Let $\mathrm{MGL}$ be the $\mathbb{P}^1$-ring spectrum over a field $k$ representing algebraic cobordism. Suppose, for simplicity, that $k$ is of characteristic 0. Let $H\mathbb{Z}$ be the motivic Eilenberg-Maclane spectrum. My question is:

Can $\mathrm{MGL}$ be given the structure of an $H\mathbb{Z}$-module?

One knows that $$ \mathrm{MGL}\wedge H\mathbb{Z}=H\mathbb{Z}[b_1,b_2,...]$$ where $b_i$ is of bidegree $(2i,i)$ in $\mathrm{MGL}_{*,*}(\mathrm{Spec}\,k)$. Thus, in view of the Morel-Hopkins isomorphism $$ \mathrm{MGL}/(b_1,b_2,...)\simeq H\mathbb{Z}$$ shown by Hoyois, is there a map $H\mathbb{Z}[b_1,b_2,...]\to\mathrm{MGL}$ in $SH(k)$. It seems to me that such a map would be an analog of Levine-Morel's generalized degree formula for $\mathrm{MGL}$.

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    $\begingroup$ Certainly not... For the same reason MU is not an HZ algebra. $\endgroup$ Commented Apr 1, 2016 at 14:45
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    $\begingroup$ MU is not an HZ-algebra because not all formal group laws are additive. I think that a similar approach will work for MGL $\endgroup$ Commented Apr 1, 2016 at 14:54
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    $\begingroup$ Even if a similar approach didn't work, $MGL$ being an $H\mathbb{Z}^{mot}$ algebra would imply that $MU$ was an $H\mathbb{Z}$ algebra (if one was working over $Spec(\mathbb{C})$). $\endgroup$ Commented Apr 1, 2016 at 15:12
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    $\begingroup$ The $b_i$'s in your first formula do not come from the homotopy groups of $MGL$. In particular, they are not the same guys that appear in the Hopkins-Morel iso! $\endgroup$ Commented Apr 1, 2016 at 16:58
  • $\begingroup$ @Marc Hoyois, is there any relation between the $b_i$s in the first equations and the ones in the Hopkins-Morel isomorphism? Do they belong to $MGL_{2i,i}$ and $\pi_{2i,i}(MGL)$ respectively? $\endgroup$ Commented Apr 2, 2016 at 4:09

1 Answer 1


$MGL$ does not admit a structure of $H\mathbb Z$-module. There are many ways to prove this. As Sean said in the comments, if it were true over $\mathbb C$, topological realization would imply that $MU$ is an $H\mathbb Z^{top}$-module, which is false. By rigidity this takes care of all characteristic zero fields. The same argument can presumably be made in positive characteristic using etale realization over a separable closure, but that would require computing the $l$-completed etale homotopy type of Thom spaces over Grassmannians which could be tricky.

Another argument is the following. If $MGL$ were an $H\mathbb Z$-algebra, $KGL$ would be too. Taking mapping spectra in $SH(k)$, this would imply that the $K$-theory spectrum $K(k)$ is an $H\mathbb Z^{top}$-module, which is false, as the first $k$-invariant of $K(k)$ is nontrivial (if the characteristic of $k$ is not $2$).

Here's an expansion of my comment about the $b_i$'s. Let's use $x_i$ for the elements in the Hopkins-Morel isomorphism. These are homogeneous generators of the Lazard ring $L$, and there is no canonical choice for them. The Hurewicz map $L \to \mathbb Z[b_1,b_2,...]$ is injective but not surjective, though it becomes surjective after tensoring with $\mathbb Q$. A complete formula for $x_i$ in terms of the $b_i$ depends of course on a specific choice of the generators $x_i$. You can see one such choice and the resulting formula in Hazewinkel Constructing formal groups II, equation (7.5.1), where $U_n$ is $x_{n-1}$ and $m_n(U)$ has degree $n-1$ and is determined by $\sum_{n\geq 1}m_n(U)(\sum_{i\geq 0}b_i)^n=1$. Usually one only cares about the image of $x_i$ modulo decomposable, which is $\pm pb_i$ if $i+1$ is a power of a prime $p$ and $\pm b_i$ otherwise (with Hazewinkel's choice of generators the signs are minus signs).

  • $\begingroup$ So using $K$-theory gets around the problem of only having Betti realization in characteristic 0? Neat! $\endgroup$ Commented Apr 4, 2016 at 10:00
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    $\begingroup$ Yeah, there's always an alternative realization functor $SH(k)\to SH$ given by the spectral enrichment of $SH(k)$. It's lax monoidal and sends $H\mathbb Z$ to $H\mathbb Z^{top}$ and $KGL$ to $K(k)$. I think it also sends $MGL$ to $H\mathbb Z^{top}$, actually. $\endgroup$ Commented Apr 4, 2016 at 14:19
  • $\begingroup$ Does this recover Betti realization when $k$ is $\mathbb{C}$? $\endgroup$ Commented Apr 4, 2016 at 16:21
  • $\begingroup$ No, $K(\mathbb C)\neq KU$. $\endgroup$ Commented Apr 4, 2016 at 23:11
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    $\begingroup$ Right, that's an interesting point. This doesn't hold in general: for $n>0$, $\pi_n\Sigma^{0,-n}(S^0_{\mathbb C})=W(\mathbb C)=\mathbb Z/2$ has nothing to do with $\pi_n(S^0)$. But I think it might hold for any slice-complete spectrum, at least if the slices have the form $\Sigma^{p,q}HA$ with $A$ an abelian group, because we know it for such slices. In particular it should hold for $MGL$ (my guess about the image of $MGL$ above was way off...). $\endgroup$ Commented Apr 5, 2016 at 15:12

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