1. ## 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}.

2. i'll assume that the base field is $\mathbb{C}.$

Originally Posted by temanratm
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 $=I,$ and $=J.$ 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.

3. Originally Posted by NonCommAlg
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 $=I,$ and $=J.$ 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

4. Originally Posted by temanratm
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 $a_j \in \mathbb{F}_q, \ j=1, \ ... \ , n. \ \ \ \square$