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Johannes Hahn
Johannes Hahn
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When is a chain complex induced up up to quasiisomorphism

I have a field extension $F/F'$ and an algebra $A'$$L'$ over $F'$. Let $L$ be the induced up $F$-algebra $F\otimes_{F'}L'$ and $C_*$ a chain complex $C_*$ over $L$.

Is there a good way to decide whether $C_*$ is quasiisomorphic to a chain complex of the form $L\otimes_{L'}D_*$ for a chain complex $D_*$ over $L'$?

For example, if $L'=F'[x]$ and thus $L=F[x]$, and we look at a chain complex of the form $0\rightarrow F[x]\stackrel{x-f}{\rightarrow}F[x]\rightarrow 0$, we get that $H_0(C_*)$ is the $F[x]$-module $F_f$. This module is $F$ where $x$ acts as multiplication by $f$. If $C_*$ is quasiisomorpic to an induced up chain complex, then $H_0(C_*)$ would also be obtained from $H_0(D_*)$ by induction, and this is the case exactly iff $f\in F'\subset F$.   

This gives a list of obstructions, but are these the only ones?

When is a chain complex induced up up to quasiisomorphism

I have a field extension $F/F'$ and an algebra $A'$ over $F'$. Let $L$ be the induced up $F$-algebra $F\otimes_{F'}L'$ and a chain complex $C_*$ over $L$.

Is there a good way to decide whether $C_*$ is quasiisomorphic to a chain complex of the form $L\otimes_{L'}D_*$ for a chain complex $D_*$ over $L'$?

For example, if $L'=F'[x]$ and thus $L=F[x]$, and we look at a chain complex of the form $0\rightarrow F[x]\stackrel{x-f}{\rightarrow}F[x]\rightarrow 0$, we get that $H_0(C_*)$ is the $F[x]$-module $F_f$. This module is $F$ where $x$ acts as multiplication by $f$. If $C_*$ is quasiisomorpic to an induced up chain complex, then $H_0(C_*)$ would also be obtained from $H_0(D_*)$ by induction, and this is the case exactly iff $f\in F'\subset F$.  This gives a list of obstructions, but are these the only ones?

When is a chain complex induced up to quasiisomorphism

I have a field extension $F/F'$ and an algebra $L'$ over $F'$. Let $L$ be the induced $F$-algebra $F\otimes_{F'}L'$ and $C_*$ a chain complex over $L$.

Is there a good way to decide whether $C_*$ is quasiisomorphic to a chain complex of the form $L\otimes_{L'}D_*$ for a chain complex $D_*$ over $L'$?

For example, if $L'=F'[x]$ and thus $L=F[x]$, and we look at a chain complex of the form $0\rightarrow F[x]\stackrel{x-f}{\rightarrow}F[x]\rightarrow 0$, we get that $H_0(C_*)$ is the $F[x]$-module $F_f$. This module is $F$ where $x$ acts as multiplication by $f$. If $C_*$ is quasiisomorpic to an induced up chain complex, then $H_0(C_*)$ would also be obtained from $H_0(D_*)$ by induction, and this is the case exactly iff $f\in F'\subset F$. 

This gives a list of obstructions, but are these the only ones?

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HenrikRüping
HenrikRüping
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When is a chain complex induced up up to quasiisomorphism

I have a field extension $F/F'$ and an algebra $A'$ over $F'$. Let $L$ be the induced up $F$-algebra $F\otimes_{F'}L'$ and a chain complex $C_*$ over $L$.

Is there a good way to decide whether $C_*$ is quasiisomorphic to a chain complex of the form $L\otimes_{L'}D_*$ for a chain complex $D_*$ over $L'$?

For example, if $L'=F'[x]$ and thus $L=F[x]$, and we look at a chain complex of the form $0\rightarrow F[x]\stackrel{x-f}{\rightarrow}F[x]\rightarrow 0$, we get that $H_0(C_*)$ is the $F[x]$-module $F_f$. This module is $F$ where $x$ acts as multiplication by $f$. If $C_*$ is quasiisomorpic to an induced up chain complex, then $H_0(C_*)$ would also be obtained from $H_0(D_*)$ by induction, and this is the case exactly iff $f\in F'\subset F$. This gives a list of obstructions, but are these the only ones?