\text{dim }V_{3L_1}=\text{dim }V_{3L_1+3L_2}=10$ and $\text{dim }V_{2L_1+L_2}=8$. If $\overline{x}=x-10$, $\overline{y}=y-8$, $\overline{z}=z-10$, then 
\begin{eqnarray}R(PSU(3))\cong\mathbb{Z}[\overline{x}, \overline{y}, \overline{z}]/(\overline{y}^3+23\overline{y}^2-2(\overline{x}+\overline{z})\overline{y}-\overline{x}\overline{z}-27(\overline{x}+\overline{z})+135\overline{y})).\end{eqnarray}
Note that $I(G)=(\overline{x}, \overline{y}, \overline{z})$. It follows that
\begin{eqnarray}I(G; \mathbb{Q})/I(G; \mathbb{Q})^2=\text{span}_\mathbb{Q}\{\overline{x}, \overline{y}, \overline{z}\}/\text{span}_\mathbb{Q}\{-27(\overline{x}+\overline{z})+135\overline{y}\}\end{eqnarray}
which is of dimension 2.
Remark: If we let 
\begin{cases}x&=X+1-2Y\\ y&=Y-1\\ z&=Z+1-2Y\end{cases}
we will get the more compact description
\begin{eqnarray}R(PSU(3))\cong\mathbb{Z}[X, Y, Z]/(Y^3-XZ)\end{eqnarray}
(see Lemma 7.1 of this paper).
Added: The following is the SAGE code which verifies the relation $y^3-y^2-xz-2y(x+z)-x-y-z=0$. A2(a, b) means the irreducible representation $V_{(a+b)L_1+bL_2}$.
sage: A2=WeylCharacterRing("A2", style="coroots")
sage: x=A2(3, 0)
sage: y=A2(1, 1)
sage: z=A2(0, 3)
sage: y^3-y^2-x*z-2*y*(x+z)-x-y-z
0

