Mixing
Algebraic dependence of $xy, xz$ and $ yz$ over $mathbb{C}$ and their symmetry
$begingroup$
- Show that $ xy, xz, yz in mathbb{C}[x,y,z]$ are algebraically dependent over $mathbb{C}$ by finding a polynomial they satisfy. Is there a more general method or approach that I should take when looking for this kind of algebraic dependence?
- Does the symmetry between $xy, xz $ and $ yz $ imply that the polynomial they satisfy must be symmetric? How to prove it without explicitly finding the polynomial?
abstract-algebra polynomials commutative-algebra
asked Jan 31 at 15:15
MCLMCL
1138
$endgroup$
add a comment |
$begingroup$
- Show that $ xy, xz, yz in mathbb{C}[x,y,z]$ are algebraically dependent over $mathbb{C}$ by finding a polynomial they satisfy. Is there a more general method or approach that I should take when looking for this kind of algebraic dependence?
- Does the symmetry between $xy, xz $ and $ yz $ imply that the polynomial they satisfy must be symmetric? How to prove it without explicitly finding the polynomial?
abstract-algebra polynomials commutative-algebra
asked Jan 31 at 15:15
MCLMCL
1138
$endgroup$
$begingroup$
What about this: $f(xy,xz,yz)=0implies f(x,y,z)=0$? If $f(x,y,z)=sum a_{ijk}x^iy^jz^k$, then $f(xy,xz,yz)=sum a_{ijk}x^{i+j}y^{i+k}z^{j+k}$ and $$i+j=i'+j',i+k=i'+k',j+k=j'+k'iff i=i',j=j',k=k'.$$
$endgroup$
– user26857
Jan 31 at 22:07
add a comment |
$begingroup$
- Show that $ xy, xz, yz in mathbb{C}[x,y,z]$ are algebraically dependent over $mathbb{C}$ by finding a polynomial they satisfy. Is there a more general method or approach that I should take when looking for this kind of algebraic dependence?
- Does the symmetry between $xy, xz $ and $ yz $ imply that the polynomial they satisfy must be symmetric? How to prove it without explicitly finding the polynomial?
abstract-algebra polynomials commutative-algebra
asked Jan 31 at 15:15
MCLMCL
1138
$endgroup$
- Show that $ xy, xz, yz in mathbb{C}[x,y,z]$ are algebraically dependent over $mathbb{C}$ by finding a polynomial they satisfy. Is there a more general method or approach that I should take when looking for this kind of algebraic dependence?
- Does the symmetry between $xy, xz $ and $ yz $ imply that the polynomial they satisfy must be symmetric? How to prove it without explicitly finding the polynomial?
abstract-algebra polynomials commutative-algebra
abstract-algebra polynomials commutative-algebra
asked Jan 31 at 15:15
MCLMCL
1138
asked Jan 31 at 15:15
MCLMCL
1138
edited Jan 31 at 16:00
MCL
asked Jan 31 at 15:15
MCLMCL
1138
asked Jan 31 at 15:15
MCLMCL
1138
asked Jan 31 at 15:15
MCLMCL
1138
1138
$begingroup$
What about this: $f(xy,xz,yz)=0implies f(x,y,z)=0$? If $f(x,y,z)=sum a_{ijk}x^iy^jz^k$, then $f(xy,xz,yz)=sum a_{ijk}x^{i+j}y^{i+k}z^{j+k}$ and $$i+j=i'+j',i+k=i'+k',j+k=j'+k'iff i=i',j=j',k=k'.$$
$endgroup$
– user26857
Jan 31 at 22:07
add a comment |
$begingroup$
What about this: $f(xy,xz,yz)=0implies f(x,y,z)=0$? If $f(x,y,z)=sum a_{ijk}x^iy^jz^k$, then $f(xy,xz,yz)=sum a_{ijk}x^{i+j}y^{i+k}z^{j+k}$ and $$i+j=i'+j',i+k=i'+k',j+k=j'+k'iff i=i',j=j',k=k'.$$
$endgroup$
– user26857
Jan 31 at 22:07
$begingroup$
What about this: $f(xy,xz,yz)=0implies f(x,y,z)=0$? If $f(x,y,z)=sum a_{ijk}x^iy^jz^k$, then $f(xy,xz,yz)=sum a_{ijk}x^{i+j}y^{i+k}z^{j+k}$ and $$i+j=i'+j',i+k=i'+k',j+k=j'+k'iff i=i',j=j',k=k'.$$
$endgroup$
– user26857
Jan 31 at 22:07
$begingroup$
What about this: $f(xy,xz,yz)=0implies f(x,y,z)=0$? If $f(x,y,z)=sum a_{ijk}x^iy^jz^k$, then $f(xy,xz,yz)=sum a_{ijk}x^{i+j}y^{i+k}z^{j+k}$ and $$i+j=i'+j',i+k=i'+k',j+k=j'+k'iff i=i',j=j',k=k'.$$
$endgroup$
– user26857
Jan 31 at 22:07
add a comment |
1 Answer
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$begingroup$
They are not algebraically dependent over $mathbb{C}$! Probably the easiest way to see this is to think about the subfield $mathbb{C}(xy,xz,yz)subseteq mathbb{C}(x,y,z)$ which they generate, since using division we can isolate one variable at a time. Notice that $(xy)(xz)/(yz)=x^2$ is in this subfield, and so by symmetry so are $y^2$ and $z^2$. But $x^2,y^2,$ and $z^2$ are algebraically independent over $mathbb{C}$, so the $mathbb{C}(xy,xz,yz)$ has transcendence degree at least $3$ over $mathbb{C}$. It follows that the three generators $xy,yz,xz$ must be algebraically independent.
answered Jan 31 at 21:38
Eric WofseyEric Wofsey
192k14220352
$endgroup$
$begingroup$
Alternatively, you can prove the algebraic independence of $yz,zx,xy $ simply by showing that any two distinct monomials in $yz,zx,xy$ will be distinct also as monomials in $x,y,z$.
$endgroup$
– darij grinberg
Jan 31 at 22:33
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
They are not algebraically dependent over $mathbb{C}$! Probably the easiest way to see this is to think about the subfield $mathbb{C}(xy,xz,yz)subseteq mathbb{C}(x,y,z)$ which they generate, since using division we can isolate one variable at a time. Notice that $(xy)(xz)/(yz)=x^2$ is in this subfield, and so by symmetry so are $y^2$ and $z^2$. But $x^2,y^2,$ and $z^2$ are algebraically independent over $mathbb{C}$, so the $mathbb{C}(xy,xz,yz)$ has transcendence degree at least $3$ over $mathbb{C}$. It follows that the three generators $xy,yz,xz$ must be algebraically independent.
answered Jan 31 at 21:38
Eric WofseyEric Wofsey
192k14220352
$endgroup$
$begingroup$
Alternatively, you can prove the algebraic independence of $yz,zx,xy $ simply by showing that any two distinct monomials in $yz,zx,xy$ will be distinct also as monomials in $x,y,z$.
$endgroup$
– darij grinberg
Jan 31 at 22:33
add a comment |
$begingroup$
They are not algebraically dependent over $mathbb{C}$! Probably the easiest way to see this is to think about the subfield $mathbb{C}(xy,xz,yz)subseteq mathbb{C}(x,y,z)$ which they generate, since using division we can isolate one variable at a time. Notice that $(xy)(xz)/(yz)=x^2$ is in this subfield, and so by symmetry so are $y^2$ and $z^2$. But $x^2,y^2,$ and $z^2$ are algebraically independent over $mathbb{C}$, so the $mathbb{C}(xy,xz,yz)$ has transcendence degree at least $3$ over $mathbb{C}$. It follows that the three generators $xy,yz,xz$ must be algebraically independent.
answered Jan 31 at 21:38
Eric WofseyEric Wofsey
192k14220352
$endgroup$
$begingroup$
Alternatively, you can prove the algebraic independence of $yz,zx,xy $ simply by showing that any two distinct monomials in $yz,zx,xy$ will be distinct also as monomials in $x,y,z$.
$endgroup$
– darij grinberg
Jan 31 at 22:33
add a comment |
$begingroup$
They are not algebraically dependent over $mathbb{C}$! Probably the easiest way to see this is to think about the subfield $mathbb{C}(xy,xz,yz)subseteq mathbb{C}(x,y,z)$ which they generate, since using division we can isolate one variable at a time. Notice that $(xy)(xz)/(yz)=x^2$ is in this subfield, and so by symmetry so are $y^2$ and $z^2$. But $x^2,y^2,$ and $z^2$ are algebraically independent over $mathbb{C}$, so the $mathbb{C}(xy,xz,yz)$ has transcendence degree at least $3$ over $mathbb{C}$. It follows that the three generators $xy,yz,xz$ must be algebraically independent.
answered Jan 31 at 21:38
Eric WofseyEric Wofsey
192k14220352
$endgroup$
They are not algebraically dependent over $mathbb{C}$! Probably the easiest way to see this is to think about the subfield $mathbb{C}(xy,xz,yz)subseteq mathbb{C}(x,y,z)$ which they generate, since using division we can isolate one variable at a time. Notice that $(xy)(xz)/(yz)=x^2$ is in this subfield, and so by symmetry so are $y^2$ and $z^2$. But $x^2,y^2,$ and $z^2$ are algebraically independent over $mathbb{C}$, so the $mathbb{C}(xy,xz,yz)$ has transcendence degree at least $3$ over $mathbb{C}$. It follows that the three generators $xy,yz,xz$ must be algebraically independent.
answered Jan 31 at 21:38
Eric WofseyEric Wofsey
192k14220352
answered Jan 31 at 21:38
Eric WofseyEric Wofsey
192k14220352
answered Jan 31 at 21:38
Eric WofseyEric Wofsey
192k14220352
answered Jan 31 at 21:38
Eric WofseyEric Wofsey
192k14220352
192k14220352
$begingroup$
Alternatively, you can prove the algebraic independence of $yz,zx,xy $ simply by showing that any two distinct monomials in $yz,zx,xy$ will be distinct also as monomials in $x,y,z$.
$endgroup$
– darij grinberg
Jan 31 at 22:33
add a comment |
$begingroup$
Alternatively, you can prove the algebraic independence of $yz,zx,xy $ simply by showing that any two distinct monomials in $yz,zx,xy$ will be distinct also as monomials in $x,y,z$.
$endgroup$
– darij grinberg
Jan 31 at 22:33
$begingroup$
Alternatively, you can prove the algebraic independence of $yz,zx,xy $ simply by showing that any two distinct monomials in $yz,zx,xy$ will be distinct also as monomials in $x,y,z$.
$endgroup$
– darij grinberg
Jan 31 at 22:33
$begingroup$
Alternatively, you can prove the algebraic independence of $yz,zx,xy $ simply by showing that any two distinct monomials in $yz,zx,xy$ will be distinct also as monomials in $x,y,z$.
$endgroup$
– darij grinberg
Jan 31 at 22:33
add a comment |
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$begingroup$
What about this: $f(xy,xz,yz)=0implies f(x,y,z)=0$? If $f(x,y,z)=sum a_{ijk}x^iy^jz^k$, then $f(xy,xz,yz)=sum a_{ijk}x^{i+j}y^{i+k}z^{j+k}$ and $$i+j=i'+j',i+k=i'+k',j+k=j'+k'iff i=i',j=j',k=k'.$$
$endgroup$
– user26857
Jan 31 at 22:07