Mixing
Fundamental ismorphism theorem
I don't understand how to apply the fundamental isomorphism theorem to polynomial quotient rings. For example is the ring $mathbb C[X,Y,Z]/langle X^2-Z,XZ-Y^3rangle$ isomorphic to $mathbb C[X,Y]/langle X^3-Y^3rangle$? Can you please elaborate a little? If yes, how is the theorem (the isomorphism theorem) applied here? Thank you!
abstract-algebra algebraic-geometry polynomial-rings ring-isomorphism
edited Nov 20 '18 at 6:22
KReiser
9,30211435
asked Nov 20 '18 at 6:11
mip
334
|
show 7 more comments
I don't understand how to apply the fundamental isomorphism theorem to polynomial quotient rings. For example is the ring $mathbb C[X,Y,Z]/langle X^2-Z,XZ-Y^3rangle$ isomorphic to $mathbb C[X,Y]/langle X^3-Y^3rangle$? Can you please elaborate a little? If yes, how is the theorem (the isomorphism theorem) applied here? Thank you!
abstract-algebra algebraic-geometry polynomial-rings ring-isomorphism
edited Nov 20 '18 at 6:22
KReiser
9,30211435
asked Nov 20 '18 at 6:11
mip
334
The first ring enforces the relation that $X^2-Z=0$, yes? Can you define a map from the first ring to the second using this information?
– KReiser
Nov 20 '18 at 6:27
I'm thinking at $x to t, z to t^2$ but the kernel of this map is $langle X^2-Z rangle$.
– mip
Nov 20 '18 at 6:31
Why not try sending $Zmapsto X^2$ and sending $Xmapsto X$ as well as $Ymapsto Y$?
– KReiser
Nov 20 '18 at 6:41
But how is this related to the fundamental theorem of isomorphism?
– mip
Nov 20 '18 at 7:03
The first isomorphism theorem says that for $f:Ato B$ surjective, we have $Bcong A/ker f$, yes? Do you see how you should fill in the blanks with $A,B,f$? If you complete the mad-lib appropriately, you should get an answer that satisfies you.
– KReiser
Nov 20 '18 at 7:05
|
show 7 more comments
I don't understand how to apply the fundamental isomorphism theorem to polynomial quotient rings. For example is the ring $mathbb C[X,Y,Z]/langle X^2-Z,XZ-Y^3rangle$ isomorphic to $mathbb C[X,Y]/langle X^3-Y^3rangle$? Can you please elaborate a little? If yes, how is the theorem (the isomorphism theorem) applied here? Thank you!
abstract-algebra algebraic-geometry polynomial-rings ring-isomorphism
edited Nov 20 '18 at 6:22
KReiser
9,30211435
asked Nov 20 '18 at 6:11
mip
334
I don't understand how to apply the fundamental isomorphism theorem to polynomial quotient rings. For example is the ring $mathbb C[X,Y,Z]/langle X^2-Z,XZ-Y^3rangle$ isomorphic to $mathbb C[X,Y]/langle X^3-Y^3rangle$? Can you please elaborate a little? If yes, how is the theorem (the isomorphism theorem) applied here? Thank you!
abstract-algebra algebraic-geometry polynomial-rings ring-isomorphism
abstract-algebra algebraic-geometry polynomial-rings ring-isomorphism
edited Nov 20 '18 at 6:22
KReiser
9,30211435
asked Nov 20 '18 at 6:11
mip
334
edited Nov 20 '18 at 6:22
KReiser
9,30211435
asked Nov 20 '18 at 6:11
mip
334
edited Nov 20 '18 at 6:22
KReiser
9,30211435
edited Nov 20 '18 at 6:22
KReiser
9,30211435
edited Nov 20 '18 at 6:22
KReiser
9,30211435
9,30211435
asked Nov 20 '18 at 6:11
mip
334
asked Nov 20 '18 at 6:11
mip
334
asked Nov 20 '18 at 6:11
mip
334
334
The first ring enforces the relation that $X^2-Z=0$, yes? Can you define a map from the first ring to the second using this information?
– KReiser
Nov 20 '18 at 6:27
I'm thinking at $x to t, z to t^2$ but the kernel of this map is $langle X^2-Z rangle$.
– mip
Nov 20 '18 at 6:31
Why not try sending $Zmapsto X^2$ and sending $Xmapsto X$ as well as $Ymapsto Y$?
– KReiser
Nov 20 '18 at 6:41
But how is this related to the fundamental theorem of isomorphism?
– mip
Nov 20 '18 at 7:03
The first isomorphism theorem says that for $f:Ato B$ surjective, we have $Bcong A/ker f$, yes? Do you see how you should fill in the blanks with $A,B,f$? If you complete the mad-lib appropriately, you should get an answer that satisfies you.
– KReiser
Nov 20 '18 at 7:05
|
show 7 more comments
The first ring enforces the relation that $X^2-Z=0$, yes? Can you define a map from the first ring to the second using this information?
– KReiser
Nov 20 '18 at 6:27
I'm thinking at $x to t, z to t^2$ but the kernel of this map is $langle X^2-Z rangle$.
– mip
Nov 20 '18 at 6:31
Why not try sending $Zmapsto X^2$ and sending $Xmapsto X$ as well as $Ymapsto Y$?
– KReiser
Nov 20 '18 at 6:41
But how is this related to the fundamental theorem of isomorphism?
– mip
Nov 20 '18 at 7:03
The first isomorphism theorem says that for $f:Ato B$ surjective, we have $Bcong A/ker f$, yes? Do you see how you should fill in the blanks with $A,B,f$? If you complete the mad-lib appropriately, you should get an answer that satisfies you.
– KReiser
Nov 20 '18 at 7:05
The first ring enforces the relation that $X^2-Z=0$, yes? Can you define a map from the first ring to the second using this information?
– KReiser
Nov 20 '18 at 6:27
The first ring enforces the relation that $X^2-Z=0$, yes? Can you define a map from the first ring to the second using this information?
– KReiser
Nov 20 '18 at 6:27
I'm thinking at $x to t, z to t^2$ but the kernel of this map is $langle X^2-Z rangle$.
– mip
Nov 20 '18 at 6:31
I'm thinking at $x to t, z to t^2$ but the kernel of this map is $langle X^2-Z rangle$.
– mip
Nov 20 '18 at 6:31
Why not try sending $Zmapsto X^2$ and sending $Xmapsto X$ as well as $Ymapsto Y$?
– KReiser
Nov 20 '18 at 6:41
Why not try sending $Zmapsto X^2$ and sending $Xmapsto X$ as well as $Ymapsto Y$?
– KReiser
Nov 20 '18 at 6:41
But how is this related to the fundamental theorem of isomorphism?
– mip
Nov 20 '18 at 7:03
But how is this related to the fundamental theorem of isomorphism?
– mip
Nov 20 '18 at 7:03
The first isomorphism theorem says that for $f:Ato B$ surjective, we have $Bcong A/ker f$, yes? Do you see how you should fill in the blanks with $A,B,f$? If you complete the mad-lib appropriately, you should get an answer that satisfies you.
– KReiser
Nov 20 '18 at 7:05
The first isomorphism theorem says that for $f:Ato B$ surjective, we have $Bcong A/ker f$, yes? Do you see how you should fill in the blanks with $A,B,f$? If you complete the mad-lib appropriately, you should get an answer that satisfies you.
– KReiser
Nov 20 '18 at 7:05
|
show 7 more comments
1 Answer
1
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FWIW a slightly different approach: It suffices to show by the UMP of $mathbb C[x,y,z]/(x^2-z,xz-y^3)$ that there exists a morphism $xi:mathbb C[x,y,z]to mathbb C[x,y]/(x^3-y^3)$ with the property that for each ring morphism $f:mathbb C[x,y,z]to R$ with $f(x^2-z)=f(xz-y^3)=0$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f=tilde{f}circ xi$.
In particular, let $k$ be the unique map $mathbb C[x,y,z]tomathbb C[x,y]$ that fixes $mathbb C,x,y$ and maps $zmapsto x^2$ and let $eta:mathbb C[x,y]tomathbb C[x,y]/(x^3-y^3)$ be the obvious quotient morphism. We claim that it is sufficient to take $xi=etacirc k$.
To that end, consider some arbitrary $f:mathbb C[x,y,z]to R$. By the UMP of polynomial rings, there exists a unique $f':mathbb C[x,y]to R$ such that $forall zetain mathbb C, f'(zeta)=f(zeta)$, $f'(x)=f(x)$ and $f'(y)=f(y)$. Furthemore, we have that $forall pin mathbb C[x,y,z], f'circ k(z)-f(z)=f'(x)-f(x)=0$. It follows (again from the UMP of polynomial rings) that $f=f'circ k$ and furthermore that $f'$ is the unique map with this property.
Now observe that $f'(x^3-y^3)=f'circ k(x^3-y^3)=f(x)f(x^2-z)+f(xz-y^3)=0$. By the UMP of $mathbb C[x,y]/(x^3,y^3)$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f'=tilde{f}circ eta$. It follows that $$f=f'circ k=(tilde{f}circ eta)circ k=tilde{f}circ (etacirc k)$$ And this $tilde{f}$ can be readily seen as unique. So we're done $blacksquare$
answered Nov 21 '18 at 0:02
Rafay Ashary
83618
add a comment |
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1 Answer
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FWIW a slightly different approach: It suffices to show by the UMP of $mathbb C[x,y,z]/(x^2-z,xz-y^3)$ that there exists a morphism $xi:mathbb C[x,y,z]to mathbb C[x,y]/(x^3-y^3)$ with the property that for each ring morphism $f:mathbb C[x,y,z]to R$ with $f(x^2-z)=f(xz-y^3)=0$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f=tilde{f}circ xi$.
In particular, let $k$ be the unique map $mathbb C[x,y,z]tomathbb C[x,y]$ that fixes $mathbb C,x,y$ and maps $zmapsto x^2$ and let $eta:mathbb C[x,y]tomathbb C[x,y]/(x^3-y^3)$ be the obvious quotient morphism. We claim that it is sufficient to take $xi=etacirc k$.
To that end, consider some arbitrary $f:mathbb C[x,y,z]to R$. By the UMP of polynomial rings, there exists a unique $f':mathbb C[x,y]to R$ such that $forall zetain mathbb C, f'(zeta)=f(zeta)$, $f'(x)=f(x)$ and $f'(y)=f(y)$. Furthemore, we have that $forall pin mathbb C[x,y,z], f'circ k(z)-f(z)=f'(x)-f(x)=0$. It follows (again from the UMP of polynomial rings) that $f=f'circ k$ and furthermore that $f'$ is the unique map with this property.
Now observe that $f'(x^3-y^3)=f'circ k(x^3-y^3)=f(x)f(x^2-z)+f(xz-y^3)=0$. By the UMP of $mathbb C[x,y]/(x^3,y^3)$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f'=tilde{f}circ eta$. It follows that $$f=f'circ k=(tilde{f}circ eta)circ k=tilde{f}circ (etacirc k)$$ And this $tilde{f}$ can be readily seen as unique. So we're done $blacksquare$
answered Nov 21 '18 at 0:02
Rafay Ashary
83618
add a comment |
FWIW a slightly different approach: It suffices to show by the UMP of $mathbb C[x,y,z]/(x^2-z,xz-y^3)$ that there exists a morphism $xi:mathbb C[x,y,z]to mathbb C[x,y]/(x^3-y^3)$ with the property that for each ring morphism $f:mathbb C[x,y,z]to R$ with $f(x^2-z)=f(xz-y^3)=0$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f=tilde{f}circ xi$.
In particular, let $k$ be the unique map $mathbb C[x,y,z]tomathbb C[x,y]$ that fixes $mathbb C,x,y$ and maps $zmapsto x^2$ and let $eta:mathbb C[x,y]tomathbb C[x,y]/(x^3-y^3)$ be the obvious quotient morphism. We claim that it is sufficient to take $xi=etacirc k$.
To that end, consider some arbitrary $f:mathbb C[x,y,z]to R$. By the UMP of polynomial rings, there exists a unique $f':mathbb C[x,y]to R$ such that $forall zetain mathbb C, f'(zeta)=f(zeta)$, $f'(x)=f(x)$ and $f'(y)=f(y)$. Furthemore, we have that $forall pin mathbb C[x,y,z], f'circ k(z)-f(z)=f'(x)-f(x)=0$. It follows (again from the UMP of polynomial rings) that $f=f'circ k$ and furthermore that $f'$ is the unique map with this property.
Now observe that $f'(x^3-y^3)=f'circ k(x^3-y^3)=f(x)f(x^2-z)+f(xz-y^3)=0$. By the UMP of $mathbb C[x,y]/(x^3,y^3)$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f'=tilde{f}circ eta$. It follows that $$f=f'circ k=(tilde{f}circ eta)circ k=tilde{f}circ (etacirc k)$$ And this $tilde{f}$ can be readily seen as unique. So we're done $blacksquare$
answered Nov 21 '18 at 0:02
Rafay Ashary
83618
add a comment |
FWIW a slightly different approach: It suffices to show by the UMP of $mathbb C[x,y,z]/(x^2-z,xz-y^3)$ that there exists a morphism $xi:mathbb C[x,y,z]to mathbb C[x,y]/(x^3-y^3)$ with the property that for each ring morphism $f:mathbb C[x,y,z]to R$ with $f(x^2-z)=f(xz-y^3)=0$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f=tilde{f}circ xi$.
In particular, let $k$ be the unique map $mathbb C[x,y,z]tomathbb C[x,y]$ that fixes $mathbb C,x,y$ and maps $zmapsto x^2$ and let $eta:mathbb C[x,y]tomathbb C[x,y]/(x^3-y^3)$ be the obvious quotient morphism. We claim that it is sufficient to take $xi=etacirc k$.
To that end, consider some arbitrary $f:mathbb C[x,y,z]to R$. By the UMP of polynomial rings, there exists a unique $f':mathbb C[x,y]to R$ such that $forall zetain mathbb C, f'(zeta)=f(zeta)$, $f'(x)=f(x)$ and $f'(y)=f(y)$. Furthemore, we have that $forall pin mathbb C[x,y,z], f'circ k(z)-f(z)=f'(x)-f(x)=0$. It follows (again from the UMP of polynomial rings) that $f=f'circ k$ and furthermore that $f'$ is the unique map with this property.
Now observe that $f'(x^3-y^3)=f'circ k(x^3-y^3)=f(x)f(x^2-z)+f(xz-y^3)=0$. By the UMP of $mathbb C[x,y]/(x^3,y^3)$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f'=tilde{f}circ eta$. It follows that $$f=f'circ k=(tilde{f}circ eta)circ k=tilde{f}circ (etacirc k)$$ And this $tilde{f}$ can be readily seen as unique. So we're done $blacksquare$
answered Nov 21 '18 at 0:02
Rafay Ashary
83618
FWIW a slightly different approach: It suffices to show by the UMP of $mathbb C[x,y,z]/(x^2-z,xz-y^3)$ that there exists a morphism $xi:mathbb C[x,y,z]to mathbb C[x,y]/(x^3-y^3)$ with the property that for each ring morphism $f:mathbb C[x,y,z]to R$ with $f(x^2-z)=f(xz-y^3)=0$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f=tilde{f}circ xi$.
In particular, let $k$ be the unique map $mathbb C[x,y,z]tomathbb C[x,y]$ that fixes $mathbb C,x,y$ and maps $zmapsto x^2$ and let $eta:mathbb C[x,y]tomathbb C[x,y]/(x^3-y^3)$ be the obvious quotient morphism. We claim that it is sufficient to take $xi=etacirc k$.
To that end, consider some arbitrary $f:mathbb C[x,y,z]to R$. By the UMP of polynomial rings, there exists a unique $f':mathbb C[x,y]to R$ such that $forall zetain mathbb C, f'(zeta)=f(zeta)$, $f'(x)=f(x)$ and $f'(y)=f(y)$. Furthemore, we have that $forall pin mathbb C[x,y,z], f'circ k(z)-f(z)=f'(x)-f(x)=0$. It follows (again from the UMP of polynomial rings) that $f=f'circ k$ and furthermore that $f'$ is the unique map with this property.
Now observe that $f'(x^3-y^3)=f'circ k(x^3-y^3)=f(x)f(x^2-z)+f(xz-y^3)=0$. By the UMP of $mathbb C[x,y]/(x^3,y^3)$ there exists a unique $tilde{f}:mathbb C[x,y]/(x^3-y^3)to R$ satisfying $f'=tilde{f}circ eta$. It follows that $$f=f'circ k=(tilde{f}circ eta)circ k=tilde{f}circ (etacirc k)$$ And this $tilde{f}$ can be readily seen as unique. So we're done $blacksquare$
answered Nov 21 '18 at 0:02
Rafay Ashary
83618
answered Nov 21 '18 at 0:02
Rafay Ashary
83618
answered Nov 21 '18 at 0:02
Rafay Ashary
83618
answered Nov 21 '18 at 0:02
Rafay Ashary
83618
83618
add a comment |
add a comment |
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The first ring enforces the relation that $X^2-Z=0$, yes? Can you define a map from the first ring to the second using this information?
– KReiser
Nov 20 '18 at 6:27
I'm thinking at $x to t, z to t^2$ but the kernel of this map is $langle X^2-Z rangle$.
– mip
Nov 20 '18 at 6:31
Why not try sending $Zmapsto X^2$ and sending $Xmapsto X$ as well as $Ymapsto Y$?
– KReiser
Nov 20 '18 at 6:41
But how is this related to the fundamental theorem of isomorphism?
– mip
Nov 20 '18 at 7:03
The first isomorphism theorem says that for $f:Ato B$ surjective, we have $Bcong A/ker f$, yes? Do you see how you should fill in the blanks with $A,B,f$? If you complete the mad-lib appropriately, you should get an answer that satisfies you.
– KReiser
Nov 20 '18 at 7:05