Prove a set $V$ is not algebraic












1














I need help showing that the set $V = {(a,b) in mathbb{A}^2(mathbb{C}) vert vert avert^2 + vert bvert^2 = 1}$ is not an algebraic subset of complex affine 2-space.



I believe that I can just choose two complex numbers $a$ and $b$ with moduli that sum to 1 and just show that $(a,b) neq 0$. For example, if I choose $a = i$ and $b=0$, then $$vert avert^2 + vert bvert^2 = vert ivert^2 + vert 0vert^2 = 1^2 = 1,$$ but $(a,b) = (i,0) neq (0,0)$, and we are done.



Is this approach correct? Thanks in advance.






affine-geometry







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  • But $- a^2+b^2=1$ is algebraic, and you can pick the same point lying on it, so not sure I understand your argument.
    – Matt Samuel
    Dec 11 '18 at 4:56






  • 4




    No, the approach is not correct. What do you think the definition of "algebraic set" is?
    – Alex Kruckman
    Dec 11 '18 at 4:56








  • 1




    @mathgeen Please don't change the question in a way that invalidates the answers you've received. If you'd like to change the question in such a fundamental way, just ask a new one.
    – André 3000
    Dec 11 '18 at 8:24
















1














I need help showing that the set $V = {(a,b) in mathbb{A}^2(mathbb{C}) vert vert avert^2 + vert bvert^2 = 1}$ is not an algebraic subset of complex affine 2-space.



I believe that I can just choose two complex numbers $a$ and $b$ with moduli that sum to 1 and just show that $(a,b) neq 0$. For example, if I choose $a = i$ and $b=0$, then $$vert avert^2 + vert bvert^2 = vert ivert^2 + vert 0vert^2 = 1^2 = 1,$$ but $(a,b) = (i,0) neq (0,0)$, and we are done.



Is this approach correct? Thanks in advance.






affine-geometry







share|cite|improve this question
























  • But $- a^2+b^2=1$ is algebraic, and you can pick the same point lying on it, so not sure I understand your argument.
    – Matt Samuel
    Dec 11 '18 at 4:56






  • 4




    No, the approach is not correct. What do you think the definition of "algebraic set" is?
    – Alex Kruckman
    Dec 11 '18 at 4:56








  • 1




    @mathgeen Please don't change the question in a way that invalidates the answers you've received. If you'd like to change the question in such a fundamental way, just ask a new one.
    – André 3000
    Dec 11 '18 at 8:24














1












1








1







I need help showing that the set $V = {(a,b) in mathbb{A}^2(mathbb{C}) vert vert avert^2 + vert bvert^2 = 1}$ is not an algebraic subset of complex affine 2-space.



I believe that I can just choose two complex numbers $a$ and $b$ with moduli that sum to 1 and just show that $(a,b) neq 0$. For example, if I choose $a = i$ and $b=0$, then $$vert avert^2 + vert bvert^2 = vert ivert^2 + vert 0vert^2 = 1^2 = 1,$$ but $(a,b) = (i,0) neq (0,0)$, and we are done.



Is this approach correct? Thanks in advance.






affine-geometry







share|cite|improve this question















I need help showing that the set $V = {(a,b) in mathbb{A}^2(mathbb{C}) vert vert avert^2 + vert bvert^2 = 1}$ is not an algebraic subset of complex affine 2-space.



I believe that I can just choose two complex numbers $a$ and $b$ with moduli that sum to 1 and just show that $(a,b) neq 0$. For example, if I choose $a = i$ and $b=0$, then $$vert avert^2 + vert bvert^2 = vert ivert^2 + vert 0vert^2 = 1^2 = 1,$$ but $(a,b) = (i,0) neq (0,0)$, and we are done.



Is this approach correct? Thanks in advance.






affine-geometry




affine-geometry






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share|cite|improve this question













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edited Dec 13 '18 at 8:52

























asked Dec 11 '18 at 4:50









mathgeen

224




224












  • But $- a^2+b^2=1$ is algebraic, and you can pick the same point lying on it, so not sure I understand your argument.
    – Matt Samuel
    Dec 11 '18 at 4:56






  • 4




    No, the approach is not correct. What do you think the definition of "algebraic set" is?
    – Alex Kruckman
    Dec 11 '18 at 4:56








  • 1




    @mathgeen Please don't change the question in a way that invalidates the answers you've received. If you'd like to change the question in such a fundamental way, just ask a new one.
    – André 3000
    Dec 11 '18 at 8:24


















  • But $- a^2+b^2=1$ is algebraic, and you can pick the same point lying on it, so not sure I understand your argument.
    – Matt Samuel
    Dec 11 '18 at 4:56






  • 4




    No, the approach is not correct. What do you think the definition of "algebraic set" is?
    – Alex Kruckman
    Dec 11 '18 at 4:56








  • 1




    @mathgeen Please don't change the question in a way that invalidates the answers you've received. If you'd like to change the question in such a fundamental way, just ask a new one.
    – André 3000
    Dec 11 '18 at 8:24
















But $- a^2+b^2=1$ is algebraic, and you can pick the same point lying on it, so not sure I understand your argument.
– Matt Samuel
Dec 11 '18 at 4:56




But $- a^2+b^2=1$ is algebraic, and you can pick the same point lying on it, so not sure I understand your argument.
– Matt Samuel
Dec 11 '18 at 4:56




4




4




No, the approach is not correct. What do you think the definition of "algebraic set" is?
– Alex Kruckman
Dec 11 '18 at 4:56






No, the approach is not correct. What do you think the definition of "algebraic set" is?
– Alex Kruckman
Dec 11 '18 at 4:56






1




1




@mathgeen Please don't change the question in a way that invalidates the answers you've received. If you'd like to change the question in such a fundamental way, just ask a new one.
– André 3000
Dec 11 '18 at 8:24




@mathgeen Please don't change the question in a way that invalidates the answers you've received. If you'd like to change the question in such a fundamental way, just ask a new one.
– André 3000
Dec 11 '18 at 8:24










2 Answers
2






active

oldest

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2














If this set were algebraic, then its intersection with the set ${b=0}$ would
be an algebraic subset of $Bbb A^1(Bbb C)$. But that intersection is
${ainBbb C:|a|^2=1}$. But apart from $Bbb A^1(Bbb C)$ itself,
the algebraic subsets of $Bbb A^1(Bbb C)$ are all finite.






share|cite|improve this answer





















  • Yes, intersections of algebraic sets are algebraic. $b=0$ is a polynomial equation.
    – Lord Shark the Unknown
    Dec 11 '18 at 7:24



















1














Whether or not $(a,b) = (0,0)$ has nothing to do with anything. For example ${xy = 1}$ is algebraic and doesn't contain $(0,0)$. Recall that a set $V$ is algebraic if it is the vanishing set of some collection of polynomials.



There are a couple strategies to show that a set is not algebraic. First, you can take a polynomial that vanishes on $V$ and show that it must always vanish at some other point not in $V$ (where "must always" means this is true for any polynomial that vanishes on $V$, not just a specific one). Think about why this is. Suppose $V = V(f_1,dots,f_r)$ and each $f_i$ vanishes at some other point not in $V$.



Another strategy is to consider dimensions and degrees and stuff like that. A common example for this is the set ${(t,sin(t)) : t in mathbf{R}} subseteq mathbf{R}^2$. Then when you intersect this with the line ${y = 0}$ you get infinitely many points. But what sort of polynomial has infinitely many zeroes?



So, what sort of polynomials vanish on your set? That is the first question you want to ask.






share|cite|improve this answer





















  • @mathgeen An example of such a property is that if $f(x,y)$ vanishes on $V$ then $f(e^{it}x,e^{is}y)$ also vanishes on $V$. You can also try the second strategy, for example $V cap {b = 0} = {a in mathbf C : |a| = 1}$. One other theme that shows up is when you have two polynomials $f(x,y)$ and $g(x,y)$ with infinitely many common zeroes then what conclusions can we draw.
    – Trevor Gunn
    Dec 11 '18 at 5:23











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2 Answers
2






active

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2 Answers
2






active

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active

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active

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votes









2














If this set were algebraic, then its intersection with the set ${b=0}$ would
be an algebraic subset of $Bbb A^1(Bbb C)$. But that intersection is
${ainBbb C:|a|^2=1}$. But apart from $Bbb A^1(Bbb C)$ itself,
the algebraic subsets of $Bbb A^1(Bbb C)$ are all finite.






share|cite|improve this answer





















  • Yes, intersections of algebraic sets are algebraic. $b=0$ is a polynomial equation.
    – Lord Shark the Unknown
    Dec 11 '18 at 7:24
















2














If this set were algebraic, then its intersection with the set ${b=0}$ would
be an algebraic subset of $Bbb A^1(Bbb C)$. But that intersection is
${ainBbb C:|a|^2=1}$. But apart from $Bbb A^1(Bbb C)$ itself,
the algebraic subsets of $Bbb A^1(Bbb C)$ are all finite.






share|cite|improve this answer





















  • Yes, intersections of algebraic sets are algebraic. $b=0$ is a polynomial equation.
    – Lord Shark the Unknown
    Dec 11 '18 at 7:24














2












2








2






If this set were algebraic, then its intersection with the set ${b=0}$ would
be an algebraic subset of $Bbb A^1(Bbb C)$. But that intersection is
${ainBbb C:|a|^2=1}$. But apart from $Bbb A^1(Bbb C)$ itself,
the algebraic subsets of $Bbb A^1(Bbb C)$ are all finite.






share|cite|improve this answer












If this set were algebraic, then its intersection with the set ${b=0}$ would
be an algebraic subset of $Bbb A^1(Bbb C)$. But that intersection is
${ainBbb C:|a|^2=1}$. But apart from $Bbb A^1(Bbb C)$ itself,
the algebraic subsets of $Bbb A^1(Bbb C)$ are all finite.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '18 at 5:22









Lord Shark the Unknown

101k958132




101k958132












  • Yes, intersections of algebraic sets are algebraic. $b=0$ is a polynomial equation.
    – Lord Shark the Unknown
    Dec 11 '18 at 7:24


















  • Yes, intersections of algebraic sets are algebraic. $b=0$ is a polynomial equation.
    – Lord Shark the Unknown
    Dec 11 '18 at 7:24
















Yes, intersections of algebraic sets are algebraic. $b=0$ is a polynomial equation.
– Lord Shark the Unknown
Dec 11 '18 at 7:24




Yes, intersections of algebraic sets are algebraic. $b=0$ is a polynomial equation.
– Lord Shark the Unknown
Dec 11 '18 at 7:24











1














Whether or not $(a,b) = (0,0)$ has nothing to do with anything. For example ${xy = 1}$ is algebraic and doesn't contain $(0,0)$. Recall that a set $V$ is algebraic if it is the vanishing set of some collection of polynomials.



There are a couple strategies to show that a set is not algebraic. First, you can take a polynomial that vanishes on $V$ and show that it must always vanish at some other point not in $V$ (where "must always" means this is true for any polynomial that vanishes on $V$, not just a specific one). Think about why this is. Suppose $V = V(f_1,dots,f_r)$ and each $f_i$ vanishes at some other point not in $V$.



Another strategy is to consider dimensions and degrees and stuff like that. A common example for this is the set ${(t,sin(t)) : t in mathbf{R}} subseteq mathbf{R}^2$. Then when you intersect this with the line ${y = 0}$ you get infinitely many points. But what sort of polynomial has infinitely many zeroes?



So, what sort of polynomials vanish on your set? That is the first question you want to ask.






share|cite|improve this answer





















  • @mathgeen An example of such a property is that if $f(x,y)$ vanishes on $V$ then $f(e^{it}x,e^{is}y)$ also vanishes on $V$. You can also try the second strategy, for example $V cap {b = 0} = {a in mathbf C : |a| = 1}$. One other theme that shows up is when you have two polynomials $f(x,y)$ and $g(x,y)$ with infinitely many common zeroes then what conclusions can we draw.
    – Trevor Gunn
    Dec 11 '18 at 5:23
















1














Whether or not $(a,b) = (0,0)$ has nothing to do with anything. For example ${xy = 1}$ is algebraic and doesn't contain $(0,0)$. Recall that a set $V$ is algebraic if it is the vanishing set of some collection of polynomials.



There are a couple strategies to show that a set is not algebraic. First, you can take a polynomial that vanishes on $V$ and show that it must always vanish at some other point not in $V$ (where "must always" means this is true for any polynomial that vanishes on $V$, not just a specific one). Think about why this is. Suppose $V = V(f_1,dots,f_r)$ and each $f_i$ vanishes at some other point not in $V$.



Another strategy is to consider dimensions and degrees and stuff like that. A common example for this is the set ${(t,sin(t)) : t in mathbf{R}} subseteq mathbf{R}^2$. Then when you intersect this with the line ${y = 0}$ you get infinitely many points. But what sort of polynomial has infinitely many zeroes?



So, what sort of polynomials vanish on your set? That is the first question you want to ask.






share|cite|improve this answer





















  • @mathgeen An example of such a property is that if $f(x,y)$ vanishes on $V$ then $f(e^{it}x,e^{is}y)$ also vanishes on $V$. You can also try the second strategy, for example $V cap {b = 0} = {a in mathbf C : |a| = 1}$. One other theme that shows up is when you have two polynomials $f(x,y)$ and $g(x,y)$ with infinitely many common zeroes then what conclusions can we draw.
    – Trevor Gunn
    Dec 11 '18 at 5:23














1












1








1






Whether or not $(a,b) = (0,0)$ has nothing to do with anything. For example ${xy = 1}$ is algebraic and doesn't contain $(0,0)$. Recall that a set $V$ is algebraic if it is the vanishing set of some collection of polynomials.



There are a couple strategies to show that a set is not algebraic. First, you can take a polynomial that vanishes on $V$ and show that it must always vanish at some other point not in $V$ (where "must always" means this is true for any polynomial that vanishes on $V$, not just a specific one). Think about why this is. Suppose $V = V(f_1,dots,f_r)$ and each $f_i$ vanishes at some other point not in $V$.



Another strategy is to consider dimensions and degrees and stuff like that. A common example for this is the set ${(t,sin(t)) : t in mathbf{R}} subseteq mathbf{R}^2$. Then when you intersect this with the line ${y = 0}$ you get infinitely many points. But what sort of polynomial has infinitely many zeroes?



So, what sort of polynomials vanish on your set? That is the first question you want to ask.






share|cite|improve this answer












Whether or not $(a,b) = (0,0)$ has nothing to do with anything. For example ${xy = 1}$ is algebraic and doesn't contain $(0,0)$. Recall that a set $V$ is algebraic if it is the vanishing set of some collection of polynomials.



There are a couple strategies to show that a set is not algebraic. First, you can take a polynomial that vanishes on $V$ and show that it must always vanish at some other point not in $V$ (where "must always" means this is true for any polynomial that vanishes on $V$, not just a specific one). Think about why this is. Suppose $V = V(f_1,dots,f_r)$ and each $f_i$ vanishes at some other point not in $V$.



Another strategy is to consider dimensions and degrees and stuff like that. A common example for this is the set ${(t,sin(t)) : t in mathbf{R}} subseteq mathbf{R}^2$. Then when you intersect this with the line ${y = 0}$ you get infinitely many points. But what sort of polynomial has infinitely many zeroes?



So, what sort of polynomials vanish on your set? That is the first question you want to ask.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 11 '18 at 5:02









Trevor Gunn

14.2k32046




14.2k32046












  • @mathgeen An example of such a property is that if $f(x,y)$ vanishes on $V$ then $f(e^{it}x,e^{is}y)$ also vanishes on $V$. You can also try the second strategy, for example $V cap {b = 0} = {a in mathbf C : |a| = 1}$. One other theme that shows up is when you have two polynomials $f(x,y)$ and $g(x,y)$ with infinitely many common zeroes then what conclusions can we draw.
    – Trevor Gunn
    Dec 11 '18 at 5:23


















  • @mathgeen An example of such a property is that if $f(x,y)$ vanishes on $V$ then $f(e^{it}x,e^{is}y)$ also vanishes on $V$. You can also try the second strategy, for example $V cap {b = 0} = {a in mathbf C : |a| = 1}$. One other theme that shows up is when you have two polynomials $f(x,y)$ and $g(x,y)$ with infinitely many common zeroes then what conclusions can we draw.
    – Trevor Gunn
    Dec 11 '18 at 5:23
















@mathgeen An example of such a property is that if $f(x,y)$ vanishes on $V$ then $f(e^{it}x,e^{is}y)$ also vanishes on $V$. You can also try the second strategy, for example $V cap {b = 0} = {a in mathbf C : |a| = 1}$. One other theme that shows up is when you have two polynomials $f(x,y)$ and $g(x,y)$ with infinitely many common zeroes then what conclusions can we draw.
– Trevor Gunn
Dec 11 '18 at 5:23




@mathgeen An example of such a property is that if $f(x,y)$ vanishes on $V$ then $f(e^{it}x,e^{is}y)$ also vanishes on $V$. You can also try the second strategy, for example $V cap {b = 0} = {a in mathbf C : |a| = 1}$. One other theme that shows up is when you have two polynomials $f(x,y)$ and $g(x,y)$ with infinitely many common zeroes then what conclusions can we draw.
– Trevor Gunn
Dec 11 '18 at 5:23


















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