Recall that a complex-oriented spectrum is a ring spectrum E with a map $MU \to E$.
Analogously, a ring with a (1-d commutative) formal group law is (represented by) a ring $R$ with a map $L \to R$ (where $L$ is the Lazard ring).
In the cases where $L \to R$ is Landweber exact, this can be made explicit:
However, the explicit lifting condition can't just be Landweber exactness. For example: the additive formal group law, though not Landweber-exact, corresponds to $H\mathbb{Z}$; the $n$th Honda formal group law, though not Landweber-exact, corresponds to $K(n)$.
My question is not "are there rings with (1-d commutative) formal group laws for which there is no corresponding map between ring spectra," but instead along the lines of "how do I build an explicit example."
Edit: A stupid explicit example of a formal group law that doesn't lift to a complex-orientable spectrum is any formal group law over $\mathbb{F}_p$ that is not isomorphic to the additive formal group law over $\mathbb{F}_p$. The only complex orientable spectra associated to formal group laws over $\mathbb{F}_p$ is $H\mathbb{F}_p$ (everything is concentrated in one degree). However, we can of course have a formal group law over $\mathbb{F}_p[[u_n, u_n^{-1}]]$ which lifts to a periodic ring spectrum.
What are explicit obstructions to realizability of formal group laws as complex-oriented ring spectra?
A vague guess is that the difference might come from the algebra of the ring $MU^*$ behaving differently from the 'homotopical' algebra of the ring spectrum $MU$, but I'm not sure how to proceed from this.
Best Answer
The other major class of cases where one can construct $E$ is when $R$ is a localised regular quotient of $L$, or in other words $R=L[S^{-1}]/I$, where the ideal $I$ can be generated by a regular sequence. There is a long history of general results of this type. I think that the sharpest versions are in my papers "Products on $MU$-modules" and "Realising formal groups" (although the methods used are not so different from earlier work). Note that the latter paper works with the periodic spectrum $MP=\bigvee_{n\in\mathbb{Z}}\Sigma^{2n}MU$ rather than $MU$ itself; this is more natural for many purposes. Note that in this context we focus on $\pi_0$ and note that $\pi_0(MP)=L$.
There is one more useful construction which is less often discussed in the literature. Choose generators $a_1,a_2,\dotsc$ for $L=\pi_0(MP)$. Let $A$ be the monoid (under multiplication) generated by these elements, and put $T=\Sigma^\infty_+A$, or equivalently $T=\bigvee_{a\in A}S^0$. Now $\pi_*(T)=\pi_*(S)\otimes L$, and there is an evident map $f\colon T\to MP$ of naive ring spectra, which is an isomorphism on $\pi_0$. If we want to make an $MP$-algebra $E$, we could hope to start by making a $T$-algebra $E'$, and then put $E=E'\wedge_TMP$. In particular, suppose that $I$ is an ideal in $L$ that is generated by some subset of monomials in the generators $a_i$. Then it is easy to construct $T/I$ as $\bigvee_{b\in B}S^0$ for a suitable subset $B\subseteq A$, and we can hope to construct $MP/I$ as $T/I\wedge_TMP$. In particular, we can choose lifts in $\pi_0(MP)$ of the chromatic generators $v_k$, and use these as some of the integral generators $a_i$; then we can form $MP/(v_1^2,v_1v_2,v_2^2)$, which is similar but not identical to the thing that @LennartMeier mentioned in his comment.
However, to make this work, we need more highly structured versions of $T$, $MP$ and $f$ (because there is no general construction of smash products of modules over unstructured ring spectra). We can construct strictly commutative versions of $T$ and $MP$, and also of $T/I$ when $I$ is generated by monomials, but unfortunately $f\colon T\to MP$ cannot be a map of strictly commutative rings, because $\pi_0(MP)$ has interesting power operations and $\pi_0(T)$ does not. There are some pitfalls with model category structures that can cause trouble here, and I have not checked all the details, but I think one can do the following, for example in the category of EKMM spectra.