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Let $MU$ be the unitary Thom spectrum, then it gives a generalized cohomology, so what is the coefficients $MU^*(point)$ like? Is it just the complex cobordism ring $\Omega_U^*?$

flag	asked Dec 2 2010 at 13:11 tiansong 121●5

Joe Johnson · Accepted Answer · 2010-12-02 13:50:32Z UTC

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It is complex cobordism. Therefore the coefficient ring is a polynomial ring on a generator in each positive even degree.

answered Dec 2 2010 at 13:50

Joe Johnson
414●6●12

It is also isomorphic to the Lazard ring, a universal one-dimensional formal group law. – Joe Johnson Dec 2 2010 at 13:53

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NB: As a *co*homology theory, the generators of the polynomial ring occur in negative cohomological degrees. – Tyler Lawson Dec 2 2010 at 16:00

what does the coefficients ring of generalized cohomology defined by the unitary Thom spectrum like?

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