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:



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



Is $MGL$ an $H\mathbb{Z}$-algebra?
$MGL$ 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](http://folk.uio.no/paularne/rigidityma.pdf) 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](http://oai.cwi.nl/oai/asset/9960/9960A.pdf), 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).


