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}.
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}.
i'll assume that the base field is
define by and
let and also for any let define the map2. Show that the circle x^{2}+y^{2}=1 is isomorphic (as an affine variety)
to the hyperbola xy=1.
by: now if then for some
but then: i.e. so is well-defined.clearly is a surjective
ring homomorphism. now suppose then i.e. for some therefore:
so hence so is injective and we're done!
1) and are not isomorphic because is not a UFD:but neither is isomorphic to A^{1}.
2) and are not isomoprphic because, if there was an isomorphism then assuming that
we'll get thus and therefore
are constant. but then 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
let and since
we'll have: therefore:
i.e.
well, this is pretty obvious: is a fixed point of if and only ifVerify that the set of fixed points of f(x) is the set of zeros of the elements
of S with coordinates in F_{q}.
so each is a root of the polynomial but every element of is also a root
of and obviously has exactly roots in thus