How to show that $phi: mathbb{A}^1(mathbb{C}) to V(Y^2 - X^3)$ is no isomorphism.
$begingroup$
Consider the polynomial function
$$phi: mathbb{A}^1(mathbb{C})to V(Y^2 - X^3): t mapsto (t^2, t^3)$$
Show that $phi$ is a bijective polynomial function, but no isomorphism (= $phi^{-1}$ is no polynomial function).
I proved that $phi$ is bijective and $phi = (X^2, X^ 3)$, thus $phi$ is a polynomial function. How would I show that $phi^{-1}$ is no polynomial function?
abstract-algebra algebraic-geometry polynomials affine-varieties
asked Dec 17 '18 at 14:14
Math_QEDMath_QED
7,64031452
$endgroup$
add a comment |
$begingroup$
Consider the polynomial function
$$phi: mathbb{A}^1(mathbb{C})to V(Y^2 - X^3): t mapsto (t^2, t^3)$$
Show that $phi$ is a bijective polynomial function, but no isomorphism (= $phi^{-1}$ is no polynomial function).
I proved that $phi$ is bijective and $phi = (X^2, X^ 3)$, thus $phi$ is a polynomial function. How would I show that $phi^{-1}$ is no polynomial function?
abstract-algebra algebraic-geometry polynomials affine-varieties
asked Dec 17 '18 at 14:14
Math_QEDMath_QED
7,64031452
$endgroup$
1
$begingroup$
Do you know about tangent spaces?
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:26
$begingroup$
@LukasKofler No didn't have differential geometry yet.
$endgroup$
– Math_QED
Dec 17 '18 at 14:27
1
$begingroup$
Me neither! You can define them purely algebraically and show that the tangent spaces at the origin are different, so the varieties are not isomorphic. I‘m sure there is another proof though.
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:28
$begingroup$
Yeah, I'm looking for something elementary. Thanks though.
$endgroup$
– Math_QED
Dec 17 '18 at 15:02
add a comment |
$begingroup$
Consider the polynomial function
$$phi: mathbb{A}^1(mathbb{C})to V(Y^2 - X^3): t mapsto (t^2, t^3)$$
Show that $phi$ is a bijective polynomial function, but no isomorphism (= $phi^{-1}$ is no polynomial function).
I proved that $phi$ is bijective and $phi = (X^2, X^ 3)$, thus $phi$ is a polynomial function. How would I show that $phi^{-1}$ is no polynomial function?
abstract-algebra algebraic-geometry polynomials affine-varieties
asked Dec 17 '18 at 14:14
Math_QEDMath_QED
7,64031452
$endgroup$
Consider the polynomial function
$$phi: mathbb{A}^1(mathbb{C})to V(Y^2 - X^3): t mapsto (t^2, t^3)$$
Show that $phi$ is a bijective polynomial function, but no isomorphism (= $phi^{-1}$ is no polynomial function).
I proved that $phi$ is bijective and $phi = (X^2, X^ 3)$, thus $phi$ is a polynomial function. How would I show that $phi^{-1}$ is no polynomial function?
abstract-algebra algebraic-geometry polynomials affine-varieties
abstract-algebra algebraic-geometry polynomials affine-varieties
asked Dec 17 '18 at 14:14
Math_QEDMath_QED
7,64031452
asked Dec 17 '18 at 14:14
Math_QEDMath_QED
7,64031452
edited Dec 17 '18 at 15:04
Math_QED
asked Dec 17 '18 at 14:14
Math_QEDMath_QED
7,64031452
asked Dec 17 '18 at 14:14
Math_QEDMath_QED
7,64031452
asked Dec 17 '18 at 14:14
Math_QEDMath_QED
7,64031452
7,64031452
1
$begingroup$
Do you know about tangent spaces?
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:26
$begingroup$
@LukasKofler No didn't have differential geometry yet.
$endgroup$
– Math_QED
Dec 17 '18 at 14:27
1
$begingroup$
Me neither! You can define them purely algebraically and show that the tangent spaces at the origin are different, so the varieties are not isomorphic. I‘m sure there is another proof though.
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:28
$begingroup$
Yeah, I'm looking for something elementary. Thanks though.
$endgroup$
– Math_QED
Dec 17 '18 at 15:02
add a comment |
1
$begingroup$
Do you know about tangent spaces?
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:26
$begingroup$
@LukasKofler No didn't have differential geometry yet.
$endgroup$
– Math_QED
Dec 17 '18 at 14:27
1
$begingroup$
Me neither! You can define them purely algebraically and show that the tangent spaces at the origin are different, so the varieties are not isomorphic. I‘m sure there is another proof though.
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:28
$begingroup$
Yeah, I'm looking for something elementary. Thanks though.
$endgroup$
– Math_QED
Dec 17 '18 at 15:02
1
1
$begingroup$
Do you know about tangent spaces?
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:26
$begingroup$
Do you know about tangent spaces?
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:26
$begingroup$
@LukasKofler No didn't have differential geometry yet.
$endgroup$
– Math_QED
Dec 17 '18 at 14:27
$begingroup$
@LukasKofler No didn't have differential geometry yet.
$endgroup$
– Math_QED
Dec 17 '18 at 14:27
1
1
$begingroup$
Me neither! You can define them purely algebraically and show that the tangent spaces at the origin are different, so the varieties are not isomorphic. I‘m sure there is another proof though.
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:28
$begingroup$
Me neither! You can define them purely algebraically and show that the tangent spaces at the origin are different, so the varieties are not isomorphic. I‘m sure there is another proof though.
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:28
$begingroup$
Yeah, I'm looking for something elementary. Thanks though.
$endgroup$
– Math_QED
Dec 17 '18 at 15:02
$begingroup$
Yeah, I'm looking for something elementary. Thanks though.
$endgroup$
– Math_QED
Dec 17 '18 at 15:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We use that fact that isomorphic varieties have isomorphic coordinate rings.
Consider the coordinate ring of $Bbb A^1, Bbb C[t]$. Note that this is a UFD.
On the other hand, the coordinate ring of $V(Y^2 - X^3)$ is $Bbb C[x, y]/(Y^2 - X^3)$. It is not a UFD since we have $Y^2 = X^3$.
Therefore $Bbb A^1 ncong V(Y^2 - X^3)$.
answered Dec 17 '18 at 15:29
Lukas KoflerLukas Kofler
1,2552520
$endgroup$
1
$begingroup$
Thanks. That was very elementary.
$endgroup$
– Math_QED
Dec 17 '18 at 15:31
4
$begingroup$
A closely related argument would be to state that the coefficient ring $Bbb{C}[t]$ is integrally closed (in its field of fractions), but the other coordinate ring is not. This is because $t=Y/X$ is integral over it ($t^2=X$). For curves smoothness is equivalent to coordinate ring being integrally closed. Apologies if you knew about this already.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 15:48
$begingroup$
I didn't, so it was surely helpful to me :)
$endgroup$
– Math_QED
Dec 17 '18 at 15:50
add a comment |
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$begingroup$
We use that fact that isomorphic varieties have isomorphic coordinate rings.
Consider the coordinate ring of $Bbb A^1, Bbb C[t]$. Note that this is a UFD.
On the other hand, the coordinate ring of $V(Y^2 - X^3)$ is $Bbb C[x, y]/(Y^2 - X^3)$. It is not a UFD since we have $Y^2 = X^3$.
Therefore $Bbb A^1 ncong V(Y^2 - X^3)$.
answered Dec 17 '18 at 15:29
Lukas KoflerLukas Kofler
1,2552520
$endgroup$
1
$begingroup$
Thanks. That was very elementary.
$endgroup$
– Math_QED
Dec 17 '18 at 15:31
4
$begingroup$
A closely related argument would be to state that the coefficient ring $Bbb{C}[t]$ is integrally closed (in its field of fractions), but the other coordinate ring is not. This is because $t=Y/X$ is integral over it ($t^2=X$). For curves smoothness is equivalent to coordinate ring being integrally closed. Apologies if you knew about this already.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 15:48
$begingroup$
I didn't, so it was surely helpful to me :)
$endgroup$
– Math_QED
Dec 17 '18 at 15:50
add a comment |
$begingroup$
We use that fact that isomorphic varieties have isomorphic coordinate rings.
Consider the coordinate ring of $Bbb A^1, Bbb C[t]$. Note that this is a UFD.
On the other hand, the coordinate ring of $V(Y^2 - X^3)$ is $Bbb C[x, y]/(Y^2 - X^3)$. It is not a UFD since we have $Y^2 = X^3$.
Therefore $Bbb A^1 ncong V(Y^2 - X^3)$.
answered Dec 17 '18 at 15:29
Lukas KoflerLukas Kofler
1,2552520
$endgroup$
1
$begingroup$
Thanks. That was very elementary.
$endgroup$
– Math_QED
Dec 17 '18 at 15:31
4
$begingroup$
A closely related argument would be to state that the coefficient ring $Bbb{C}[t]$ is integrally closed (in its field of fractions), but the other coordinate ring is not. This is because $t=Y/X$ is integral over it ($t^2=X$). For curves smoothness is equivalent to coordinate ring being integrally closed. Apologies if you knew about this already.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 15:48
$begingroup$
I didn't, so it was surely helpful to me :)
$endgroup$
– Math_QED
Dec 17 '18 at 15:50
add a comment |
$begingroup$
We use that fact that isomorphic varieties have isomorphic coordinate rings.
Consider the coordinate ring of $Bbb A^1, Bbb C[t]$. Note that this is a UFD.
On the other hand, the coordinate ring of $V(Y^2 - X^3)$ is $Bbb C[x, y]/(Y^2 - X^3)$. It is not a UFD since we have $Y^2 = X^3$.
Therefore $Bbb A^1 ncong V(Y^2 - X^3)$.
answered Dec 17 '18 at 15:29
Lukas KoflerLukas Kofler
1,2552520
$endgroup$
We use that fact that isomorphic varieties have isomorphic coordinate rings.
Consider the coordinate ring of $Bbb A^1, Bbb C[t]$. Note that this is a UFD.
On the other hand, the coordinate ring of $V(Y^2 - X^3)$ is $Bbb C[x, y]/(Y^2 - X^3)$. It is not a UFD since we have $Y^2 = X^3$.
Therefore $Bbb A^1 ncong V(Y^2 - X^3)$.
answered Dec 17 '18 at 15:29
Lukas KoflerLukas Kofler
1,2552520
answered Dec 17 '18 at 15:29
Lukas KoflerLukas Kofler
1,2552520
answered Dec 17 '18 at 15:29
Lukas KoflerLukas Kofler
1,2552520
answered Dec 17 '18 at 15:29
Lukas KoflerLukas Kofler
1,2552520
1,2552520
1
$begingroup$
Thanks. That was very elementary.
$endgroup$
– Math_QED
Dec 17 '18 at 15:31
4
$begingroup$
A closely related argument would be to state that the coefficient ring $Bbb{C}[t]$ is integrally closed (in its field of fractions), but the other coordinate ring is not. This is because $t=Y/X$ is integral over it ($t^2=X$). For curves smoothness is equivalent to coordinate ring being integrally closed. Apologies if you knew about this already.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 15:48
$begingroup$
I didn't, so it was surely helpful to me :)
$endgroup$
– Math_QED
Dec 17 '18 at 15:50
add a comment |
1
$begingroup$
Thanks. That was very elementary.
$endgroup$
– Math_QED
Dec 17 '18 at 15:31
4
$begingroup$
A closely related argument would be to state that the coefficient ring $Bbb{C}[t]$ is integrally closed (in its field of fractions), but the other coordinate ring is not. This is because $t=Y/X$ is integral over it ($t^2=X$). For curves smoothness is equivalent to coordinate ring being integrally closed. Apologies if you knew about this already.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 15:48
$begingroup$
I didn't, so it was surely helpful to me :)
$endgroup$
– Math_QED
Dec 17 '18 at 15:50
1
1
$begingroup$
Thanks. That was very elementary.
$endgroup$
– Math_QED
Dec 17 '18 at 15:31
$begingroup$
Thanks. That was very elementary.
$endgroup$
– Math_QED
Dec 17 '18 at 15:31
4
4
$begingroup$
A closely related argument would be to state that the coefficient ring $Bbb{C}[t]$ is integrally closed (in its field of fractions), but the other coordinate ring is not. This is because $t=Y/X$ is integral over it ($t^2=X$). For curves smoothness is equivalent to coordinate ring being integrally closed. Apologies if you knew about this already.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 15:48
$begingroup$
A closely related argument would be to state that the coefficient ring $Bbb{C}[t]$ is integrally closed (in its field of fractions), but the other coordinate ring is not. This is because $t=Y/X$ is integral over it ($t^2=X$). For curves smoothness is equivalent to coordinate ring being integrally closed. Apologies if you knew about this already.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 15:48
$begingroup$
I didn't, so it was surely helpful to me :)
$endgroup$
– Math_QED
Dec 17 '18 at 15:50
$begingroup$
I didn't, so it was surely helpful to me :)
$endgroup$
– Math_QED
Dec 17 '18 at 15:50
add a comment |
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1
$begingroup$
Do you know about tangent spaces?
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:26
$begingroup$
@LukasKofler No didn't have differential geometry yet.
$endgroup$
– Math_QED
Dec 17 '18 at 14:27
1
$begingroup$
Me neither! You can define them purely algebraically and show that the tangent spaces at the origin are different, so the varieties are not isomorphic. I‘m sure there is another proof though.
$endgroup$
– Lukas Kofler
Dec 17 '18 at 14:28
$begingroup$
Yeah, I'm looking for something elementary. Thanks though.
$endgroup$
– Math_QED
Dec 17 '18 at 15:02