A module $B$ is flat if Tor $= 0$












1












$begingroup$


From Weibel's "An Introduction to Homological Algebra":



Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.



It seems to me that the obvious way to do this would be to use the definition:



Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.



We need to show that for an exact sequence:



$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,



$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.



Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.



Any help is appreciated.






homological-algebra exact-sequence







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
    $endgroup$
    – Arthur
    Jan 7 at 14:53












  • $begingroup$
    Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
    $endgroup$
    – Daven
    Jan 7 at 14:59






  • 1




    $begingroup$
    Check the horseshoe lemma, then the snake lemma.
    $endgroup$
    – Arthur
    Jan 7 at 15:01










  • $begingroup$
    That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
    $endgroup$
    – Daven
    Jan 7 at 15:05










  • $begingroup$
    It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
    $endgroup$
    – Arthur
    Jan 7 at 15:08


















1












$begingroup$


From Weibel's "An Introduction to Homological Algebra":



Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.



It seems to me that the obvious way to do this would be to use the definition:



Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.



We need to show that for an exact sequence:



$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,



$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.



Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.



Any help is appreciated.






homological-algebra exact-sequence







share|cite|improve this question









$endgroup$












  • $begingroup$
    I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
    $endgroup$
    – Arthur
    Jan 7 at 14:53












  • $begingroup$
    Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
    $endgroup$
    – Daven
    Jan 7 at 14:59






  • 1




    $begingroup$
    Check the horseshoe lemma, then the snake lemma.
    $endgroup$
    – Arthur
    Jan 7 at 15:01










  • $begingroup$
    That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
    $endgroup$
    – Daven
    Jan 7 at 15:05










  • $begingroup$
    It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
    $endgroup$
    – Arthur
    Jan 7 at 15:08
















1












1








1


1



$begingroup$


From Weibel's "An Introduction to Homological Algebra":



Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.



It seems to me that the obvious way to do this would be to use the definition:



Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.



We need to show that for an exact sequence:



$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,



$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.



Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.



Any help is appreciated.






homological-algebra exact-sequence







share|cite|improve this question









$endgroup$




From Weibel's "An Introduction to Homological Algebra":



Exercise 3.2.1: An $R$-module $B$ is flat if Tor$_i^R(A,B) = 0$ for every $R$-module A.



It seems to me that the obvious way to do this would be to use the definition:



Tor$_i^R(A,B) = $H$_i(P.otimes B)$ where $P.$ is a projective resolution of A.



We need to show that for an exact sequence:



$...rightarrow A_{n+1} rightarrow A_{n} rightarrow A_{n-1} rightarrow ...$ ,



$...rightarrow A_{n+1} otimes B rightarrow A_{n} otimes B rightarrow A_{n-1} otimes B rightarrow ...$ is exact.



Since we can only access exactness via the definition of Tor, it seems to me that we might have to construct a projective resolution around this point in the sequence, however I am currently struggling.



Any help is appreciated.






homological-algebra exact-sequence




homological-algebra exact-sequence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 14:40









DavenDaven

38319




38319












  • $begingroup$
    I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
    $endgroup$
    – Arthur
    Jan 7 at 14:53












  • $begingroup$
    Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
    $endgroup$
    – Daven
    Jan 7 at 14:59






  • 1




    $begingroup$
    Check the horseshoe lemma, then the snake lemma.
    $endgroup$
    – Arthur
    Jan 7 at 15:01










  • $begingroup$
    That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
    $endgroup$
    – Daven
    Jan 7 at 15:05










  • $begingroup$
    It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
    $endgroup$
    – Arthur
    Jan 7 at 15:08




















  • $begingroup$
    I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
    $endgroup$
    – Arthur
    Jan 7 at 14:53












  • $begingroup$
    Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
    $endgroup$
    – Daven
    Jan 7 at 14:59






  • 1




    $begingroup$
    Check the horseshoe lemma, then the snake lemma.
    $endgroup$
    – Arthur
    Jan 7 at 15:01










  • $begingroup$
    That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
    $endgroup$
    – Daven
    Jan 7 at 15:05










  • $begingroup$
    It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
    $endgroup$
    – Arthur
    Jan 7 at 15:08


















$begingroup$
I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
$endgroup$
– Arthur
Jan 7 at 14:53






$begingroup$
I think it's enough to show that for any short exact sequence $$0to A_1to A_2to A_3to 0$$ we have $$0to A_1otimes Bto A_2otimes Bto A_3otimes Bto 0$$exact. At least, that is the definition of "flat" that I know (I seem to recall that the translation between the two isn't all that difficult, and short exact sequences are often nicer to work with)..
$endgroup$
– Arthur
Jan 7 at 14:53














$begingroup$
Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
$endgroup$
– Daven
Jan 7 at 14:59




$begingroup$
Yeah you're right, we can work with short exact sequences. Although I still have a problem! Thanks
$endgroup$
– Daven
Jan 7 at 14:59




1




1




$begingroup$
Check the horseshoe lemma, then the snake lemma.
$endgroup$
– Arthur
Jan 7 at 15:01




$begingroup$
Check the horseshoe lemma, then the snake lemma.
$endgroup$
– Arthur
Jan 7 at 15:01












$begingroup$
That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
$endgroup$
– Daven
Jan 7 at 15:05




$begingroup$
That looks like it should do it! Every time I saw an instance of the question I asked it was always phrased in such a way that made it look like a nice simple question, however this proof is certainly not a simple rearranging of definitions.
$endgroup$
– Daven
Jan 7 at 15:05












$begingroup$
It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
$endgroup$
– Arthur
Jan 7 at 15:08






$begingroup$
It is possible that there are more elementary solutions, but if those lemmas are available to use why not use them?
$endgroup$
– Arthur
Jan 7 at 15:08












1 Answer
1






active

oldest

votes


















2












$begingroup$

We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is the kind of nice answer I was looking for, thanks so much!
    $endgroup$
    – Daven
    Jan 7 at 17:04










  • $begingroup$
    you are very welcome! one just has to love long exact sequences!
    $endgroup$
    – Enkidu
    Jan 8 at 8:06






  • 1




    $begingroup$
    @Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
    $endgroup$
    – Arthur
    Jan 8 at 11:44













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is the kind of nice answer I was looking for, thanks so much!
    $endgroup$
    – Daven
    Jan 7 at 17:04










  • $begingroup$
    you are very welcome! one just has to love long exact sequences!
    $endgroup$
    – Enkidu
    Jan 8 at 8:06






  • 1




    $begingroup$
    @Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
    $endgroup$
    – Arthur
    Jan 8 at 11:44


















2












$begingroup$

We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is the kind of nice answer I was looking for, thanks so much!
    $endgroup$
    – Daven
    Jan 7 at 17:04










  • $begingroup$
    you are very welcome! one just has to love long exact sequences!
    $endgroup$
    – Enkidu
    Jan 8 at 8:06






  • 1




    $begingroup$
    @Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
    $endgroup$
    – Arthur
    Jan 8 at 11:44
















2












2








2





$begingroup$

We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)






share|cite|improve this answer











$endgroup$



We want to show that $_ otimes B$ is exact. Now Lets take for this a short exact sequence $$0 to X to Y to Z to 0$$
Now lets apply the functors $mathrm{Tor}^i(_,B)$ to that sequence. Now this gives by the definition of $mathrm{Tor}^i(_,B)$ as a homological functor a long exact sequence $$ ... to mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) tomathrm{Tor}^0(X,B) to mathrm{Tor}^0(Y,B) to mathrm{Tor}^0(Z,B) to 0 $$
Now since we have a natural isomorphism $mathrm{Tor}^0(Z,B) cong Zotimes B$ we may rewrite the top sequence as:
$$ mathrm{Tor}^1(X,B) to mathrm{Tor}^1(Y,B) to mathrm{Tor}^1(Z,B) to Xotimes B to Yotimes B to Z otimes B to 0 $$
But since $ mathrm{Tor}^1(Z,B)=0$ this becomes:
$$0to Xotimes B to Yotimes B to Z otimes B to 0 $$
as desired.
(for the other direction of the implication just observe that if $B$ is flat, the projective resolution stays exact after tensoring and hence the higher $mathrm{Tor}$-terms vanish)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 8:06

























answered Jan 7 at 15:48









EnkiduEnkidu

1,36719




1,36719












  • $begingroup$
    This is the kind of nice answer I was looking for, thanks so much!
    $endgroup$
    – Daven
    Jan 7 at 17:04










  • $begingroup$
    you are very welcome! one just has to love long exact sequences!
    $endgroup$
    – Enkidu
    Jan 8 at 8:06






  • 1




    $begingroup$
    @Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
    $endgroup$
    – Arthur
    Jan 8 at 11:44




















  • $begingroup$
    This is the kind of nice answer I was looking for, thanks so much!
    $endgroup$
    – Daven
    Jan 7 at 17:04










  • $begingroup$
    you are very welcome! one just has to love long exact sequences!
    $endgroup$
    – Enkidu
    Jan 8 at 8:06






  • 1




    $begingroup$
    @Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
    $endgroup$
    – Arthur
    Jan 8 at 11:44


















$begingroup$
This is the kind of nice answer I was looking for, thanks so much!
$endgroup$
– Daven
Jan 7 at 17:04




$begingroup$
This is the kind of nice answer I was looking for, thanks so much!
$endgroup$
– Daven
Jan 7 at 17:04












$begingroup$
you are very welcome! one just has to love long exact sequences!
$endgroup$
– Enkidu
Jan 8 at 8:06




$begingroup$
you are very welcome! one just has to love long exact sequences!
$endgroup$
– Enkidu
Jan 8 at 8:06




1




1




$begingroup$
@Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
$endgroup$
– Arthur
Jan 8 at 11:44






$begingroup$
@Daven In some sense, the $operatorname{Tor}$ functor's purpose is to measure (an to some extent fix) the fact that tensoring isn't left-exact. So, when you have a module where tensoring in fact is left exact, $operatorname{Tor}$ gives you $0$. Similarily, the $operatorname{Ext}$ functor measures the non-exactness of $operatorname{hom}(B, -)$, in the other direction (and even for $operatorname{hom}(-, B)$, if you just remember to turn all the arrows the right way).
$endgroup$
– Arthur
Jan 8 at 11:44




















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Human spaceflight

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Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
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Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
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File:DeusFollowingSea.jpg

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File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
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