## Prime Ideals (Pt. II)

**Point of Post: **This is a continuation of this post.

Now that we have some basic machinery we give a few examples of prime rings by finding (in full or partially) for some rings:

**Proposition: ***.*

**Proof: **There are many obvious equivalent proofs of this statement, probably the most succinct is the fact that every ideal of is of the form for some and since is an integral domain (field even) if and only if is prime (this is a common fact) the proposition follows from the previous characterization of primes ideals.

The less highfalutin way to put this is that obvious is not prime for composite since, if , then and yet . Moreover, the fact that is prime is really just a restatement of the celebrated lemma of Euclid which states that if is prime then .

Let’s move on to a slightly more difficult example:

**Proposition: ***.*

**Proof: **Evidently is prime since we know is an integral domain. To see that for each we merely note that the kernel of the evaluation homomorphism is (this follows from the classic property of complex polynomials [as we shall see more generally for polynomials over a field] that a polynomial has a root if and only if it can be factored as ) and so and so the conclusion follows. To see these are the only elements of one must take my word, for now, that every ideal of is principal (i.e. that any ideal of is of the form for ) from where the fact easily follows since if is not of the form then it factors as (this is just the statement of the fundamental theorem of algebra) and so evidently even though .

**Proposition: **

**Proof: **We know from the fourth ring isomorphism theorem that the ideals of are precisely with . But, by the third ring isomorphism theorem we know that and so, in particular, we have that is an integral domain if and only if is prime. The conclusion follows.

More difficult still is the following:

**Proposition: **

**Proof: **To see that is a prime ideal for each prime we define is an epimorphism (being the composition of the natural epimorphisms and ) and evidently so that . To see that for prime we merely note that evidently if then are constant (this is because is an integral domain so that we know ) and so the conclusion follows from reducing to a previous case. To see that we merely note that and so by previous theorem we know that is prime. Lastly we know that is prime since is an integral domain.

To see that the stated inclusion is proper we merely note, for example, that the maps and are both morphisms with and . We may therefore conclude that and are prime since the quotients and are isomorphic to subrings of and thus integral domains. The full classification of eludes us for now.

**An Innocuous Lemma and Some Important Consequences**

Now that we have a better feel for prime ideals by getting our hands dirty with some real live ideals we get back to the abstraction for which this blog is named. The strange thing we shall note in this section is that from a ‘simple lemma’ we are able to derive a lot of important theory, seemingly branching in several directions. The idea for this lemma starts with the following (silly) observation: if is multiplicative and is disjoint from then need not be prime. Indeed, the set of odd numbers is multiplicative and the ideal is disjoint from yet is not prime. Another example could be given by noting that the set of non-zero constants sitting inside is multiplicative, the set is an ideal disjoint from yet is not prime since yet . That said, in both these cases if the question had not been “Given a multiplicative set and a disjoint ideal, is the ideal prime?” but instead “Given a multiplicative set, does there exists a disjoint prime ideal?” the answer would have been yes. Indeed, in the first case we clearly have that is disjoint from and is prime and similarly in the second case we have that is disjoint from and is prime. In both cases we see that the error of our ways was in picking the disjoint ideal ‘too small’ in terms of containment. So, one may be led to conjecture that if we are given some multiplicative set that there exists (if we choose ‘large enough’) a prime ideal disjoint from . This is make precise by the following theorem:

**Theorem (The Innocuous Lemma): ***Let be a commutative unital ring and a multiplicative set. Then, if is maximal (in terms of containment) among all ideals disjoint from , then is prime.*

The key thing to note about this theorem is that it does not assert that such a maximal ideal exists, it only asserts that if there were such an ideal then it would be prime. The idea to the proof is simple, namely we’ll suppose we have such a ‘largest’ ideal and show that if and then by showing that is canonically in some bigger ideal . We shall then show that by showing that is disjoint from (and so equal to by a simple maximality argument). So, let’s get to it:

**Proof: **Suppose that then and are both ideals strictly containing . By assumption we then have that and intersect , say that they contain and respectively. By assumption that is multiplicative we have that , that said

Note though that and so if then so are and thus so are . Thus, we have that , which contradicts our assumption. Thus, we may conclude that and so is multiplicative (it’s nonempty since ).

**References:**

1. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. *Basic Abstract Algebra*. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

3. Reid, Miles. *Undergraduate Commutative Algebra*. Cambridge: Cambridge UP, 1995. Print.

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