Number of homomorphisms from a stem field to a given field The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Field extensions and irreducible polynomial questionNormal field extension implies splitting fieldHow to find the degree of an extension field?Understanding the connection of the roots of an irreducible polynomial and a basis for field extensionsWhy is $mathbbQ(sqrt[3]2)$ not a stem field for the polynomial $X^3-2 in mathbbQ[X]$?The general form of (algebraic) number field?Number of homomorphisms from a number field to $mathbbR$Degree of a field extension $K/mathbbQ$How to determine whether the images of two real embeddings of a number field are equal?Number of ring homomorphisms from a finite field to its extension
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Number of homomorphisms from a stem field to a given field
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Field extensions and irreducible polynomial questionNormal field extension implies splitting fieldHow to find the degree of an extension field?Understanding the connection of the roots of an irreducible polynomial and a basis for field extensionsWhy is $mathbbQ(sqrt[3]2)$ not a stem field for the polynomial $X^3-2 in mathbbQ[X]$?The general form of (algebraic) number field?Number of homomorphisms from a number field to $mathbbR$Degree of a field extension $K/mathbbQ$How to determine whether the images of two real embeddings of a number field are equal?Number of ring homomorphisms from a finite field to its extension
$begingroup$
This is a homework, but I've generalized it as possible in order not to have exact answer rather that to understand the very principle of solution.
The problem is following: consider $mathbbK$ a field and $E$ an extension field of $mathbbK$. For a given irreducible polynomial $P(x)$ from the ring $mathbb K[x]$ find the number of homomorphisms from the stem field for $P(x)$ to the field $E$.
A stem field for an irreducible polynomial $P$ in $mathbbK[x]$ is a pair $(F,alpha)$, where $alpha$ is a root of $P$ and $F$ is an extension of $mathbbK$, i.e. $F = mathbbK[alpha]$ and $P(alpha)$=0
My understanding is following:
Any stem field $F$ is isomorphic to $
dfracmathbb K[x](P(x))$The number of homomorphisms from $dfracmathbb K[x](P(x))$ to $E$ is equal to the number of roots of this particular polynomial in $E$.
Example: if $mathbbK = mathbbQ$ and $P(x$) has $n$ roots in $mathbb R$ (real roots) and $m$ complex (strictly non-real) roots, then the number of homomorphisms to $mathbb R$ is n and number of homomorphisms to $mathbb C$ is n+m.
Is my understanding correct at all? If not, can you give me a hint in what direction I should look for.
field-theory extension-field
edited Mar 31 at 13:47
polettix
17810
asked Mar 6 '16 at 1:28
shabuncshabunc
1486
$endgroup$
|
show 1 more comment
$begingroup$
This is a homework, but I've generalized it as possible in order not to have exact answer rather that to understand the very principle of solution.
The problem is following: consider $mathbbK$ a field and $E$ an extension field of $mathbbK$. For a given irreducible polynomial $P(x)$ from the ring $mathbb K[x]$ find the number of homomorphisms from the stem field for $P(x)$ to the field $E$.
A stem field for an irreducible polynomial $P$ in $mathbbK[x]$ is a pair $(F,alpha)$, where $alpha$ is a root of $P$ and $F$ is an extension of $mathbbK$, i.e. $F = mathbbK[alpha]$ and $P(alpha)$=0
My understanding is following:
Any stem field $F$ is isomorphic to $
dfracmathbb K[x](P(x))$The number of homomorphisms from $dfracmathbb K[x](P(x))$ to $E$ is equal to the number of roots of this particular polynomial in $E$.
Example: if $mathbbK = mathbbQ$ and $P(x$) has $n$ roots in $mathbb R$ (real roots) and $m$ complex (strictly non-real) roots, then the number of homomorphisms to $mathbb R$ is n and number of homomorphisms to $mathbb C$ is n+m.
Is my understanding correct at all? If not, can you give me a hint in what direction I should look for.
field-theory extension-field
edited Mar 31 at 13:47
polettix
17810
asked Mar 6 '16 at 1:28
shabuncshabunc
1486
$endgroup$
5
$begingroup$
This is my first time hearing of the term "stem field" out of the context of "Science, Engineering, Technology and Mathematics". That's certainly an interesting word choice.
$endgroup$
– Cameron Williams
Mar 6 '16 at 1:37
3
$begingroup$
@CameronWilliams I've found this particular term for instance, in this book - jmilne.org/math/CourseNotes/FT.pdf - if this term is not clear for English speaker - just let me know, and I'll add the definition of stem field to the question. Or if there's more appropriate term I will use it instead )
$endgroup$
– shabunc
Mar 6 '16 at 1:41
1
$begingroup$
This is the second time in less than 8 hours that the term "stem field" appears in the site. The first time was from a finnish user, who said it was close to the term in finnish language but actually means "splitting field" . I couldn't find the term in Milne's book.
$endgroup$
– DonAntonio
Mar 6 '16 at 6:20
2
$begingroup$
The term is mentioned in Milne's Fields and Galois Theory book and he in turn attributes it along with term root field(i.e. splitting field) to M.Albert's Modern Higher Algebra book from 1937. I guess its an arcane term not much in vogue now.
$endgroup$
– Vishesh
Mar 6 '16 at 7:39
1
$begingroup$
I have edited the question a bit and added the definition of a stem field. Are you referring to $E$ as an extension field of $mathbbK$ or is it arbitrary? I am not sure.
$endgroup$
– Vishesh
Mar 6 '16 at 7:59
|
show 1 more comment
$begingroup$
This is a homework, but I've generalized it as possible in order not to have exact answer rather that to understand the very principle of solution.
The problem is following: consider $mathbbK$ a field and $E$ an extension field of $mathbbK$. For a given irreducible polynomial $P(x)$ from the ring $mathbb K[x]$ find the number of homomorphisms from the stem field for $P(x)$ to the field $E$.
A stem field for an irreducible polynomial $P$ in $mathbbK[x]$ is a pair $(F,alpha)$, where $alpha$ is a root of $P$ and $F$ is an extension of $mathbbK$, i.e. $F = mathbbK[alpha]$ and $P(alpha)$=0
My understanding is following:
Any stem field $F$ is isomorphic to $
dfracmathbb K[x](P(x))$The number of homomorphisms from $dfracmathbb K[x](P(x))$ to $E$ is equal to the number of roots of this particular polynomial in $E$.
Example: if $mathbbK = mathbbQ$ and $P(x$) has $n$ roots in $mathbb R$ (real roots) and $m$ complex (strictly non-real) roots, then the number of homomorphisms to $mathbb R$ is n and number of homomorphisms to $mathbb C$ is n+m.
Is my understanding correct at all? If not, can you give me a hint in what direction I should look for.
field-theory extension-field
edited Mar 31 at 13:47
polettix
17810
asked Mar 6 '16 at 1:28
shabuncshabunc
1486
$endgroup$
This is a homework, but I've generalized it as possible in order not to have exact answer rather that to understand the very principle of solution.
The problem is following: consider $mathbbK$ a field and $E$ an extension field of $mathbbK$. For a given irreducible polynomial $P(x)$ from the ring $mathbb K[x]$ find the number of homomorphisms from the stem field for $P(x)$ to the field $E$.
A stem field for an irreducible polynomial $P$ in $mathbbK[x]$ is a pair $(F,alpha)$, where $alpha$ is a root of $P$ and $F$ is an extension of $mathbbK$, i.e. $F = mathbbK[alpha]$ and $P(alpha)$=0
My understanding is following:
Any stem field $F$ is isomorphic to $
dfracmathbb K[x](P(x))$The number of homomorphisms from $dfracmathbb K[x](P(x))$ to $E$ is equal to the number of roots of this particular polynomial in $E$.
Example: if $mathbbK = mathbbQ$ and $P(x$) has $n$ roots in $mathbb R$ (real roots) and $m$ complex (strictly non-real) roots, then the number of homomorphisms to $mathbb R$ is n and number of homomorphisms to $mathbb C$ is n+m.
Is my understanding correct at all? If not, can you give me a hint in what direction I should look for.
field-theory extension-field
field-theory extension-field
edited Mar 31 at 13:47
polettix
17810
asked Mar 6 '16 at 1:28
shabuncshabunc
1486
edited Mar 31 at 13:47
polettix
17810
asked Mar 6 '16 at 1:28
shabuncshabunc
1486
edited Mar 31 at 13:47
polettix
17810
edited Mar 31 at 13:47
polettix
17810
edited Mar 31 at 13:47
polettix
17810
17810
asked Mar 6 '16 at 1:28
shabuncshabunc
1486
asked Mar 6 '16 at 1:28
shabuncshabunc
1486
asked Mar 6 '16 at 1:28
shabuncshabunc
1486
1486
5
$begingroup$
This is my first time hearing of the term "stem field" out of the context of "Science, Engineering, Technology and Mathematics". That's certainly an interesting word choice.
$endgroup$
– Cameron Williams
Mar 6 '16 at 1:37
3
$begingroup$
@CameronWilliams I've found this particular term for instance, in this book - jmilne.org/math/CourseNotes/FT.pdf - if this term is not clear for English speaker - just let me know, and I'll add the definition of stem field to the question. Or if there's more appropriate term I will use it instead )
$endgroup$
– shabunc
Mar 6 '16 at 1:41
1
$begingroup$
This is the second time in less than 8 hours that the term "stem field" appears in the site. The first time was from a finnish user, who said it was close to the term in finnish language but actually means "splitting field" . I couldn't find the term in Milne's book.
$endgroup$
– DonAntonio
Mar 6 '16 at 6:20
2
$begingroup$
The term is mentioned in Milne's Fields and Galois Theory book and he in turn attributes it along with term root field(i.e. splitting field) to M.Albert's Modern Higher Algebra book from 1937. I guess its an arcane term not much in vogue now.
$endgroup$
– Vishesh
Mar 6 '16 at 7:39
1
$begingroup$
I have edited the question a bit and added the definition of a stem field. Are you referring to $E$ as an extension field of $mathbbK$ or is it arbitrary? I am not sure.
$endgroup$
– Vishesh
Mar 6 '16 at 7:59
|
show 1 more comment
5
$begingroup$
This is my first time hearing of the term "stem field" out of the context of "Science, Engineering, Technology and Mathematics". That's certainly an interesting word choice.
$endgroup$
– Cameron Williams
Mar 6 '16 at 1:37
3
$begingroup$
@CameronWilliams I've found this particular term for instance, in this book - jmilne.org/math/CourseNotes/FT.pdf - if this term is not clear for English speaker - just let me know, and I'll add the definition of stem field to the question. Or if there's more appropriate term I will use it instead )
$endgroup$
– shabunc
Mar 6 '16 at 1:41
1
$begingroup$
This is the second time in less than 8 hours that the term "stem field" appears in the site. The first time was from a finnish user, who said it was close to the term in finnish language but actually means "splitting field" . I couldn't find the term in Milne's book.
$endgroup$
– DonAntonio
Mar 6 '16 at 6:20
2
$begingroup$
The term is mentioned in Milne's Fields and Galois Theory book and he in turn attributes it along with term root field(i.e. splitting field) to M.Albert's Modern Higher Algebra book from 1937. I guess its an arcane term not much in vogue now.
$endgroup$
– Vishesh
Mar 6 '16 at 7:39
1
$begingroup$
I have edited the question a bit and added the definition of a stem field. Are you referring to $E$ as an extension field of $mathbbK$ or is it arbitrary? I am not sure.
$endgroup$
– Vishesh
Mar 6 '16 at 7:59
5
5
$begingroup$
This is my first time hearing of the term "stem field" out of the context of "Science, Engineering, Technology and Mathematics". That's certainly an interesting word choice.
$endgroup$
– Cameron Williams
Mar 6 '16 at 1:37
$begingroup$
This is my first time hearing of the term "stem field" out of the context of "Science, Engineering, Technology and Mathematics". That's certainly an interesting word choice.
$endgroup$
– Cameron Williams
Mar 6 '16 at 1:37
3
3
$begingroup$
@CameronWilliams I've found this particular term for instance, in this book - jmilne.org/math/CourseNotes/FT.pdf - if this term is not clear for English speaker - just let me know, and I'll add the definition of stem field to the question. Or if there's more appropriate term I will use it instead )
$endgroup$
– shabunc
Mar 6 '16 at 1:41
$begingroup$
@CameronWilliams I've found this particular term for instance, in this book - jmilne.org/math/CourseNotes/FT.pdf - if this term is not clear for English speaker - just let me know, and I'll add the definition of stem field to the question. Or if there's more appropriate term I will use it instead )
$endgroup$
– shabunc
Mar 6 '16 at 1:41
1
1
$begingroup$
This is the second time in less than 8 hours that the term "stem field" appears in the site. The first time was from a finnish user, who said it was close to the term in finnish language but actually means "splitting field" . I couldn't find the term in Milne's book.
$endgroup$
– DonAntonio
Mar 6 '16 at 6:20
$begingroup$
This is the second time in less than 8 hours that the term "stem field" appears in the site. The first time was from a finnish user, who said it was close to the term in finnish language but actually means "splitting field" . I couldn't find the term in Milne's book.
$endgroup$
– DonAntonio
Mar 6 '16 at 6:20
2
2
$begingroup$
The term is mentioned in Milne's Fields and Galois Theory book and he in turn attributes it along with term root field(i.e. splitting field) to M.Albert's Modern Higher Algebra book from 1937. I guess its an arcane term not much in vogue now.
$endgroup$
– Vishesh
Mar 6 '16 at 7:39
$begingroup$
The term is mentioned in Milne's Fields and Galois Theory book and he in turn attributes it along with term root field(i.e. splitting field) to M.Albert's Modern Higher Algebra book from 1937. I guess its an arcane term not much in vogue now.
$endgroup$
– Vishesh
Mar 6 '16 at 7:39
1
1
$begingroup$
I have edited the question a bit and added the definition of a stem field. Are you referring to $E$ as an extension field of $mathbbK$ or is it arbitrary? I am not sure.
$endgroup$
– Vishesh
Mar 6 '16 at 7:59
$begingroup$
I have edited the question a bit and added the definition of a stem field. Are you referring to $E$ as an extension field of $mathbbK$ or is it arbitrary? I am not sure.
$endgroup$
– Vishesh
Mar 6 '16 at 7:59
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Your understanding is correct, except for one important assumption you've left out. What you say is correct provided that $mathbbK$ is a subfield of $E$ and you are only considering homomorphisms $Fto E$ which are the identity of $mathbbK$. In general, to define a homomorphism $mathbbK[x]/(P(x))to E$, you have to first choose a homomorphism $varphi:mathbbKto E$, and then choose an element of $E$ which is a root of the polynomial obtained by applying $varphi$ to the coefficients of $P$. When you require $varphi$ to be the inclusion map, this means your only choice is an element of $E$ which is a root of $P$. But if you do not assume this, there may be many more homomorphisms $mathbbK[x]/(P(x))to E$ that do something else on $mathbbK$.
answered Dec 20 '17 at 0:24
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
Thanks for pointing out. Milne (in jmilne.org/math/CourseNotes/FT.pdf section on "Stem Fields" in chapter 1) actually talks about homomorphism of F-algebras (in this case, $mathbbK$-algebras) that is defined as a homomorphism whose restriction on F is the identity (see section "Fields" in chapter 1). Similarly, I guess the homework comes from a Coursera course on Galois Theory, and even there the isomorphism in bullet 1 of the original post is "of $mathbbK$-algebras".
$endgroup$
– polettix
Apr 2 at 21:44
add a comment |
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$begingroup$
Your understanding is correct, except for one important assumption you've left out. What you say is correct provided that $mathbbK$ is a subfield of $E$ and you are only considering homomorphisms $Fto E$ which are the identity of $mathbbK$. In general, to define a homomorphism $mathbbK[x]/(P(x))to E$, you have to first choose a homomorphism $varphi:mathbbKto E$, and then choose an element of $E$ which is a root of the polynomial obtained by applying $varphi$ to the coefficients of $P$. When you require $varphi$ to be the inclusion map, this means your only choice is an element of $E$ which is a root of $P$. But if you do not assume this, there may be many more homomorphisms $mathbbK[x]/(P(x))to E$ that do something else on $mathbbK$.
answered Dec 20 '17 at 0:24
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
Thanks for pointing out. Milne (in jmilne.org/math/CourseNotes/FT.pdf section on "Stem Fields" in chapter 1) actually talks about homomorphism of F-algebras (in this case, $mathbbK$-algebras) that is defined as a homomorphism whose restriction on F is the identity (see section "Fields" in chapter 1). Similarly, I guess the homework comes from a Coursera course on Galois Theory, and even there the isomorphism in bullet 1 of the original post is "of $mathbbK$-algebras".
$endgroup$
– polettix
Apr 2 at 21:44
add a comment |
$begingroup$
Your understanding is correct, except for one important assumption you've left out. What you say is correct provided that $mathbbK$ is a subfield of $E$ and you are only considering homomorphisms $Fto E$ which are the identity of $mathbbK$. In general, to define a homomorphism $mathbbK[x]/(P(x))to E$, you have to first choose a homomorphism $varphi:mathbbKto E$, and then choose an element of $E$ which is a root of the polynomial obtained by applying $varphi$ to the coefficients of $P$. When you require $varphi$ to be the inclusion map, this means your only choice is an element of $E$ which is a root of $P$. But if you do not assume this, there may be many more homomorphisms $mathbbK[x]/(P(x))to E$ that do something else on $mathbbK$.
answered Dec 20 '17 at 0:24
Eric WofseyEric Wofsey
193k14221352
$endgroup$
$begingroup$
Thanks for pointing out. Milne (in jmilne.org/math/CourseNotes/FT.pdf section on "Stem Fields" in chapter 1) actually talks about homomorphism of F-algebras (in this case, $mathbbK$-algebras) that is defined as a homomorphism whose restriction on F is the identity (see section "Fields" in chapter 1). Similarly, I guess the homework comes from a Coursera course on Galois Theory, and even there the isomorphism in bullet 1 of the original post is "of $mathbbK$-algebras".
$endgroup$
– polettix
Apr 2 at 21:44
add a comment |
$begingroup$
Your understanding is correct, except for one important assumption you've left out. What you say is correct provided that $mathbbK$ is a subfield of $E$ and you are only considering homomorphisms $Fto E$ which are the identity of $mathbbK$. In general, to define a homomorphism $mathbbK[x]/(P(x))to E$, you have to first choose a homomorphism $varphi:mathbbKto E$, and then choose an element of $E$ which is a root of the polynomial obtained by applying $varphi$ to the coefficients of $P$. When you require $varphi$ to be the inclusion map, this means your only choice is an element of $E$ which is a root of $P$. But if you do not assume this, there may be many more homomorphisms $mathbbK[x]/(P(x))to E$ that do something else on $mathbbK$.
answered Dec 20 '17 at 0:24
Eric WofseyEric Wofsey
193k14221352
$endgroup$
Your understanding is correct, except for one important assumption you've left out. What you say is correct provided that $mathbbK$ is a subfield of $E$ and you are only considering homomorphisms $Fto E$ which are the identity of $mathbbK$. In general, to define a homomorphism $mathbbK[x]/(P(x))to E$, you have to first choose a homomorphism $varphi:mathbbKto E$, and then choose an element of $E$ which is a root of the polynomial obtained by applying $varphi$ to the coefficients of $P$. When you require $varphi$ to be the inclusion map, this means your only choice is an element of $E$ which is a root of $P$. But if you do not assume this, there may be many more homomorphisms $mathbbK[x]/(P(x))to E$ that do something else on $mathbbK$.
answered Dec 20 '17 at 0:24
Eric WofseyEric Wofsey
193k14221352
answered Dec 20 '17 at 0:24
Eric WofseyEric Wofsey
193k14221352
answered Dec 20 '17 at 0:24
Eric WofseyEric Wofsey
193k14221352
answered Dec 20 '17 at 0:24
Eric WofseyEric Wofsey
193k14221352
193k14221352
$begingroup$
Thanks for pointing out. Milne (in jmilne.org/math/CourseNotes/FT.pdf section on "Stem Fields" in chapter 1) actually talks about homomorphism of F-algebras (in this case, $mathbbK$-algebras) that is defined as a homomorphism whose restriction on F is the identity (see section "Fields" in chapter 1). Similarly, I guess the homework comes from a Coursera course on Galois Theory, and even there the isomorphism in bullet 1 of the original post is "of $mathbbK$-algebras".
$endgroup$
– polettix
Apr 2 at 21:44
add a comment |
$begingroup$
Thanks for pointing out. Milne (in jmilne.org/math/CourseNotes/FT.pdf section on "Stem Fields" in chapter 1) actually talks about homomorphism of F-algebras (in this case, $mathbbK$-algebras) that is defined as a homomorphism whose restriction on F is the identity (see section "Fields" in chapter 1). Similarly, I guess the homework comes from a Coursera course on Galois Theory, and even there the isomorphism in bullet 1 of the original post is "of $mathbbK$-algebras".
$endgroup$
– polettix
Apr 2 at 21:44
$begingroup$
Thanks for pointing out. Milne (in jmilne.org/math/CourseNotes/FT.pdf section on "Stem Fields" in chapter 1) actually talks about homomorphism of F-algebras (in this case, $mathbbK$-algebras) that is defined as a homomorphism whose restriction on F is the identity (see section "Fields" in chapter 1). Similarly, I guess the homework comes from a Coursera course on Galois Theory, and even there the isomorphism in bullet 1 of the original post is "of $mathbbK$-algebras".
$endgroup$
– polettix
Apr 2 at 21:44
$begingroup$
Thanks for pointing out. Milne (in jmilne.org/math/CourseNotes/FT.pdf section on "Stem Fields" in chapter 1) actually talks about homomorphism of F-algebras (in this case, $mathbbK$-algebras) that is defined as a homomorphism whose restriction on F is the identity (see section "Fields" in chapter 1). Similarly, I guess the homework comes from a Coursera course on Galois Theory, and even there the isomorphism in bullet 1 of the original post is "of $mathbbK$-algebras".
$endgroup$
– polettix
Apr 2 at 21:44
add a comment |
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This is my first time hearing of the term "stem field" out of the context of "Science, Engineering, Technology and Mathematics". That's certainly an interesting word choice.
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– Cameron Williams
Mar 6 '16 at 1:37
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@CameronWilliams I've found this particular term for instance, in this book - jmilne.org/math/CourseNotes/FT.pdf - if this term is not clear for English speaker - just let me know, and I'll add the definition of stem field to the question. Or if there's more appropriate term I will use it instead )
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– shabunc
Mar 6 '16 at 1:41
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This is the second time in less than 8 hours that the term "stem field" appears in the site. The first time was from a finnish user, who said it was close to the term in finnish language but actually means "splitting field" . I couldn't find the term in Milne's book.
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– DonAntonio
Mar 6 '16 at 6:20
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The term is mentioned in Milne's Fields and Galois Theory book and he in turn attributes it along with term root field(i.e. splitting field) to M.Albert's Modern Higher Algebra book from 1937. I guess its an arcane term not much in vogue now.
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– Vishesh
Mar 6 '16 at 7:39
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I have edited the question a bit and added the definition of a stem field. Are you referring to $E$ as an extension field of $mathbbK$ or is it arbitrary? I am not sure.
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– Vishesh
Mar 6 '16 at 7:59