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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 $\Omega_U^*?$

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

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It is also isomorphic to the Lazard ring, a universal one-dimensional formal group law. –  Joe Johnson Dec 2 '10 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 '10 at 16:00
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