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I am a beginner in this field. My question is

(1) Is the existence of $E_\infty$ ring structure not closed under weak equivalence of ring spectra?

(2) If (1) is true, what is the risk of replacing a ring spectrum by a weakly equivalent $E_\infty$ ring?

I am a little confused about the construction of $BP$ in Tilman Bauer's notes. According to section 3 of this note, the Brown-Peterson spectrum $BP$ can be constructed by taking the quotient of $MU_p$ by an ideal $I=ker(MU_* \to BP_*)$.

This quotient is given by theorem V.2.6. in Rings, Modules, and Algebras in Stable Homotopy Theory. By theorem V.3.2. in this book, this quotient ring $BP':=MU_p/I$ is an associated and commutative $MU$-algebras for $p>2$. Lemma VII.1.3. shows a commutative $MU$-algebras $BP'$ is a commutative $S$-algebra with a map $MU\to MP'$ of $S$-algebras. So lemma II.3.4. tells us $BP'$ is an $E_\infty$-ring which is also an $S$-algebra. In section V.4., $BP'$ is called a model of $BP$.

On the other hand, it is known that $BP$ is not $E_\infty$ for all primes $p$. So $BP'$ can not be exactly $BP$ in this sense.

I guess Tilman Bauer is just talking about "an $E_\infty$ model of $BP$" in his notes because this note is for a workshop of topological modular forms where we are interested in lifting commutative homotopy ring spectra to $E_\infty$ rings and the existence of $E_\infty$ ring structure is not closed under weak equivalence of ring spectra.

If this is the case, what is the risk of doing such replacement?

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    $\begingroup$ If I remember correctly, the EKMM book uses the term "$MU$-algebra" to mean an $A_\infty$ $MU$-algebra, while "associative $MU$-ring spectrum" is used in that book to mean something weaker, where associativity is only required up to homotopy. Something similar for "commutative $MU$-ring spectrum" vs. "commutative $MU$-algebra". The results V.2.6 and V.3.2 you refer to are only claiming you get $MU$-ring spectrum structures on $BP$, not $MU$-algebra structures--not quite what you wrote in your question. I think that's where your confusion is coming from. $\endgroup$
    – user509184
    Commented Feb 2 at 13:50
  • $\begingroup$ Thank you very much for your answer. I should have checked the definitions more carefully. $\endgroup$
    – Miso
    Commented Feb 2 at 14:21

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