Bidegree completion of two symmetric-function rings

Let $\mathrm{\Lambda }(x)$ and $\mathrm{\Lambda }(y)$ be two copies of the graded ring The stable graded ring of symmetric functions, with degree pieces ${\mathrm{\Lambda }}^a(x)$ and ${\mathrm{\Lambda }}^b(y)$. Define the bidegree-completed tensor product by $\mathrm{\Lambda }(x)\widehat{\otimes }\mathrm{\Lambda }(y):={\prod }_{a,b\ge 0}({\mathrm{\Lambda }}^a(x){\otimes }_{\mathbb{Z}}{\mathrm{\Lambda }}^b(y))\mathrm{.}$

Addition is componentwise. If $u=(u_{a,b})$ and $v=(v_{a,b})$, their product has component $(uv{)}_{a,b}:={\sum }_{i=0}^a{\sum }_{j=0}^b{u}_{i,j}{v}_{a-i,b-j},$ where multiplication in each tensor factor is the graded multiplication of $\mathrm{\Lambda }$. This is a finite sum for each fixed $(a,b)$, so it defines a commutative ring; the unit has component $1\otimes 1$ at $(0,0)$ and zero elsewhere.

The Cauchy kernel is the element $\mathrm{\Omega }(x,y):={\prod }_{r,s\ge 1}(1-x_r{y}_s{)}^{-1}$ interpreted bidegree by bidegree: its degree-$(d,d)$ component is the stable degree-$d$ coefficient of the finite products, and all components $(a,b)$ with $a\mathrm{\ne }b$ are zero. Thus $\mathrm{\Omega }$ belongs to the diagonal bidegrees of this completion, not to the finite-support graded ring $\mathrm{\Lambda }$ itself.