Prove that is a maximal ideal in . i.e.
Proof: Let us use the reductio ad absurdum technique, and assume that . Then,
Since is a general polynomial, we can use the remainder theorem to obtain:
Now, since we're free to choose , choose . Then must be of a lower degree, and hence a constant .
There are two cases to consider: either or is a nonzero constant. In the first case, since , our assumption that breaks, and we have a contradiction. In the second, would contain the unit ideal, and hence , breaking the assumption that . This leads to another contradiction, finishing off the proof
Prove that is a maximal ideal in . i.e.
Proof: Let us again use proof by contradiction, and assume that . Then,
Since is a general polynomial, we can use the remainder theorem to obtain:
Now, choose ; can only be a zero, because if it were a nonzero constant, would contain , and hence otherwise:
Since from our assumption, we get . Note that , from our hypothesis:
But is irreducible in ; assume that it factors:
This cannot happen in , and we hence arrive at a contradiction
Provide an example of an irreducible element which is not prime, in a ring of your choosing.
Solution: Consider the ring of Gauss integers . Then,
and this ring is not a unique factorization domain. Let us inspect : it belongs to the multiplicative set containing , but it does not divide either or .
Prime ideals are defined as , where is a multiplicative set such that:
Now, does not cleave into and cleanly when . Since or are prime, and are disjoint from the set containing them, and so are in S. Hence, is not prime, but is irreducible, as required
Provide examples of prime ideals that are not maximal in .
Solution: are prime ideals, but only the last one is maximal; the first two are irreducible.
Prove that every maximal ideal is a prime ideal in ring .
Proof: The definition of a prime ideal is:
The definition of a maximal ideal is:
From this definition, it should be clear that maximal ideals can never be the product of two different maximal ideals and , for if it were, then
contradicting the definition of a maximal ideal. i.e. for
This is precisely the definition of a prime ideal
Give an example of an algebraically closed ring, and an example of one that is not.
Solution: is algebraically closed, because all polynomials in one variable in the ring have solutions in the ring. is not algebraically closed, because the solution of, for instance, does not lie in the ring
List all the maximal ideals of and .
Solution: The maximal ideals of are , while those of are and , where is a real number, and is a complex number. Notice that all maximal ideals of can be factorized into monic factors, while in the case of , quadratic factors are required
Prove that is a prime ideal in ring iff is an integral domain.
An integral domain is defined by
Proof: Given that is an integral domain, show that is prime. Consider elements . Elements of are then of the form . Now, use the definition of an integral domain to write:
This is exactly the definition of a prime ideal, as required.
Proof, the other way: To show is an integral domain, given that is a prime ideal. First, consider the surjective homomorphism
From isomorphism theorem, . We know that:
This gives us the required condition for integral domain
Prove that is a maximal ideal in ring iff is a field.
A field is defined by
and
Proof: To show that is maximal given that is a field. For the zerodivisor proof, follow exactly. For the next part, assume :
hence is unit, which implies that as required.
Proof, the other way: To show that is a field, given is maximal. Consider the surjective homomorphism:
Now, since is a field, the only ideals are and ; by isomorphism theorem, there is a bijective correspondence between ideals of containing and those of .
Since , we have shown that is maximal, as required
For ideals and , show that is contained within .
Proof: From isomorphism theorem,
Since , we see that unless the gcd is , is strictly smaller than