Determining ideals, isomorphic rings of $Bbb C[x, y]/(y^2 - x^3)$?
$begingroup$
I've been having a substantial amount of trouble trying to understand the workings of $Bbb C[x, y]$ mod... anything really. I figure this particular example is a good one to ask here because I probably understand it the least.
As I guiding problem, I'd like to understand what $Bbb C[x, y]/(y^2 - x^3)$ is isomorphic to and why, but also what $Bbb C[x, y]/(y^2 - x^3)$ mod the ideal generated by the coset of x is isomorphic to.
The first time I encountered this type of ring it was introduced with long division and some business about a function sending things to t. I hope this vague and clearly confused description paints a picture how how little I understand about this. For the second problem above, I tried to circumvent this by saying the following, where $x`$ is the coset of $x$ in $Bbb C[x, y]/(y^2 - x^3)$:
$(Bbb C[x, y]/(y^2 - x^3))/(x`)$ is equivalent to $(Bbb C[x, y]/(y^2 - x^3))/((x)/(y^2 - x^3))$ -is this even true?
And then by the Third Isomorphism Theorem
$(Bbb C[x, y]/(y^2 - x^3))/((x)/(y^2 - x^3))$ $cong$ $Bbb C[x, y]/(x)$
I feel like there's no way this is correct, but also I understand it so little that I can't even be sure of that.
What's with the $t$, $Bbb C[t]$, $Bbb C[t^2, t^3]$ stuff? It seems necessary to working with this but I've yet to encounter a clear description of what's going on with it. And there are also things using $f(x, y) = g(x, y)(y^2-x^3) + h(y)$ by long division... how does knowing this help?
abstract-algebra algebraic-geometry ring-theory
asked Dec 9 '18 at 0:06
Doug MacArthurDoug MacArthur
61
$endgroup$
add a comment |
$begingroup$
I've been having a substantial amount of trouble trying to understand the workings of $Bbb C[x, y]$ mod... anything really. I figure this particular example is a good one to ask here because I probably understand it the least.
As I guiding problem, I'd like to understand what $Bbb C[x, y]/(y^2 - x^3)$ is isomorphic to and why, but also what $Bbb C[x, y]/(y^2 - x^3)$ mod the ideal generated by the coset of x is isomorphic to.
The first time I encountered this type of ring it was introduced with long division and some business about a function sending things to t. I hope this vague and clearly confused description paints a picture how how little I understand about this. For the second problem above, I tried to circumvent this by saying the following, where $x`$ is the coset of $x$ in $Bbb C[x, y]/(y^2 - x^3)$:
$(Bbb C[x, y]/(y^2 - x^3))/(x`)$ is equivalent to $(Bbb C[x, y]/(y^2 - x^3))/((x)/(y^2 - x^3))$ -is this even true?
And then by the Third Isomorphism Theorem
$(Bbb C[x, y]/(y^2 - x^3))/((x)/(y^2 - x^3))$ $cong$ $Bbb C[x, y]/(x)$
I feel like there's no way this is correct, but also I understand it so little that I can't even be sure of that.
What's with the $t$, $Bbb C[t]$, $Bbb C[t^2, t^3]$ stuff? It seems necessary to working with this but I've yet to encounter a clear description of what's going on with it. And there are also things using $f(x, y) = g(x, y)(y^2-x^3) + h(y)$ by long division... how does knowing this help?
abstract-algebra algebraic-geometry ring-theory
asked Dec 9 '18 at 0:06
Doug MacArthurDoug MacArthur
61
$endgroup$
$begingroup$
The application of Third Isomorphism Theorem is correct; and the quotient ring is isomorphic to $mathbb{C}[y]$.
$endgroup$
– Armando j18eos
Dec 9 '18 at 13:10
add a comment |
$begingroup$
I've been having a substantial amount of trouble trying to understand the workings of $Bbb C[x, y]$ mod... anything really. I figure this particular example is a good one to ask here because I probably understand it the least.
As I guiding problem, I'd like to understand what $Bbb C[x, y]/(y^2 - x^3)$ is isomorphic to and why, but also what $Bbb C[x, y]/(y^2 - x^3)$ mod the ideal generated by the coset of x is isomorphic to.
The first time I encountered this type of ring it was introduced with long division and some business about a function sending things to t. I hope this vague and clearly confused description paints a picture how how little I understand about this. For the second problem above, I tried to circumvent this by saying the following, where $x`$ is the coset of $x$ in $Bbb C[x, y]/(y^2 - x^3)$:
$(Bbb C[x, y]/(y^2 - x^3))/(x`)$ is equivalent to $(Bbb C[x, y]/(y^2 - x^3))/((x)/(y^2 - x^3))$ -is this even true?
And then by the Third Isomorphism Theorem
$(Bbb C[x, y]/(y^2 - x^3))/((x)/(y^2 - x^3))$ $cong$ $Bbb C[x, y]/(x)$
I feel like there's no way this is correct, but also I understand it so little that I can't even be sure of that.
What's with the $t$, $Bbb C[t]$, $Bbb C[t^2, t^3]$ stuff? It seems necessary to working with this but I've yet to encounter a clear description of what's going on with it. And there are also things using $f(x, y) = g(x, y)(y^2-x^3) + h(y)$ by long division... how does knowing this help?
abstract-algebra algebraic-geometry ring-theory
asked Dec 9 '18 at 0:06
Doug MacArthurDoug MacArthur
61
$endgroup$
I've been having a substantial amount of trouble trying to understand the workings of $Bbb C[x, y]$ mod... anything really. I figure this particular example is a good one to ask here because I probably understand it the least.
As I guiding problem, I'd like to understand what $Bbb C[x, y]/(y^2 - x^3)$ is isomorphic to and why, but also what $Bbb C[x, y]/(y^2 - x^3)$ mod the ideal generated by the coset of x is isomorphic to.
The first time I encountered this type of ring it was introduced with long division and some business about a function sending things to t. I hope this vague and clearly confused description paints a picture how how little I understand about this. For the second problem above, I tried to circumvent this by saying the following, where $x`$ is the coset of $x$ in $Bbb C[x, y]/(y^2 - x^3)$:
$(Bbb C[x, y]/(y^2 - x^3))/(x`)$ is equivalent to $(Bbb C[x, y]/(y^2 - x^3))/((x)/(y^2 - x^3))$ -is this even true?
And then by the Third Isomorphism Theorem
$(Bbb C[x, y]/(y^2 - x^3))/((x)/(y^2 - x^3))$ $cong$ $Bbb C[x, y]/(x)$
I feel like there's no way this is correct, but also I understand it so little that I can't even be sure of that.
What's with the $t$, $Bbb C[t]$, $Bbb C[t^2, t^3]$ stuff? It seems necessary to working with this but I've yet to encounter a clear description of what's going on with it. And there are also things using $f(x, y) = g(x, y)(y^2-x^3) + h(y)$ by long division... how does knowing this help?
abstract-algebra algebraic-geometry ring-theory
abstract-algebra algebraic-geometry ring-theory
asked Dec 9 '18 at 0:06
Doug MacArthurDoug MacArthur
61
asked Dec 9 '18 at 0:06
Doug MacArthurDoug MacArthur
61
asked Dec 9 '18 at 0:06
Doug MacArthurDoug MacArthur
61
asked Dec 9 '18 at 0:06
Doug MacArthurDoug MacArthur
61
asked Dec 9 '18 at 0:06
Doug MacArthurDoug MacArthur
61
61
$begingroup$
The application of Third Isomorphism Theorem is correct; and the quotient ring is isomorphic to $mathbb{C}[y]$.
$endgroup$
– Armando j18eos
Dec 9 '18 at 13:10
add a comment |
$begingroup$
The application of Third Isomorphism Theorem is correct; and the quotient ring is isomorphic to $mathbb{C}[y]$.
$endgroup$
– Armando j18eos
Dec 9 '18 at 13:10
$begingroup$
The application of Third Isomorphism Theorem is correct; and the quotient ring is isomorphic to $mathbb{C}[y]$.
$endgroup$
– Armando j18eos
Dec 9 '18 at 13:10
$begingroup$
The application of Third Isomorphism Theorem is correct; and the quotient ring is isomorphic to $mathbb{C}[y]$.
$endgroup$
– Armando j18eos
Dec 9 '18 at 13:10
add a comment |
1 Answer
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For the last part of your question, you have a $mathbf C$-algebra homomorphism
begin{align}
mathbf C[x,y]&longrightarrow mathbf C[t]\
x & longmapsto t^2,\
y & longmapsto t^3
end{align}
The kernel is the prime ideal generated by $y^2-x^3$. The maximal ideals of $mathbf C[x,y]$, by the Nullstellensatz, have the form $(x-alpha, y-beta)$. Those which contain the kernel satisfy the equation $;beta^2-alpha^3=0$, and the above homomorphism corresponds to a parametrisation of the cubic curve with equation $y^2-x^3=0$, obtained by determining the non-trivial intersections of the curve with the lines $y=tx$.
answered Dec 9 '18 at 0:33
BernardBernard
119k740113
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1 Answer
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1 Answer
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$begingroup$
For the last part of your question, you have a $mathbf C$-algebra homomorphism
begin{align}
mathbf C[x,y]&longrightarrow mathbf C[t]\
x & longmapsto t^2,\
y & longmapsto t^3
end{align}
The kernel is the prime ideal generated by $y^2-x^3$. The maximal ideals of $mathbf C[x,y]$, by the Nullstellensatz, have the form $(x-alpha, y-beta)$. Those which contain the kernel satisfy the equation $;beta^2-alpha^3=0$, and the above homomorphism corresponds to a parametrisation of the cubic curve with equation $y^2-x^3=0$, obtained by determining the non-trivial intersections of the curve with the lines $y=tx$.
answered Dec 9 '18 at 0:33
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
For the last part of your question, you have a $mathbf C$-algebra homomorphism
begin{align}
mathbf C[x,y]&longrightarrow mathbf C[t]\
x & longmapsto t^2,\
y & longmapsto t^3
end{align}
The kernel is the prime ideal generated by $y^2-x^3$. The maximal ideals of $mathbf C[x,y]$, by the Nullstellensatz, have the form $(x-alpha, y-beta)$. Those which contain the kernel satisfy the equation $;beta^2-alpha^3=0$, and the above homomorphism corresponds to a parametrisation of the cubic curve with equation $y^2-x^3=0$, obtained by determining the non-trivial intersections of the curve with the lines $y=tx$.
answered Dec 9 '18 at 0:33
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
For the last part of your question, you have a $mathbf C$-algebra homomorphism
begin{align}
mathbf C[x,y]&longrightarrow mathbf C[t]\
x & longmapsto t^2,\
y & longmapsto t^3
end{align}
The kernel is the prime ideal generated by $y^2-x^3$. The maximal ideals of $mathbf C[x,y]$, by the Nullstellensatz, have the form $(x-alpha, y-beta)$. Those which contain the kernel satisfy the equation $;beta^2-alpha^3=0$, and the above homomorphism corresponds to a parametrisation of the cubic curve with equation $y^2-x^3=0$, obtained by determining the non-trivial intersections of the curve with the lines $y=tx$.
answered Dec 9 '18 at 0:33
BernardBernard
119k740113
$endgroup$
For the last part of your question, you have a $mathbf C$-algebra homomorphism
begin{align}
mathbf C[x,y]&longrightarrow mathbf C[t]\
x & longmapsto t^2,\
y & longmapsto t^3
end{align}
The kernel is the prime ideal generated by $y^2-x^3$. The maximal ideals of $mathbf C[x,y]$, by the Nullstellensatz, have the form $(x-alpha, y-beta)$. Those which contain the kernel satisfy the equation $;beta^2-alpha^3=0$, and the above homomorphism corresponds to a parametrisation of the cubic curve with equation $y^2-x^3=0$, obtained by determining the non-trivial intersections of the curve with the lines $y=tx$.
answered Dec 9 '18 at 0:33
BernardBernard
119k740113
edited Dec 9 '18 at 13:12
answered Dec 9 '18 at 0:33
BernardBernard
119k740113
answered Dec 9 '18 at 0:33
BernardBernard
119k740113
answered Dec 9 '18 at 0:33
BernardBernard
119k740113
119k740113
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$begingroup$
The application of Third Isomorphism Theorem is correct; and the quotient ring is isomorphic to $mathbb{C}[y]$.
$endgroup$
– Armando j18eos
Dec 9 '18 at 13:10