Let $Ell^*(X)$ be the elliptic cohomology theory (associated to a given elliptic curve $E$) of a nice space $X$.

Recall the Landweber-Ravenel-Stong construction: $MU^*(X) \otimes_{MU^*} R \simeq Ell^*(X)$, where $R \simeq Ell^*$.

How does the Landweber-Ravenel-Stong construction of $Ell^*(-)$ tell us how to actually compute $Ell^*(X)$ with the Atiyah-Hirzebruch spectral sequence?

Let $L$ be the Lazard ring, and $R$ is flat over $L$. It seems like the Landweber-Ravenel-Stong construction tells us that:

1. Given a map $L \to R$, $$MU^* \simeq L \longrightarrow R \simeq Ell^*$$ This gives us a map between their Atiyah-Hirzebruch SS? $$MU^*-AHSS(X) \longrightarrow Ell^*-AHSS(X)$$
2. The formal group law associated to our elliptic curve $\hat{E} \simeq \text{Spf } Ell^*(\mathbb{CP}^\infty)$.

It was told that $Ell^*(X)$ is computationally simpler than computing $MU^*(X)$. But we seem to need to know $MU^*(X)$ before we can compute $Ell^*(X)$, so I must be missing something!

Do we need to know the AHSS of $MU^*(X)$ before we can compute the AHSS of $Ell^*(X)$?

Let $Ell^*(X)$ be the elliptic cohomology theory (associated to a given elliptic curve $E$) of a nice space $X$.

Recall the Landweber-Ravenel-Stong construction: $MU^*(X) \otimes_{MU^*} R \simeq Ell^*(X)$, where $R \simeq Ell^*$.

How does the Landweber-Ravenel-Stong construction of $Ell^*(-)$ tell us how to actually compute $Ell^*(X)$ with the Atiyah-Hirzebruch spectral sequence?

Let $L$ be the Lazard ring, and $R$ is flat over $L$. It seems like the Landweber-Ravenel-Stong construction tells us that:

1. Given a map $L \to R$, $$MU^* \simeq L \longrightarrow R \simeq Ell^*$$ This gives us a map between their Atiyah-Hirzebruch SS? $$MU^*-AHSS(X) \longrightarrow Ell^*-AHSS(X)$$
2. The formal group law associated to our elliptic curve $\hat{E} \simeq \text{Spf } Ell^*(\mathbb{CP}^\infty)$.

It was told that $Ell^*(X)$ is computationally simpler than computing $MU^*(X)$. But we seem to need to know $MU^*(X)$ before we can compute $Ell^*(X)$, so I must be missing something!

Do we need to know the AHSS of $MU^*(X)$ before we can compute the AHSS of $Ell^*(X)$?

Edit wrt computing the AHSS of $Ell^*(X)$: It was explained to me that we don't need $MU^*(X)$, we need $Ell^*(pt)$ and $H^*(X)$ (plus differentials) to compute $Ell^*(X)$. We know $MU^*(pt)$ and thus our choice of elliptic curve (or better, it's ground ring) determines $Ell^*(pt)$.

Let $Ell^*(X)$ be the elliptic cohomology theory (associated to a given elliptic curve $E$) of a nice space $X$.

Recall the Landweber-Stong construction: $MU^*(X) \otimes_{MU^*} R \simeq Ell^*(X)$, where $R \simeq Ell^*$.

How does the Landweber-Stong construction of $Ell^*(-)$ tell us how to actually compute $Ell^*(X)$ with the Atiyah-Hirzebruch spectral sequence?

Let $L$ be the Lazard ring, and $R$ is flat over $L$. It seems like the Landweber-Stong construction tells us that:

1. Given a map $L \to R$, $$MU^* \simeq L \longrightarrow R \simeq Ell^*$$ This gives us a map between their Atiyah-Hirzebruch SS? $$MU^*-AHSS(X) \longrightarrow Ell^*-AHSS(X)$$
2. The formal group law associated to our elliptic curve $\hat{E} \simeq \text{Spf } Ell^*(\mathbb{CP}^\infty)$.

It was told that $Ell^*(X)$ is computationally simpler than computing $MU^*(X)$. But we seem to need to know $MU^*(X)$ before we can compute $Ell^*(X)$, so I must be missing something!

Do we need to know the AHSS of $MU^*(X)$ before we can compute the AHSS of $Ell^*(X)$?

Let $Ell^*(X)$ be the elliptic cohomology theory (associated to a given elliptic curve $E$) of a nice space $X$.

Recall the Landweber-Ravenel-Stong construction: $MU^*(X) \otimes_{MU^*} R \simeq Ell^*(X)$, where $R \simeq Ell^*$.

How does the Landweber-Ravenel-Stong construction of $Ell^*(-)$ tell us how to actually compute $Ell^*(X)$ with the Atiyah-Hirzebruch spectral sequence?

Let $L$ be the Lazard ring, and $R$ is flat over $L$. It seems like the Landweber-Ravenel-Stong construction tells us that:

1. Given a map $L \to R$, $$MU^* \simeq L \longrightarrow R \simeq Ell^*$$ This gives us a map between their Atiyah-Hirzebruch SS? $$MU^*-AHSS(X) \longrightarrow Ell^*-AHSS(X)$$
2. The formal group law associated to our elliptic curve $\hat{E} \simeq \text{Spf } Ell^*(\mathbb{CP}^\infty)$.

It was told that $Ell^*(X)$ is computationally simpler than computing $MU^*(X)$. But we seem to need to know $MU^*(X)$ before we can compute $Ell^*(X)$, so I must be missing something!

Do we need to know the AHSS of $MU^*(X)$ before we can compute the AHSS of $Ell^*(X)$?