Prove that a variety is irreducible if is a prime ideal. The definition of an irreducible variety is as follows. Let and be two varieties.
Proof: Let us proceed using proof by contradiction, and assume that is not prime. Let and be two prime ideals; i.e. , for ring . Now, define:
which yields , a reducible, as required
If is nilpotent is unit.
Proof: By binomial expansion,
Prove that is unit iff are nilpotent, and is unit.
Proof: Let , where . Clearly, are units, and by collecting terms, we get:
Since , it follows that , which means is nilpotent. By induction, we can prove that all s for are nilpotent
Prove that, in , the Jacobson radical is equal to the nilradical.
Proof: Let . Then, is unit, and by the previous problem, are all nilpotent, implying
Let be a ring, and be its nilradical. Then, show that the following are equivalent:
has exactly one prime ideal.
Every element of is either unit or nilpotent.
is a field.
Proof: (i) implies that there is exactly one maximal ideal, since every ideal must be contained within some maximal ideal: then, let be this prime/maximal ideal. (iii) follows immediately, since, is a field (see ra/hw).
It follows from (i) that is a local ring with maximal ideal . Let . The ideal generated by and is , since is maximal. Let ,
and hence . This implies (ii), since every element is invertible.
Starting from (iii), we know that the only ideals of are and , by definition of a field. Hence, elements are of the form and ; the second coset generates the entire ring, while the first coset generates the single ideal in , and (i) follows from this
Let ring be such that every ideal not contained in the nilradical contains a nonzero idempotent (i.e. ). Prove that .
Proof: is unit, for some , but . Inverting logic, we get is not unit. Now, plugging , we get is not unit must be contained in some maximal ideal must be contained in every maximal ideal, and we arrive at our contradiction
Let be an ideal in ring that is . Then, show that is an intersection of prime ideals.
Proof: By definition,
Now suppose, for , . Then, , by definition of an ideal, and the chain would not terminate at . Hence, , and this is exactly the definition of the set of nilpotent elements.
Proof, the other way: is the set of nilpotent elements. Let be any element; then, . Since every element of is of this form,
Prove that a local ring contains no idempotent .
Proof: A local ring is defined as having exactly one maximal ideal, . Consider idempotent is unit, and is not contained within . Since is idempotent, is unit as well. Now, since and must generate ,
A ring is boolean if for all . Show that in a Boolean ring :
, for all .
Every prime ideal is maximal, and is a field with two elements.
Every finitely generated ideal in is principal.
Proof: To prove (ii), let . Then, . Since is prime, and , implying that is invertible, and hence is a field.
Let be a ring, and the set of all prime ideals of . For each subset of , let denote the set of all prime ideals of which contain . Prove that:
If is the ideal generated by , then .
If is the family of subsets of then:
For ideals and ,
These results show that satisfy the axioms for a closed set in a topological space. The resulting topology is called a Zariski topology, and is called the prime spectrum of , written .
Prove that is irreducible iff of is a prime ideal.
Prove that if are coprime.
Let be a ring, and ideal, and a module. Prove that is isomorphic to .
Proof: Tensor exact sequence with .
Let be finitely-generated -modules. If show that or .
Proof: Let be a maximal ideal, and the reside field. Let . By Nakayama's lemma, . But:
which in turn implies or since are vector spaces over a field.
Let be a family of -modules, and be their direct sum. Then, prove is flat each is flat.
If is a prime ideal in , prove that is a prime ideal in .
Rings of fractions
If is a multiplicatively closed set of ring , and a finitely-generated -module, prove that iff there exists such that .
Let be an ideal of ring , and . Show that is contained within the Jacobson radical of . Use this and Nakayama's lemma to show that there exists such that , when .
Proof: To prove the second part, notice that if , then , hence by Nakayama's lemma, we have . Now, from the previous exercise,