Abstract Nonsense

Crushing one theorem at a time

Prime Ideals (Pt. II)


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

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

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Proposition: \text{Spec}(\mathbb{Z})=\{(p):p\text{ is prime}\}\cup\{(0)\}.

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

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The less highfalutin way to put this is that obvious n\mathbb{Z} is not prime for composite n since, if n=ab, then a\notin n\mathbb{Z} and b\notin\mathbb{Z} yet ab=n\in n\mathbb{Z}. Moreover, the fact that p\mathbb{Z} is prime is really just a restatement of the celebrated lemma of Euclid which states that if p is prime then p\mid ab\implies p\mid a\text{ or }p\mid b. \blacksquare

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

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Proposition: \text{Spec}\left(\mathbb{C}[x]\right)=\left\{(x-z):z\in\mathbb{C}\right\}\cup\{(0)\}.

Proof: Evidently (0) is prime since we know \mathbb{C}[x] is an integral domain. To see that (x-z)\in\text{Spec}(\mathbb{C}[x]) for each z\in\mathbb{C} we merely note that the kernel of the evaluation homomorphism \text{ev}_z is (x-z) (this follows from the classic property of complex polynomials [as we shall see more generally for polynomials over a field] that a polynomial p has a root \alpha if and only if it can be factored as p(x)=(x-\alpha)q(x)) and so \mathbb{C}[x]/(x-z)\cong\mathbb{C} and so the conclusion follows. To see these are the only elements of \text{Spec}(\mathbb{C}[x]) one must take my word, for now, that every ideal of \mathbb{C}[x] is principal (i.e. that any ideal of \mathbb{C}[x] is of the form (p(x)) for p(x)\in\mathbb{C}[x]) from where the fact easily follows since if p(x) is not of the form x-z then it factors as (x-z)q(x) (this is just the statement of the fundamental theorem of algebra) and so evidently (x-z)q(x)\in (p(x)) even though (x-z),q(x)\notin p(x). \blacksquare

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Proposition: \text{Spec}\left(\mathbb{Z}/n\mathbb{Z}\right)=\left\{p\mathbb{Z}/n\mathbb{Z}:p\text{ is prime}\right\}

Proof: We know from the fourth ring isomorphism theorem that the ideals of \mathbb{Z}/n\mathbb{Z} are precisely m\mathbb{Z}/n\mathbb{Z} with m\mid n. But, by the third ring isomorphism theorem we know that \left(\mathbb{Z}/n\mathbb{Z}\right)/\left(m\mathbb{Z}/n\mathbb{Z}\right)\cong\mathbb{Z}/m\mathbb{Z} and so, in particular, we have that \left(\mathbb{Z}/n\mathbb{Z}\right)/\left(m\mathbb{Z}/n\mathbb{Z}\right) is an integral domain if and only if m is prime. The conclusion follows. \blacksquare

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More difficult still is the following:

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Proposition: \left\{(p,x):p\text{ is prime}\right\}\cup\left\{(p):p\text{ is prime}\right\}\cup\{(x)\}\cup\{(0)\}\subsetneq\mathbb{Z}[x]

Proof: To see that (p,x) is a prime ideal for each prime p we define f:\mathbb{Z}[x]\to\mathbb{Z}/p\mathbb{Z}:p(x)\mapsto p(0)+p\mathbb{Z} is an epimorphism (being the composition of the natural epimorphisms \mathbb{Z}[x]\to\mathbb{Z} and \mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}) and evidently \ker f=(p,x) so that (p,x)\in\text{Spec}(\mathbb{Z}[x]). To see that (p)\in\text{Spec}(\mathbb{Z}[x]) for p prime we merely note that evidently if p(x)q(x)\in (p) then p(x),q(x) are constant (this is because \mathbb{Z} is an integral domain so that we know \deg(p(x)q(x))=\deg(p(x))+\deg(q(x))) and so the conclusion follows from reducing to a previous case. To see that (x)\in\text{Spec}(\mathbb{Z}[x]) we merely note that \mathbb{Z}[x]/(x)\cong\mathbb{Z} and so by previous theorem we know that (x) is prime. Lastly we know that (0) is prime since \mathbb{Z}[x] is an integral domain.

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To see that the stated inclusion is proper we merely note, for example, that the maps f:\mathbb{Z}[x]\to\mathbb{C}:p(x)\mapsto p(i) and g:\mathbb{Z}[x]\to\mathbb{C}:p(x)\mapsto p(\sqrt{2}) are both morphisms with \ker f=(x^2-2) and \ker g=(x^2+1). We may therefore conclude that (x^2-2) and (x^2+1) are prime since the quotients \mathbb{Z}[x]/(x^2+1) and \mathbb{Z}[x]/(x^2-2) are isomorphic to subrings of \mathbb{C} and thus integral domains. The full classification of \text{Spec}(\mathbb{Z}[x]) eludes us for now. \blacksquare

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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 S\subseteq R is multiplicative and \mathfrak{a}\in\mathfrak{L}(R) is disjoint from S then \mathfrak{a} need not be prime. Indeed, the set O\subseteq\mathbb{Z} of odd numbers is multiplicative and the ideal 4\mathbb{Z} is disjoint from O yet 4\mathbb{Z} is not prime. Another example could be given by noting that the set \mathbb{Z}^\times of non-zero constants sitting inside \mathbb{Z}[x] is multiplicative, the set (x^2)=\left\{p(x):\deg(p(x))\geqslant 2\text{ or }p(x)=0\right\} is an ideal disjoint from \mathbb{Z}^\times yet (x^2) is not prime since x\notin(x^2) yet x\cdot x=x^2\in(x^2). 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 2\mathbb{Z} is disjoint from O and is prime and similarly in the second case we have that (x) is disjoint from \mathbb{Z}^\times 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 S that there exists (if we choose ‘large enough’) a prime ideal \mathfrak{p} disjoint from S. This is make precise by the following theorem:

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Theorem (The Innocuous Lemma): Let R be a commutative unital ring and S\subseteq R a multiplicative set. Then, if \mathfrak{p} is maximal (in terms of containment) among all ideals disjoint from S, then \mathfrak{p} is prime.

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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 \mathfrak{l} and show that if ab\in\mathfrak{l} and b\notin\mathfrak{l} then a\in\mathfrak{l} by showing that a is canonically in some bigger ideal \mathfrak{h}\supseteq\mathfrak{l}. We shall then show that \mathfrak{h}=\mathfrak{l} by showing that \mathfrak{h} is disjoint from S (and so equal to \mathfrak{l} by a simple maximality argument). So, let’s get to it:

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Proof: Suppose that a,b\notin\mathfrak{p} then \mathfrak{p}+Ra and \mathfrak{p}+Rb are both ideals strictly containing \mathfrak{p}. By assumption we then have that \mathfrak{p}+Ra and \mathfrak{p}+Rb intersect S, say that they contain p+ra and p'+r'b respectively. By assumption that S is multiplicative we have that (p+ra)(p'+r'b)\in S, that said

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\left(p+ra\right)\left(p'+r'b\right)=\underbrace{pp'+pr'b+p'ra}_{P}+rr'ab

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Note though that P\in\mathfrak{p} and so if ab\in\mathfrak{p} then so are rr'ab and thus so are P+rr'ab\in S. Thus, we have that \mathfrak{p}\cap S\ne\varnothing, which contradicts our assumption. Thus, we may conclude that ab\notin\mathfrak{p} and so R-\mathfrak{p} is multiplicative (it’s nonempty since S\subseteq R-\mathfrak{p}). \blacksquare

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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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August 17, 2011 - Posted by | Algebra, Ring Theory | , , , , , , ,

1 Comment »

  1. […] Prime Ideals (Pt. III) Point of Post: This is a continuation of this post. […]

    Pingback by Prime Ideals (Pt. III) « Abstract Nonsense | August 17, 2011 | Reply


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