Mixing
$mathbb{F}_{4096}$,$mathbb{F}_{16}$ and $mathbb{F}_{2}$
$begingroup$
I'm stuck at these two questions:
i)
How many distinct elements $a in mathbb{F}_{4096}$ exist, such that $mathbb{F}_{4096}=mathbb{F}_2 left[aright]$
ii)
Find an irreducible $g in mathbb{F}_4 left[xright]$ such that $mathbb{F}_{16}=mathbb{F}_4/(g)$
As for i) I know that the degree of the minimal polynomial of $a$ has to be $12$, since $2$ is prime and $4096=2^{12}$.
As for ii) I'm lost: Does $deg(g)$ equal to $2$? If so, how would I show that $mathbb{F}_4/(g)$ indeed is $mathbb{F}_{16}$? Is $g=x^2+1$ a viable guess?
finite-fields extension-field
asked Feb 3 at 10:33
C. BrendelC. Brendel
638
$endgroup$
add a comment |
$begingroup$
I'm stuck at these two questions:
i)
How many distinct elements $a in mathbb{F}_{4096}$ exist, such that $mathbb{F}_{4096}=mathbb{F}_2 left[aright]$
ii)
Find an irreducible $g in mathbb{F}_4 left[xright]$ such that $mathbb{F}_{16}=mathbb{F}_4/(g)$
As for i) I know that the degree of the minimal polynomial of $a$ has to be $12$, since $2$ is prime and $4096=2^{12}$.
As for ii) I'm lost: Does $deg(g)$ equal to $2$? If so, how would I show that $mathbb{F}_4/(g)$ indeed is $mathbb{F}_{16}$? Is $g=x^2+1$ a viable guess?
finite-fields extension-field
asked Feb 3 at 10:33
C. BrendelC. Brendel
638
$endgroup$
add a comment |
$begingroup$
I'm stuck at these two questions:
i)
How many distinct elements $a in mathbb{F}_{4096}$ exist, such that $mathbb{F}_{4096}=mathbb{F}_2 left[aright]$
ii)
Find an irreducible $g in mathbb{F}_4 left[xright]$ such that $mathbb{F}_{16}=mathbb{F}_4/(g)$
As for i) I know that the degree of the minimal polynomial of $a$ has to be $12$, since $2$ is prime and $4096=2^{12}$.
As for ii) I'm lost: Does $deg(g)$ equal to $2$? If so, how would I show that $mathbb{F}_4/(g)$ indeed is $mathbb{F}_{16}$? Is $g=x^2+1$ a viable guess?
finite-fields extension-field
asked Feb 3 at 10:33
C. BrendelC. Brendel
638
$endgroup$
I'm stuck at these two questions:
i)
How many distinct elements $a in mathbb{F}_{4096}$ exist, such that $mathbb{F}_{4096}=mathbb{F}_2 left[aright]$
ii)
Find an irreducible $g in mathbb{F}_4 left[xright]$ such that $mathbb{F}_{16}=mathbb{F}_4/(g)$
As for i) I know that the degree of the minimal polynomial of $a$ has to be $12$, since $2$ is prime and $4096=2^{12}$.
As for ii) I'm lost: Does $deg(g)$ equal to $2$? If so, how would I show that $mathbb{F}_4/(g)$ indeed is $mathbb{F}_{16}$? Is $g=x^2+1$ a viable guess?
finite-fields extension-field
finite-fields extension-field
asked Feb 3 at 10:33
C. BrendelC. Brendel
638
asked Feb 3 at 10:33
C. BrendelC. Brendel
638
asked Feb 3 at 10:33
C. BrendelC. Brendel
638
asked Feb 3 at 10:33
C. BrendelC. Brendel
638
asked Feb 3 at 10:33
C. BrendelC. Brendel
638
638
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
i) Hint: By a bad choice for $a$, you would end up with $Bbb F_{2048}$ or a sub-field thereof
ii) Yes, we need $deg g=2$. However, $x^2+1$ is not irreducible: $x^2+1=(x+1)(x+1)$. In fact, any $x^2+ax+b$ with $a,binBbb F_2$ will fail to work as it can at most produce a degree 2 extension of $Bbb F_2$, i.e., it can already be factored in $Bbb F_4$.
answered Feb 3 at 10:40
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
If $a$ is in a sub-ring $K$ of $mathbb{F}_{4096}$ isomorphic to $mathbb{F}_{2048}$ does that imply that $mathbb{F}_2 left[aright] cong mathbb{F}_{2048}$ ? If so: Are the elements that satisfy this condition exactly those of $mathbb{F}_{4096} setminus K$ ? Also: I don't see how $2x=0$ in $mathbb{F}_4$. Thank you for your help!
$endgroup$
– C. Brendel
Feb 3 at 11:06
$begingroup$
Figured it out myself! Thanks!
$endgroup$
– C. Brendel
Feb 3 at 16:04
$begingroup$
Hmm. $Bbb{F}_{2048}$ intersects trivially with $Bbb{F}_{4096}$.
$endgroup$
– Jyrki Lahtonen
Feb 4 at 6:35
add a comment |
$begingroup$
Because $12=2^2cdot3$, the field $Bbb{F}_{4096}$ has the following subfields
$$
Bbb{F}_2,Bbb{F}_4,Bbb{F}_8,Bbb{F}_{16},Bbb{F}_{64}.
$$
The exponents $m$ of $2^m$ here are the proper factors of $12$, namely $1,2,3,4$ and $6$.
Of the listed subfields $K_1=Bbb{F}_{16}$ and $K_2=Bbb{F}_{64}$ contain the smaller ones. We further observe that $K_1cap K_2=Bbb{F}_4$. This lets us answer your first question:
$Bbb{F}_2[alpha]=Bbb{F}_{4096}$ unless $alphain K_1cup K_2$. We need to exclude $$|K_1cup K_2|=|K_1|+|K_2|-|K_1cap K_2|=64+16-4=76$$
elements. Implying that $4096-76=4020$ choices of $alpha$ will work.
To answer your second question see the bottom quarter of this answer I prepared for referrals. You will find that with
$Bbb{F}_4={0,1,beta,beta+1=beta^2}$ an irreducible quadratic is $g(x)=x^2+x+beta$. There are a total of six irreducible monic quadratics in $Bbb{F}_4[x]$. Anyone of them can serve in the role of $g(x)$.
answered Feb 4 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172392
$endgroup$
$begingroup$
Oh! Neat! Thanks. Already handed the assignment in... I guess in i) I goofed a little :P as for ii) in the end I just went with multiplying every possible linear factor with another to rule out any reducibles of degree 2 :) Very much appreciated!
$endgroup$
– C. Brendel
Feb 5 at 16:29
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
i) Hint: By a bad choice for $a$, you would end up with $Bbb F_{2048}$ or a sub-field thereof
ii) Yes, we need $deg g=2$. However, $x^2+1$ is not irreducible: $x^2+1=(x+1)(x+1)$. In fact, any $x^2+ax+b$ with $a,binBbb F_2$ will fail to work as it can at most produce a degree 2 extension of $Bbb F_2$, i.e., it can already be factored in $Bbb F_4$.
answered Feb 3 at 10:40
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
If $a$ is in a sub-ring $K$ of $mathbb{F}_{4096}$ isomorphic to $mathbb{F}_{2048}$ does that imply that $mathbb{F}_2 left[aright] cong mathbb{F}_{2048}$ ? If so: Are the elements that satisfy this condition exactly those of $mathbb{F}_{4096} setminus K$ ? Also: I don't see how $2x=0$ in $mathbb{F}_4$. Thank you for your help!
$endgroup$
– C. Brendel
Feb 3 at 11:06
$begingroup$
Figured it out myself! Thanks!
$endgroup$
– C. Brendel
Feb 3 at 16:04
$begingroup$
Hmm. $Bbb{F}_{2048}$ intersects trivially with $Bbb{F}_{4096}$.
$endgroup$
– Jyrki Lahtonen
Feb 4 at 6:35
add a comment |
$begingroup$
i) Hint: By a bad choice for $a$, you would end up with $Bbb F_{2048}$ or a sub-field thereof
ii) Yes, we need $deg g=2$. However, $x^2+1$ is not irreducible: $x^2+1=(x+1)(x+1)$. In fact, any $x^2+ax+b$ with $a,binBbb F_2$ will fail to work as it can at most produce a degree 2 extension of $Bbb F_2$, i.e., it can already be factored in $Bbb F_4$.
answered Feb 3 at 10:40
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
If $a$ is in a sub-ring $K$ of $mathbb{F}_{4096}$ isomorphic to $mathbb{F}_{2048}$ does that imply that $mathbb{F}_2 left[aright] cong mathbb{F}_{2048}$ ? If so: Are the elements that satisfy this condition exactly those of $mathbb{F}_{4096} setminus K$ ? Also: I don't see how $2x=0$ in $mathbb{F}_4$. Thank you for your help!
$endgroup$
– C. Brendel
Feb 3 at 11:06
$begingroup$
Figured it out myself! Thanks!
$endgroup$
– C. Brendel
Feb 3 at 16:04
$begingroup$
Hmm. $Bbb{F}_{2048}$ intersects trivially with $Bbb{F}_{4096}$.
$endgroup$
– Jyrki Lahtonen
Feb 4 at 6:35
add a comment |
$begingroup$
i) Hint: By a bad choice for $a$, you would end up with $Bbb F_{2048}$ or a sub-field thereof
ii) Yes, we need $deg g=2$. However, $x^2+1$ is not irreducible: $x^2+1=(x+1)(x+1)$. In fact, any $x^2+ax+b$ with $a,binBbb F_2$ will fail to work as it can at most produce a degree 2 extension of $Bbb F_2$, i.e., it can already be factored in $Bbb F_4$.
answered Feb 3 at 10:40
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
i) Hint: By a bad choice for $a$, you would end up with $Bbb F_{2048}$ or a sub-field thereof
ii) Yes, we need $deg g=2$. However, $x^2+1$ is not irreducible: $x^2+1=(x+1)(x+1)$. In fact, any $x^2+ax+b$ with $a,binBbb F_2$ will fail to work as it can at most produce a degree 2 extension of $Bbb F_2$, i.e., it can already be factored in $Bbb F_4$.
answered Feb 3 at 10:40
Hagen von EitzenHagen von Eitzen
283k23273508
answered Feb 3 at 10:40
Hagen von EitzenHagen von Eitzen
283k23273508
answered Feb 3 at 10:40
Hagen von EitzenHagen von Eitzen
283k23273508
answered Feb 3 at 10:40
Hagen von EitzenHagen von Eitzen
283k23273508
283k23273508
$begingroup$
If $a$ is in a sub-ring $K$ of $mathbb{F}_{4096}$ isomorphic to $mathbb{F}_{2048}$ does that imply that $mathbb{F}_2 left[aright] cong mathbb{F}_{2048}$ ? If so: Are the elements that satisfy this condition exactly those of $mathbb{F}_{4096} setminus K$ ? Also: I don't see how $2x=0$ in $mathbb{F}_4$. Thank you for your help!
$endgroup$
– C. Brendel
Feb 3 at 11:06
$begingroup$
Figured it out myself! Thanks!
$endgroup$
– C. Brendel
Feb 3 at 16:04
$begingroup$
Hmm. $Bbb{F}_{2048}$ intersects trivially with $Bbb{F}_{4096}$.
$endgroup$
– Jyrki Lahtonen
Feb 4 at 6:35
add a comment |
$begingroup$
If $a$ is in a sub-ring $K$ of $mathbb{F}_{4096}$ isomorphic to $mathbb{F}_{2048}$ does that imply that $mathbb{F}_2 left[aright] cong mathbb{F}_{2048}$ ? If so: Are the elements that satisfy this condition exactly those of $mathbb{F}_{4096} setminus K$ ? Also: I don't see how $2x=0$ in $mathbb{F}_4$. Thank you for your help!
$endgroup$
– C. Brendel
Feb 3 at 11:06
$begingroup$
Figured it out myself! Thanks!
$endgroup$
– C. Brendel
Feb 3 at 16:04
$begingroup$
Hmm. $Bbb{F}_{2048}$ intersects trivially with $Bbb{F}_{4096}$.
$endgroup$
– Jyrki Lahtonen
Feb 4 at 6:35
$begingroup$
If $a$ is in a sub-ring $K$ of $mathbb{F}_{4096}$ isomorphic to $mathbb{F}_{2048}$ does that imply that $mathbb{F}_2 left[aright] cong mathbb{F}_{2048}$ ? If so: Are the elements that satisfy this condition exactly those of $mathbb{F}_{4096} setminus K$ ? Also: I don't see how $2x=0$ in $mathbb{F}_4$. Thank you for your help!
$endgroup$
– C. Brendel
Feb 3 at 11:06
$begingroup$
If $a$ is in a sub-ring $K$ of $mathbb{F}_{4096}$ isomorphic to $mathbb{F}_{2048}$ does that imply that $mathbb{F}_2 left[aright] cong mathbb{F}_{2048}$ ? If so: Are the elements that satisfy this condition exactly those of $mathbb{F}_{4096} setminus K$ ? Also: I don't see how $2x=0$ in $mathbb{F}_4$. Thank you for your help!
$endgroup$
– C. Brendel
Feb 3 at 11:06
$begingroup$
Figured it out myself! Thanks!
$endgroup$
– C. Brendel
Feb 3 at 16:04
$begingroup$
Figured it out myself! Thanks!
$endgroup$
– C. Brendel
Feb 3 at 16:04
$begingroup$
Hmm. $Bbb{F}_{2048}$ intersects trivially with $Bbb{F}_{4096}$.
$endgroup$
– Jyrki Lahtonen
Feb 4 at 6:35
$begingroup$
Hmm. $Bbb{F}_{2048}$ intersects trivially with $Bbb{F}_{4096}$.
$endgroup$
– Jyrki Lahtonen
Feb 4 at 6:35
add a comment |
$begingroup$
Because $12=2^2cdot3$, the field $Bbb{F}_{4096}$ has the following subfields
$$
Bbb{F}_2,Bbb{F}_4,Bbb{F}_8,Bbb{F}_{16},Bbb{F}_{64}.
$$
The exponents $m$ of $2^m$ here are the proper factors of $12$, namely $1,2,3,4$ and $6$.
Of the listed subfields $K_1=Bbb{F}_{16}$ and $K_2=Bbb{F}_{64}$ contain the smaller ones. We further observe that $K_1cap K_2=Bbb{F}_4$. This lets us answer your first question:
$Bbb{F}_2[alpha]=Bbb{F}_{4096}$ unless $alphain K_1cup K_2$. We need to exclude $$|K_1cup K_2|=|K_1|+|K_2|-|K_1cap K_2|=64+16-4=76$$
elements. Implying that $4096-76=4020$ choices of $alpha$ will work.
To answer your second question see the bottom quarter of this answer I prepared for referrals. You will find that with
$Bbb{F}_4={0,1,beta,beta+1=beta^2}$ an irreducible quadratic is $g(x)=x^2+x+beta$. There are a total of six irreducible monic quadratics in $Bbb{F}_4[x]$. Anyone of them can serve in the role of $g(x)$.
answered Feb 4 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172392
$endgroup$
$begingroup$
Oh! Neat! Thanks. Already handed the assignment in... I guess in i) I goofed a little :P as for ii) in the end I just went with multiplying every possible linear factor with another to rule out any reducibles of degree 2 :) Very much appreciated!
$endgroup$
– C. Brendel
Feb 5 at 16:29
add a comment |
$begingroup$
Because $12=2^2cdot3$, the field $Bbb{F}_{4096}$ has the following subfields
$$
Bbb{F}_2,Bbb{F}_4,Bbb{F}_8,Bbb{F}_{16},Bbb{F}_{64}.
$$
The exponents $m$ of $2^m$ here are the proper factors of $12$, namely $1,2,3,4$ and $6$.
Of the listed subfields $K_1=Bbb{F}_{16}$ and $K_2=Bbb{F}_{64}$ contain the smaller ones. We further observe that $K_1cap K_2=Bbb{F}_4$. This lets us answer your first question:
$Bbb{F}_2[alpha]=Bbb{F}_{4096}$ unless $alphain K_1cup K_2$. We need to exclude $$|K_1cup K_2|=|K_1|+|K_2|-|K_1cap K_2|=64+16-4=76$$
elements. Implying that $4096-76=4020$ choices of $alpha$ will work.
To answer your second question see the bottom quarter of this answer I prepared for referrals. You will find that with
$Bbb{F}_4={0,1,beta,beta+1=beta^2}$ an irreducible quadratic is $g(x)=x^2+x+beta$. There are a total of six irreducible monic quadratics in $Bbb{F}_4[x]$. Anyone of them can serve in the role of $g(x)$.
answered Feb 4 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172392
$endgroup$
$begingroup$
Oh! Neat! Thanks. Already handed the assignment in... I guess in i) I goofed a little :P as for ii) in the end I just went with multiplying every possible linear factor with another to rule out any reducibles of degree 2 :) Very much appreciated!
$endgroup$
– C. Brendel
Feb 5 at 16:29
add a comment |
$begingroup$
Because $12=2^2cdot3$, the field $Bbb{F}_{4096}$ has the following subfields
$$
Bbb{F}_2,Bbb{F}_4,Bbb{F}_8,Bbb{F}_{16},Bbb{F}_{64}.
$$
The exponents $m$ of $2^m$ here are the proper factors of $12$, namely $1,2,3,4$ and $6$.
Of the listed subfields $K_1=Bbb{F}_{16}$ and $K_2=Bbb{F}_{64}$ contain the smaller ones. We further observe that $K_1cap K_2=Bbb{F}_4$. This lets us answer your first question:
$Bbb{F}_2[alpha]=Bbb{F}_{4096}$ unless $alphain K_1cup K_2$. We need to exclude $$|K_1cup K_2|=|K_1|+|K_2|-|K_1cap K_2|=64+16-4=76$$
elements. Implying that $4096-76=4020$ choices of $alpha$ will work.
To answer your second question see the bottom quarter of this answer I prepared for referrals. You will find that with
$Bbb{F}_4={0,1,beta,beta+1=beta^2}$ an irreducible quadratic is $g(x)=x^2+x+beta$. There are a total of six irreducible monic quadratics in $Bbb{F}_4[x]$. Anyone of them can serve in the role of $g(x)$.
answered Feb 4 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172392
$endgroup$
Because $12=2^2cdot3$, the field $Bbb{F}_{4096}$ has the following subfields
$$
Bbb{F}_2,Bbb{F}_4,Bbb{F}_8,Bbb{F}_{16},Bbb{F}_{64}.
$$
The exponents $m$ of $2^m$ here are the proper factors of $12$, namely $1,2,3,4$ and $6$.
Of the listed subfields $K_1=Bbb{F}_{16}$ and $K_2=Bbb{F}_{64}$ contain the smaller ones. We further observe that $K_1cap K_2=Bbb{F}_4$. This lets us answer your first question:
$Bbb{F}_2[alpha]=Bbb{F}_{4096}$ unless $alphain K_1cup K_2$. We need to exclude $$|K_1cup K_2|=|K_1|+|K_2|-|K_1cap K_2|=64+16-4=76$$
elements. Implying that $4096-76=4020$ choices of $alpha$ will work.
To answer your second question see the bottom quarter of this answer I prepared for referrals. You will find that with
$Bbb{F}_4={0,1,beta,beta+1=beta^2}$ an irreducible quadratic is $g(x)=x^2+x+beta$. There are a total of six irreducible monic quadratics in $Bbb{F}_4[x]$. Anyone of them can serve in the role of $g(x)$.
answered Feb 4 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172392
answered Feb 4 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172392
answered Feb 4 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172392
answered Feb 4 at 6:34
Jyrki LahtonenJyrki Lahtonen
110k13172392
110k13172392
$begingroup$
Oh! Neat! Thanks. Already handed the assignment in... I guess in i) I goofed a little :P as for ii) in the end I just went with multiplying every possible linear factor with another to rule out any reducibles of degree 2 :) Very much appreciated!
$endgroup$
– C. Brendel
Feb 5 at 16:29
add a comment |
$begingroup$
Oh! Neat! Thanks. Already handed the assignment in... I guess in i) I goofed a little :P as for ii) in the end I just went with multiplying every possible linear factor with another to rule out any reducibles of degree 2 :) Very much appreciated!
$endgroup$
– C. Brendel
Feb 5 at 16:29
$begingroup$
Oh! Neat! Thanks. Already handed the assignment in... I guess in i) I goofed a little :P as for ii) in the end I just went with multiplying every possible linear factor with another to rule out any reducibles of degree 2 :) Very much appreciated!
$endgroup$
– C. Brendel
Feb 5 at 16:29
$begingroup$
Oh! Neat! Thanks. Already handed the assignment in... I guess in i) I goofed a little :P as for ii) in the end I just went with multiplying every possible linear factor with another to rule out any reducibles of degree 2 :) Very much appreciated!
$endgroup$
– C. Brendel
Feb 5 at 16:29
add a comment |
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