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Math Help - Affine varieties

  1. #1
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    Affine varieties

    1. Show that a map between affine varieties can be continuous for the
    Zariski topology without being regular.
    2. Show that the circle x^{2}+y^{2}=1 is isomorphic (as an affine variety)
    to the hyperbola xy=1, but neither is isomorphic to A^{1}.
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  2. #2
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    i'll assume that the base field is \mathbb{C}.

    Quote Originally Posted by temanratm View Post
    1. Show that a map between affine varieties can be continuous for the
    Zariski topology without being regular.[/tex]
    define f: \mathbb{A}^1 \rightarrow \mathbb{A}^1 by f(0)=0, and f(t)=\frac{1}{t}, \ \forall \ t \neq 0.

    2. Show that the circle x^{2}+y^{2}=1 is isomorphic (as an affine variety)
    to the hyperbola xy=1.
    let <x^2+y^2-1>=I, and <xy-1>=J. also for any f(x,y) \in \mathbb{C}[x,y] let f^{*}(x,y)=f \left(\frac{x+y}{2}, \frac{i(y-x)}{2} \right). define the map

    \phi: \frac{\mathbb{C}[x,y]}{I} \rightarrow \frac{\mathbb{C}[x,y]}{J}, by: \phi(f(x,y) + I)=f^{*}(x,y)+J. now if f(x,y) \in I, then f(x,y)=(x^2+y^2-1)g(x,y), for some

    g(x,y) \in \mathbb{C}[x,y]. but then: f^{*}(x,y)=(xy-1)g \left(\frac{x+y}{2}, \frac{i(y-x)}{2} \right), i.e. f^{*}(x,y) \in J. so \phi is well-defined.clearly \phi is a surjective

    ring homomorphism. now suppose f \in \ker \phi. then f^{*}(x,y) \in J, i.e. f^{*}(x,y)= (xy-1)h(x,y), for some h(x,y) \in \mathbb{C}[x,y]. therefore:

    f(x,y)=f^{*}(x+iy,x-iy)=(x^2+y^2-1)h(x+iy,x-iy). so f(x,y) \in I. hence \ker \phi =0. so \phi is injective and we're done!


    but neither is isomorphic to A^{1}.
    1) R_1=\frac{\mathbb{C}[x,y]}{I} and \mathbb{C}[t] are not isomorphic because R_1 is not a UFD: (y + I)(y+I)=(1-x + I)(1+x + I).

    2) R_2=\frac{\mathbb{C}[x,y]}{J} and \mathbb{C}[t] are not isomoprphic because, if there was an isomorphism \psi: R_2 \rightarrow \mathbb{C}[t], then assuming that

    \psi(x+J)=p(t), \ \psi(y+J) = q(t), we'll get 0=\psi(0)=\psi(xy-1+J)=p(t)q(t)-1. thus p(t)q(t)=1, and therefore p(t), q(t)

    are constant. but then \psi would not be surjective. Q.E.D.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    i'll assume that the base field is \mathbb{C}.



    define f: \mathbb{A}^1 \rightarrow \mathbb{A}^1 by f(0)=0, and f(t)=\frac{1}{t}, \ \forall \ t \neq 0.



    let <x^2+y^2-1>=I, and <xy-1>=J. also for any f(x,y) \in \mathbb{C}[x,y] let f^{*}(x,y)=f \left(\frac{x+y}{2}, \frac{i(y-x)}{2} \right). define the map

    \phi: \frac{\mathbb{C}[x,y]}{I} \rightarrow \frac{\mathbb{C}[x,y]}{J}, by: \phi(f(x,y) + I)=f^{*}(x,y)+J. now if f(x,y) \in I, then f(x,y)=(x^2+y^2-1)g(x,y), for some

    g(x,y) \in \mathbb{C}[x,y]. but then: f^{*}(x,y)=(xy-1)g \left(\frac{x+y}{2}, \frac{i(y-x)}{2} \right), i.e. f^{*}(x,y) \in J. so \phi is well-defined.clearly \phi is a surjective

    ring homomorphism. now suppose f \in \ker \phi. then f^{*}(x,y) \in J, i.e. f^{*}(x,y)= (xy-1)h(x,y), for some h(x,y) \in \mathbb{C}[x,y]. therefore:

    f(x,y)=f^{*}(x+iy,x-iy)=(x^2+y^2-1)h(x+iy,x-iy). so f(x,y) \in I. hence \ker \phi =0. so \phi is injective and we're done!




    1) R_1=\frac{\mathbb{C}[x,y]}{I} and \mathbb{C}[t] are not isomorphic because R_1 is not a UFD: (y + I)(y+I)=(1-x + I)(1+x + I).

    2) R_2=\frac{\mathbb{C}[x,y]}{J} and \mathbb{C}[t] are not isomoprphic because, if there was an isomorphism \psi: R_2 \rightarrow \mathbb{C}[t], then assuming that

    \psi(x+J)=p(t), \ \psi(y+J) = q(t), we'll get 0=\psi(0)=\psi(xy-1+J)=p(t)q(t)-1. thus p(t)q(t)=1, and therefore p(t), q(t)

    are constant. but then \psi would not be surjective. Q.E.D.
    Thank you a lot
    I've one more problem

    Let q be a power of a prime p and let F_{q} be the field with q elements.
    S is a subset of F_{q}[X_{1},...,X_{n}] and let V be ots zero set in k^{n}.
    where k is alg-c closure of F_{q}.
    Show that the map (a_{1}, .... a_{n})->((a_{1})^q,...,(a_{n})^q) is a regular map
    f:V->V (i.e., f(v)subset of V)
    Verify that the set of fixed points of f(x) is the set of zeros of the elements
    of S with coordinates in F_{q}.

    I would be very grateful
    thx in advance
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  4. #4
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    Quote Originally Posted by temanratm View Post
    Thank you a lot
    I've one more problem

    Let q be a power of a prime p and let F_{q} be the field with q elements.
    S is a subset of F_{q}[X_{1},...,X_{n}] and let V be its zero set in k^{n}.
    where k is alg-c closure of F_{q}.
    Show that the map (a_{1}, .... a_{n})->((a_{1})^q,...,(a_{n})^q) is a regular map
    f:V->V (i.e., f(v)subset of V)
    let v = (a_1, \ ... \ ,a_n) \in V, and g(x_1, \ ... \ ,x_n)=\sum \beta_{i_1, \ ... , \ i_n}x_1^{i_1} \ ... \ x_n^{i_n} \in S. since \beta_{i_1, \ ... \ ,i_n} \in \mathbb{F}_q,

    we'll have: (g(x_1, \ ... \ ,x_n))^q=\sum \beta_{i_1, \ ... \ ,i_n} x_1^{qi_1} \ ... \ , x_n^{qi_n}=g(x_1^q, \ ... \ , x_n^q). therefore:

    g(a_1^q, \ ... \ , a_n^q) = (g(a_1, \ ... \ , a_n))^q = (g(v))^q=0, i.e. f(v)=(a_1^q, \ ... \ ,a_n^q) \in V. \ \ \ \square

    Verify that the set of fixed points of f(x) is the set of zeros of the elements
    of S with coordinates in F_{q}.
    well, this is pretty obvious: (a_1, \ ... \ , a_n) is a fixed point of f if and only if a_j^q=a_j, \ j=1, \ ... \ , n.

    so each a_j is a root of the polynomial h(x)=x^q - x \in k[x]. but every element of \mathbb{F}_q is also a root

    of h(x), and obviously h(x) has exactly q=|\mathbb{F}_q| roots in k. thus a_j \in \mathbb{F}_q, \ j=1, \ ... \ , n. \ \ \ \square
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