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How weird can a ring spectrum be if all of its modules are free?
You're absolutely right, that's the better way to say this. But to get any control over how $x$ acts on $R/x$ from the left then, I think you need an $E_2$ structure. Otherwise you don't have any reason to expect left multiplication with $x$ to act nilpotently on $R/x$. (and I guess something like a free associative algebra on multiple generators shows that this does indeed occur)
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How weird can a ring spectrum be if all of its modules are free?
(I deleted the nonsensical comment since MO doesn't let you edit comments older than 5 minutes for some reason...)
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How weird can a ring spectrum be if all of its modules are free?
Ah wait, I mixed something up, sorry! The statement is that if $R/x$ itself admits a ring structure, then $x$ acts trivially on it. In general this can fail even if $R$ is $\mathbb{E}_\infty$, as $R=\mathbb{S}$, $x=2$ shows. What I mixed up is that you need at least $\mathbb{E}_2$ to even get a module structure on $R/x$. So my second comment is nonsense and the first applies to $\mathbb{E}_2$ and higher only.
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How weird can a ring spectrum be if all of its modules are free?
If any module over $R$ is free, then the cofiber $R/x$ is free for any $x\in\pi_*(R)$. If $x$ is not a unit, then $R/x$ is necessarily nontrivial. Now I am unable to come up with an example of $R$ and $x$ where $x$ is nonzero and $R/x$ is a nontrivial free module (I think this can't happen), but it would necessarily be weird. For example, note that $x^2$ acts trivially on $R/x$, thus also on the summand $R$, so your $R$ has the property that every element in homotopy is either a unit or has zero square.
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Does a non-singular matrix have a large minor with disjoint rows and columns and full rank?
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Is $\operatorname{Hom}(F,G)$ finite if $F$ and $G$ are endofunctors of the category of finite sets?
Is $\phi_k$ really a natural transformation? Doesn't $\phi_k(f(S)) = f(\phi_k(S))$ fail if $S$ has size bigger than $k$ and $f(S)$ has size smaller than $k$?
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The Eilenberg-MacLane spectrum and retractions
You can avoid $2$-localizing and quoting Wilson by observing that the third space in the spectrum for $ku$ is $SU$, and that $K(\mathbb{Z},3)$ can't be a retract of that because, as you said, there's torsion in the homology of $K(\mathbb{Z},3)$ but not in the one of $SU$.
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The Eilenberg-MacLane spectrum and retractions
I don't know what $\tau_2$ is, but the mod $2$ homology of $ku$ is definitely $0$. For example, it is known that its homology is, as a comodule algebra over the dual Steenrod algebra, given by $\mathbb{F}_2[\xi_1^2, \xi_2^2,\xi_3,\xi_4,\ldots]$, which doesn't have anything in degree $3$. Maybe you're thinking about the $\mathbb{Z}$-homology of $ku$?
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Space of algebraic closures of $\mathbb{Q}$
The collection of all algebraic closures naturally forms a groupoid, to get a "space of algebraic closures" you could take the nerve of that. Since the groupoid is connected, that is going to give you a $K(G_\mathbb{Q}, 1)$.
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Doing scheme theory with Hausdorff spaces
The constant functor from schemes to Hausdorff spaces that sends everything to the empty space is full and has a right adjoint which is the constant functor that sends everything to Spec $\mathbb{Z}$. Similarly, the constant functor with value the one-point space is full and has as left adjoint the constant functor with value the empty scheme. So I don't think "full + adjoint" is anywhere close to what you want.
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Integration over a Surface without using Partition of Unity
By the characteristic function of $V_n$ I just mean the function that is constantly 1 on $V_n$ and 0 outside of it. That's a partition of 1 if the $V_n$ cover the whole space and don't overlap.
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Integration over a Surface without using Partition of Unity
Isn't this a special case of a partition of unity, namely the characteristic functions of the $V_n$?
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Is the cohomology ring $H^*(BG,\mathbb{Z})$ generated by Euler classes?
Sadly, I don't know one, but I'm not really an expert in this, so there might very well exist a good one.
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homotopy and (co)filtered limits
No. For example, take all your $X_i$ to be the $1$-sphere, and all the maps the degree $2$ map. The limit is actually not pathconnected.
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