Fulton 1.2
Posted on November 15, 20171.8.* Show that the algebraic subsets of are just the finite subsets, together with itself.
Suppose for some If then Otherwise let Note Since has finitely many roots, is finite. Thus is finite.
Conversely, by (4) any finite subset of is algebraic.
1.9. If is a finite field, show that every subset of is algebraic.
Since is finite, is finite. Thus every subset of is finite and therefore algebraic by (4).
1.10. Give an example of a countable collection of algebraic sets whose union is not algebraic.
Let We claim is not algebraic in
Suppose is algebraic. Then for some Observe Thus is infinite, a contradiction.
1.11. Show that the following are algebraic sets:
(a) ;
Observe
(b) ;
It is easy to check
(c) the set of points in whose polar coordinates satisfy the equation
It is can be verified that this is the set
1.12. Suppose is an affine plane curve, and is a line in Suppose a polynomial of degree Show that is a finite set of no more than points.
Suppose where and at least one of or is nonzero. Without loss of generality, suppose Then where and
If then Thus if then Observe with degree Thus has at most roots and contains at most points.
1.13. Show that each of the following sets is not algebraic:
(a) ;
Suppose Clearly so Let Then But is infinite. Thus is infinite, a contradiction to Problem 1.12.
(b) ;
Suppose Clearly Let Observe an infinite set. Thus is infinite, a contradiction to Problem 1.12.
(c)
Suppose Clearly Let Note So for all Thus for any for any But fixing this gives with for any This is a contradiction as must have finitely many roots in
1.14.* Let be a nonconstant polynomial in algebraically closed. Show that is infinite if
Suppose Then and has finitely many roots. Since is algebraically closed, is infinite by Problem 1.6. Thus is infinite.
Suppose Let Then Note has finitely many roots Observe Therefore is infinite.
Show is infinite if
Since is nonconstant there exists such that is nonconstant, with arbitrary Since is algebraically closed and nonconstant, has a root in Thus for each selection of infinite, there is a distinct root of in Thus is infinite.
1.15.*Let be algebraic sets. Show that is an algebraic set.
Let with Let us define by Similarly define by It is simple to see