Abstract Nonsense

Crushing one theorem at a time

Prime Ideals (Pt. I)

Point of Post: In this post we define the notion of a prime ideal and \text{Spec}(R) for a ring R, characterize prime ideals in commutative unital rings via their quotient rings, and find \text{Spec}(R) for a few specific rings.

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We now come to what shall be one of the most important topics not only in the ring theory to come but in the application of ring theory to other subjects. Namely, in this post we shall discuss the notion of prime ideals. With a statement like that one would hope that there is some serious intuition to back it up. Thankfully, not only is the intuitive motivation for prime ideals abounding but it is accessible to us even with our current knowledge. In fact, there are two natural ways to motivate prime ideals, both with their own appeal. While either way is equally effective depending upon the goals of the readers, it is the bias of the author in saying that the first (we discuss)  is ‘more important’. Whereas the second approach has the quality that it answers a ‘natural’ question the second answers a ‘deep’ question. Anyways, enough of the pre-preamble:

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The first motivation for prime ideals is the one suggested from their name. In particular, I have said before that in a lot of ways the integers \mathbb{Z} provide a guiding example for a lot of the ring-theoretic concepts that we shall discuss in these beginning posts. To this end it does us some serious good to examine the concepts of \mathbb{Z} which proved most fruitful and see how precisely we may generalize these to fit our ring-theoretic purposes. Well, doubtlessly obvious to anyone who has done any kind of basic number theory (in fact, to anyone who has done any kind of mathematics) that one of the most striking and deeply important objects in \mathbb{Z} are the primes. In particular, the fact that every integer may be written, up to units, uniquely as a product of primes. It would be fantastic if we could find a more general sort of ring for which this unique factorization into is possible. The first step in this is the necessity to define what precisely prime should mean in a general ring. While there are many equivalent ways of defining primes in \mathbb{Z}, after some experimentation though one discovers that the key characterization of primes is Euclid’s lemma. Said differently, the primes of \mathbb{Z} are precisely the non-unit numbers p with the property that whenever p\mid ab one has p\mid a or p\mid b. That said, since most of our focus has been on ideals it would be nice to phrase this in such a language. In particular, it’s easy to see that a number p\in\mathbb{Z} is prime if the generated ideal  (p) is proper (i.e. not equal to \mathbb{Z}) and has the property that ab\in(p) implies a\in(p) or b\in(p). Or, taking it one step further the equivalent definition that (p)\ne\mathbb{Z} and (a)(b)\subseteq(p) (recalling the product of ideals) implies (a)\subseteq(p) or (b)\subseteq(p). Thus, we are lead to the notion of prime ideals in a general ring R which are ideals \mathfrak{p}\subsetneq R with the property that \mathfrak{a}\mathfrak{b}\subseteq\mathfrak{p} implies \mathfrak{a}\subseteq\mathfrak{p} or \mathfrak{b}\subseteq\mathfrak{p}. Therefore, we are led, intuitively for now, to define a prime element p\in R to be one such that (p) is prime. Moreover, we shall see that prime ideals shall take the place of primes in a very real sense which shall be made clear later.

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For the sake of  convenience we assume for this second motivational point that we are dealing with a commutative unital ring R. We have seen that given an ideal \mathfrak{a}\in\mathfrak{L}(R) we can form the quotient ring R/\mathfrak{a}. While the existence of this ring is all nice-and-well (especially in consideration of the first isomorphism theorem) it’s not immediate what properties one can expect R/\mathfrak{a} to have besides the ones it inherits being a homomorphic image of R. For example, we may wish to know when R/\mathfrak{a} is an integral domain. Exploring this further we see this is the same as asking when (a+\mathfrak{a})(b+\mathfrak{a})=ab+\mathfrak{a}=\mathfrak{a} implies a+\mathfrak{a}=\mathfrak{a} or b+\mathfrak{a}=\mathfrak{a}. Of course, this is equivalent to having that ab\in\mathfrak{a} implies a\in\mathfrak{a} or b\in\mathfrak{a}. Thus, also noting the fact that 1+\mathfrak{a}\ne\mathfrak{a} if and only if \mathfrak{a}\ne R we see that for commutative unital ring R one has that the prime ideals \mathfrak{p} of R are precisely those for which R/\mathfrak{p} is an integral domain.

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Prime Ideals

In accordance with the above definitions we define a prime ideal in a ring R to be an ideal \mathfrak{p}\subsetneq R with the property that if \mathfrak{a},\mathfrak{b}\in\mathfrak{L}(R) are such that \mathfrak{a}\mathfrak{b}\subseteq\mathfrak{p} then \mathfrak{a}\subseteq\mathfrak{p} or \mathfrak{b}\subseteq\mathfrak{p}. We call the set of all prime ideals of a ring R the spectrum of R and denote it \text{Spec}(R).

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For commutative rings this definition simplifies considerably, but first some notation. We call a subset S\subseteq R multiplicative if it is non-empty and closed under multiplication. What we now claim is that if R is commutative then a proper ideal \mathfrak{p} is prime if and only if R-\mathfrak{p}. In particular, if R is furthermore unital then \mathfrak{p} is a prime ideal if and only if R-\mathfrak{p} is a submonoid of (R,\cdot). Indeed:

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Lemma: Let R be a commutative ring, then for any a,b\in R one has that (a)(b)=(ab).

Proof: Evidently (ab)\subseteq (a)(b) since ab\in(a)(b). Conversely, every element of (a)(b) is of the form

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with n_j,m_j\in\mathbb{Z} and r_j,s_j\in R. But, then we have that (n_ja+r_ja)(m_jb+s_jb)=n_jm_jab+m_jr_jab+n_js_jab+r_js_jab\in (ab) from where the reverse inclusion and thus the lemma follows. \blacksquare

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Theorem: Let R be a commutative ring. Then, \mathfrak{p}\in\text{Spec}(R) if and only if R-\mathfrak{p} is multiplicative.

Proof: Suppose first that R-\mathfrak{p} is multiplicative and let \mathfrak{a},\mathfrak{b} be such that \mathfrak{a}\mathfrak{b}\subseteq\mathfrak{p}. Now, if neither \mathfrak{a}\subseteq\mathfrak{p} or \mathfrak{b}\subseteq\mathfrak{p} then we could find a\in\mathfrak{a}-\mathfrak{p} and b\in\mathfrak{b}-\mathfrak{p} and so by assumption we have that ab\notin \mathfrak{p} but since ab\in\mathfrak{a}\mathfrak{b} this is a contradiction.

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Conversely, suppose that \mathfrak{p} is prime and ab\in\mathfrak{p}. If we knew that this implied (a)(b)\in\mathfrak{p} we’d be done since this would imply either a\in \mathfrak{p} or b\in\mathfrak{p}. But, this is precisely what the lemma preceeding this theorem says. The conclusion follows. \blacksquare

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With this we are able to formally relate our two motivations by the following important theorem whose proof is really what we discussed in the motivation):

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Theorem: Let R be a commutative unital ring. Then, \mathfrak{p}\in\text{Spec}(R) if and only if R/\mathfrak{p} is an integral domain.

Proof: Suppose first that \mathfrak{p}\in\text{Spec}(R) then we have, from basic discussions of quotient rings, that R/\mathfrak{p} is a commutative unital ring with 1\ne 0 (this follows since \mathfrak{p}\subsetneq R). Thus, it suffices to prove that R/\mathfrak{p} has no zero divisors. To do this we merely note that if \mathfrak{p}=(a+\mathfrak{p})(b+\mathfrak{p})=ab+\mathfrak{p} then by definition we must have that ab\in\mathfrak{p} and so by the previous theorem this is equivalent to a\in\mathfrak{p} or b\in\mathfrak{p} which is equivalent to a+\mathfrak{p}=\mathfrak{a} or b+\mathfrak{p}=\mathfrak{p}. Since a,b\in R were arbitrary the fact that R/\mathfrak{p} is an integral domain follows.

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Conversely, suppose that R/\mathfrak{p} then since 1\ne 0 in R/\mathfrak{p} we must have that \mathfrak{p}\subsetneq R and moreover we see that since R/\mathfrak{p} has no zero divisors then a,b\notin\mathfrak{p} implies a+\mathfrak{p},b+\mathfrak{p}\ne\mathfrak{p} and so by definition ab+\mathfrak{p}=(a+\mathfrak{p})(b+\mathfrak{p})\ne\mathfrak{p} and so by definition ab\notin\mathfrak{p}. Since a,b\notin\mathfrak{p} were arbitrary we may conclude that R-\mathfrak{p} is multiplicative from where the conclusion follows. \blacksquare

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This theorem turns out to be a double-edged sword. On one hand, we have a theoretical factory for churning out integral domains, namely just finding commutative unital rings and prime ideals of such rings and modding out. On the other hand, this theorem takes the sometimes formidable problem of proving an ideal is or is not prime and reducing it to checking a property of its quotient ring. But, why is this easier? Well, there are two reasons really–firstly modding out often ‘takes away clutter’ in the sense that a lot of unimportant stuff ‘goes away’, secondly in the most natural cases we are provided with a nice description of the quotient ring via the first isomorphism theorem (e.g. the fact that R[x]/(x)\cong R).

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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.


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


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