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This is related to this posting:

complex cobordism from formal group laws?

but not entirely the same. I'd like to know if there are any proofs of Quillen's theorem that $\pi_* (MU)$ is the Lazard ring (home of the universal formal group law) other than Quillen's. Specifically, is it possible to prove the result without a deep understanding of the structure of $H^*(MU)$ as a module over the Steenrod algebra?

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    $\begingroup$ I'm not sure I understand the question. I thought Quillen's proof (in his "Elementary proofs..." article) was notable in that it does not use the Steenrod algebra, Adams spectral sequences, and so on? $\endgroup$
    – Mark Grant
    Feb 20, 2012 at 8:13
  • $\begingroup$ Thanks, Mark, this is very helpful; I was unaware of this proof (and was really only thinking of Milnor's computation of $\pi_* MU$ using the Adams SS). $\endgroup$
    – Craig Westerland
    Feb 20, 2012 at 9:17

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Quillen's proof of this in the paper

Elementary proofs of some results of cobordism theory using Steenrod operations. Advances in Math. 7 1971 29–56 (1971)

does not make use of Adams spectral sequences or the structure of $H^\ast(MU)$ over the Steenrod algebra. In fact the only result from homotopy which is assumed is that the groups $MU^\ast(X)$ are finitely generated in each dimension.

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Jacob Lurie taught a course at Harvard in the spring of 2010 which included a slick proof of Quillen's theorem. I'm pretty sure Steenrod operations are not mentioned at all. The course page is here, and the notes which include this proof are in lecture 7. Also, the proof of Lazard's theorem (on the structure of the Lazard ring) in lectures 2 and 3 is very cool.

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    $\begingroup$ Hi David, thanks for linking to these notes! However, it seems what is proved in lecture 7 is that the map $L\to \pi_\ast(MU)$ classifying the formal group law of complex cobordism is a rational isomorphism (this is indeed very nice, using just the Hurewicz homomorphism). Then Adams spectral sequence arguments are used in subsequent lectures to prove the result at each prime. $\endgroup$
    – Mark Grant
    Feb 20, 2012 at 15:31

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