We prove that $A_{R}(G)\otimes _{R}A_{R}(H)\cong A_{R}(G\times H)$ if $G$ and $H$ are Hausdorff ample groupoids. As part of the proof, we give a new universal property of Steinberg algebras. We then consider the isomorphism problem for tensor products of Leavitt algebras, and show that no diagonal-preserving isomorphism exists between $L_{2,R}\otimes L_{3,R}$ and $L_{2,R}\otimes L_{2,R}$ . In fact, there are no unexpected diagonal-preserving isomorphisms between tensor products of finitely many Leavitt algebras. We give an easy proof that every $\ast$ -isomorphism of Steinberg algebras over the integers preserves the diagonal, and it follows that $L_{2,\mathbb{Z}}\otimes L_{3,\mathbb{Z}}\not \cong L_{2,\mathbb{Z}}\otimes L_{2,\mathbb{Z}}$ (as $\ast$ -rings).

中文翻译：

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STEINBERG 代数的张量积

我们证明 $A_{R}(G)\otimes _{R}A_{R}(H)\cong A_{R}(G\times H)$ 如果 $G$ 和 $H$ 是 Hausdorff 充足群。作为证明的一部分，我们给出了 Steinberg 代数的一个新的普遍性质。然后我们考虑 Leavitt 代数的张量积的同构问题，并证明在 $L_{2,R}\otimes L_{3,R}$ 和 $L_{2,R}\otimes L_{2,R}$ . 事实上，在有限多个 Leavitt 代数的张量积之间没有意外的保对角同构。我们给出一个简单的证明，每个 $\ast$ - Steinberg 代数在整数上的同构保留了对角线，并且它遵循 $L_{2,\mathbb{Z}}\otimes L_{3,\mathbb{Z}}\not \cong L_{2,\mathbb{Z}}\otimes L_{2,\mathbb{Z}}$ （作为 $\ast$ -戒指）。