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\nIs the notion of stable equivalence the same as tensor equivalence for K-theory?
\nIs the notion of stable equivalence the same as tensor equivalence for K-theory?\nOf course it implies tensor equivalence, but is it equivalent?\nI am not sure I understand K-theory well enough for this question.
\nA:
\nYes, stable equivalence is equivalent to tensor (or descent) equivalence. Tensor equivalence of two $K$-theory spectra is equivalent to the category of $K$-theory spaces being a symmetric monoidal category (i.e. this is true in Morita equivalence, but also often in stable or descent equivalences). The essential content of an $E_{\\infty}$ ring spectrum is that it is a commutative monoid in the category of spectra with tensor product. For the extension of this to commutative ring spectra, one firstly has to extend the equivalence of categories of $K$-theory spectra to commutative ring spectra, which should be the case (I haven't checked!). It then follows from Brown representability (or something related) that a commutative monoid in the category of commutative ring spectra must be equivalently represented by a commutative ring spectrum, i.e. it is still the same as an $E_{\\infty}$ ring spectrum.\n(This is what I discussed in the comments with BenWithers.)
\nA:
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