# How to prove that pZ is a maximal ideal for the ring of integers?

I know that Z/pZ is a field therefore pZ must be a maximal ideal because of the theorem that states "R/I is a field if and only if I is a maximal ideal" but I want to see a direct proof of it because I hope it would give me an idea how to prove something is a maximal ideal in a general field. Suppose that I've been given an Ideal in a ring R, how can I prove that it's maximal? What are the general strategies that I should consider first? (except the definition of a maximal ideal of course!). I'd like to see a direct proof of pZ being a maximal ideal and also learn how to deal with the situation in a general case.

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Consider an ideal $p\mathbb{Z}\subseteq I$. Take $x\in I\setminus p\mathbb{Z}$. Then p does not divide x, hence gcd(p,x)=1.
By Bezout's theorem, we can find a and b such that

$$ap+bx=1$$

The left-hand side is in I, so the right hand side is too. So I contains 1, an invertible element. This means that $I=\mathbb{Z}$ necessarily. This implies that $p\mathbb{Z}$ is maximal.

morphism
Homework Helper
To slightly generalize micromass's answer, note that ##aR \subseteq bR \iff b\mid a##. So if your ring is a PID, it's not difficult to determine when an ideal is maximal.

By the way, what is the relation between an ideal being maximal and an ideal being principal? Does one imply the other?

By the way, what is the relation between an ideal being maximal and an ideal being principal? Does one imply the other?
Not in general.

Not in general.
OK, then under what circumstances does one imply the other?

Consider an ideal $p\mathbb{Z}\subseteq I$. Take $x\in I\setminus p\mathbb{Z}$. Then p does not divide x, hence gcd(p,x)=1.
By Bezout's theorem, we can find a and b such that

$$ap+bx=1$$

The left-hand side is in I, so the right hand side is too. So I contains 1, an invertible element. This means that $I=\mathbb{Z}$ necessarily. This implies that $p\mathbb{Z}$ is maximal.

To slightly generalize micromass's answer, note that ##aR \subseteq bR \iff b\mid a##. So if your ring is a PID, it's not difficult to determine when an ideal is maximal.
Is every P.I.D ring a Euclidean domain? Because if it's not a Euclidean domain and we don't have the division algorithm, how can we find if b|a in the ring?
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Anyway, I used micromass's answer to find a class of maximal ideals of Gaussian integers I think. Please verify if what I've obtained is correct.

The set Ip={ a+bi : p|a and p|b} is a maximal ideal of Gaussian integers if p is of the form p=4k+3. The proof is attached.
Is the quotient ring R/Ip an interesting field in number theory?

Thanks.

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Is every P.I.D ring a Euclidean domain? Because if it's not a Euclidean domain and we don't have the division algorithm, how can we find if b|a in the ring?

The ring $Z\left[\frac{1+sqr{-19}}{2}\right]$ is a PID that is not an Euclidean domain, though the proof isn't that easy and, undoubtedly, not that trivial.

Anyway, in ANY ring we can use $b|a$ to denote "b divides a" iff there exist an element r in the ring s.t. $a=br$ .

DonAntonio

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Anyway, I used micromass's answer to find a class of maximal ideals of Gaussian integers I think. Please verify if what I've obtained is correct.

The set Ip={ a+bi : p|a and p|b} is a maximal ideal of Gaussian integers if p is of the form p=4k+3. The proof is attached.
Is the quotient ring R/Ip an interesting field in number theory?

Thanks.
....

The ring Z[(1+√19)/2] is a PID that is not an Euclidean domain, though the proof isn't that easy and, undoubtedly, not that trivial.

Anyway, in ANY ring we can use b|a to denote "b divides a" iff there exist an element r in the ring s.t. a=br .

DonAntonio
Thanks for the answer, I knew what b|a meant but I wondered how we could find b|a if there's no division algorithm. I mean for polynomials and integers and other rings that have a division algorithm it's easy, but for other rings it must be hard to show that a=br. Am I right or we can always find a way to show if b|a or not in a general ring?

Thanks for the answer, I knew what b|a meant but I wondered how we could find b|a if there's no division algorithm. I mean for polynomials and integers and other rings that have a division algorithm it's easy, but for other rings it must be hard to show that a=br. Am I right or we can always find a way to show if b|a or not in a general ring?

I think you're confusing things here: under the definition I gave it is ALWAYS, for any ring (even non-commutative and/or unitary) to define b|a as already explained.

What you probably want is a way that guarantees the possibility of DIVIDING an element by another with residue (and with some other properties, which for example is what the euclidean algorithm shows us to do in the integers Z, pol. rings K[x] over a field K and etc.). For this then you need an Euclidean Domain.

DonAntonio

I think you're confusing things here: under the definition I gave it is ALWAYS, for any ring (even non-commutative and/or unitary) to define b|a as already explained.

What you probably want is a way that guarantees the possibility of DIVIDING an element by another with residue (and with some other properties, which for example is what the euclidean algorithm shows us to do in the integers Z, pol. rings K[x] over a field K and etc.). For this then you need an Euclidean Domain.

DonAntonio
No, I know the definition, the definition is definition, I have nothing to say about that. Here's my question: Suppose I have given to you two elements of a ring. like b and a. How do you find if b|a or not? you must find an element like r that rb=a, but how do you find that particular r? If we have the Euclidean division algorithm, it's easy, but when there's no such algorithm then I guess it won't always be easy to show that b|a because we must look for an element like r which might be found with extreme hardship. My question precisely is, how do we find that particular r such that br=a in a general ring?

No, I know the definition, the definition is definition, I have nothing to say about that. Here's my question: Suppose I have given to you two elements of a ring. like b and a. How do you find if b|a or not? you must find an element like r that rb=a, but how do you find that particular r?

*** That is ANOTHER problem: the definition is sound and appliable ALWAYS! The very same question you could ask even in an Euclidean domain: we know we can divide with remainder any two elements, but...what's the remainder? Well, what's the remainder is a question DIFFERENT from the one asking whether we can actually divide with remainder.

Again, it is ALWAYS possible to define b|a as already noted. How will you know whether one specific element divides another specific element in some specific ring is another matter.

How is it done in a general ring? I don't know and I'm not sure whether there's a general answer for this, and you're right: if we have an Euclidean Domain then there's an algorithm to do this, otherwise...I can't tell.

Anyway, I think I read not long ago that there're rings that are not Euclidean Domains yet they still have the division with remainder property (I suppose without the added property that the remainder is less, in the Euclidean function, than the divided element's...). Perhaps a search in Google will help.

DonAntonio ***

If we have the Euclidean division algorithm, it's easy, but when there's no such algorithm then I guess it won't always be easy to show that b|a because we must look for an element like r which might be found with extreme hardship. My question precisely is, how do we find that particular r such that br=a in a general ring?
....

Hurkyl
Staff Emeritus
Gold Member
By the way, what is the relation between an ideal being maximal and an ideal being principal? Does one imply the other?
Usually, maximal ideals cannot be principal. The exceptions only occur in situations that are "one-dimensional" (or zero-dimensional) in an appropriate sense.

The integers (or more generally, any subring of a number field other than the number field itself) is a one-dimensional ring, so maximal ideals can be principal there.

mathwonk
Homework Helper
rings have a "dimension". e.g. the polynomial ring in n variables over a field k say algebraically closed, like the complex numbers, has dimension n. the dimension has several manifestations, transcendence degree over k, length of a chain of prime ideals, s well as a cohomological version in terms of lengths of "resolutions of modules".

If we think of an ideal of polynomials in k[X1,...,Xn] geometrically, in terms of the common zero set in k^n of all polynomials in the ideal, then the minimal number of generators of the ideal equals the minimal number of equations needed tod efine that common zero set.

Now the more equations, the smaller is the set of common zeroes, so a principal ideal corresponds to a "hypersurface" or a zero set of dimension n-1, while a maximal ideal, like (X1-a1,....,Xn-an), corresponds to a single point, the point (a1,....,an).

So a principal ideal is more or less the opposite of a maximal ideal, i.e. a principal ideal is sort of a "minimal" ideal.

Of course if the ring has dimension one, like k[t], or Z, or Z, then minimal and maximal are the same.

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morphism
Homework Helper
The set Ip={ a+bi : p|a and p|b} is a maximal ideal of Gaussian integers if p is of the form p=4k+3. The proof is attached.
Is the quotient ring R/Ip an interesting field in number theory?
Your set Ip is perhaps better known as the principal ideal (p) in Z. :)

Your observation that (p) is maximal if p is an integer prime of the form 4k+3 is a very important one from the point of view of algebraic number theory. One could further observe that (p) is maximal only if p=3 mod 4 (p an integer prime), and this is in some sense the starting point of the whole subject. The fact that (p) isn't maximal if p=2 or p=1 (mod 4) is a manifestation of Fermat's theorem on primes that are sums of two squares.

The field R/Ip is indeed interesting (well, maybe it's not the field itself that's interesting, but more the fact that there is such a field 'attached' to p). First off, note that it's just Z/(p) = Z[x]/(x^2+1,p) = (Z/pZ)[x]/(x^2+1). Your observation that (p) is maximal when p=3(mod 4) translates into the statement that x^2+1 is irreducible mod p when p=3(mod 4), which in turn translates into the (well-known) statement that -1 is not a quadratic residue mod p if p=4k+3. In any case, the irreducibility of x^2+1 means that R/Ip=(Z/pZ)[x]/(x^2+1) is a degree 2 extension of the field Z/pZ -- i.e., it's the field of nine elements.

In algebraic number theory, the degree of this extension of finite fields is called the inertial degree of the prime p in the extension Z/Z. Now let me blabber on a bit more to put this in proper context.

Suppose we have a finite-degree extension of fields $K/\mathbb Q$ (e.g. ##K = \mathbb Q##). Then we would perhaps like to do number theory in K, so we would like to find an analogue of ##\mathbb Z \subset \mathbb Q## in K; such an analogue exists, and is called the ring of integers of K. For example, Z is the ring of integers of Q.

Seeing how prime numbers in Z are important, one might wonder if the same is true in the ring of integers of K (which I hereafter denote by R). This leads us to a very fascinating investigation, which is in some sense what basic algebraic number theory is all about. I will restrict myself to one key point of the theory, namely: it turns out it isn't the primes of R that are important, but the prime ideals. This is because it frequently happens that R need not be a ufd - so there is no unique factorization into primes like in Z. However, there is another type of factorization available in R: a unique factorization of ideals into prime ideals! (This is in fact how the notion of an "ideal" in a ring came to be. Dedekind defined "ideal numbers" to circumvent the lack of unique factorization in the elements of R.)

In particular, suppose you start with an integer prime p, and consider the principal ideal pR it generates in R. This ideal will factor into a direct product of prime ideals of R; let's say this factorization looks like ##pR = P_1^{e_1} \ldots P_r^{e_r}## where the P_i's are distinct prime ideals of R. The integers e_i showing up are called the ramification indices of p (more precisely, e_i is the ramification index of P_i over p). Now one can make the following important observation: each P_i turns out to be not just a prime ideal, but in fact a maximal ideal. So we have a field R/P_i called a residue field (your R/Ip is an example of such), which is in fact always a finite field. This field naturally contains Z/pZ. The degree f_i of the extension R/P_i over Z/pZ is called the inertial degree of P_i over p.

Going back to the case where K=Q and R=Z, we see that a prime of the form p=4k+3 generates a prime ideal pR in R - i.e., the prime factorization of pR contains one prime ideal P (namely, pR itself...), and the residue field R/pR is of degree 2 over Z/pZ. So e=1 and f=2 in this case.

If p on the other hand were of the form 4k+1, then it will turn out that pR factors into a product of two distinct prime ideals in R, say p=P_1P_2. And one has that R/P_i=Z/pZ. So e_1=e_2=1 and f_1=f_2=1.

What about p=2? It factors as p=P^2, with R/PR=Z/pZ. So e=2 and f=1.

If you look at the above three cases, you might notice that there is some kind of relationship between the numbers e and f. There is indeed such a relationship. In symbols, it's: ref=2, where r is the number of prime ideals occurring in the factorization of pR. This is a special case of a general theorem that states that if K/Q is a galois extension (which Q/Q is), and ##pR = P_1^{e_1} \ldots P_r^{e_r}## is the prime factorization of an integer prime in R, then e_1=...=e_r (call this number e), f_1=...=f_r (call this number f) and ref=[K:Q].

Apologies for this overly lengthy post, but I hope it will be of some value.

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mathwonk
Homework Helper
I am not expert here like morphism, but his post inspires me to speculate as follows.

In algebraic geometry one studies the duality between points of a space and ideals of functions vanishing at those points. Thus the point of the x axis with coordinate c, corresponds to the maximal ideal (x-c) in k[x], the ideal of functions that vanish at c, and equivalently are divisible by x-c, in the ring of functions ion the x axis.

the parabola y=x^2 has function ring k[x,y]/(y-x^2) which is isomorphic to k[x], corresponding to the fact that the vertical projection is an isomorphism from the x axis to the parabola.

The map from the x axis to the y axis sending x=c to y = c^2, is the composition of that vertical projection with horizontal projection on the y axis, and corresponds to the ring map in the other direction k[y]-->k[x] sending the function y on the y axis, to the function x^2 on the x axis. i.e. a function on the y axis yields a function on the x axis by preceding it with the previous map from the x to the y axis.

(Assume k is algebraically closed if need be.) This ring map can recover the map on points as follows: if c is a point of the x axis, corresponding to the maximal ideal (x-c), then the preimage of that ideal in the ring k[y] under the map k[y]-->k[x] sending y to x^2, is the ideal (y-c^2), corresponding to the point c^2 on the y axis, i.e. the image of the point c under the geometric map from x axis to y axis.

Now given the point c^2 on the y axis, the inverse image of it under that geometric map would be points corresponding to all ideals in k[x] whose preimage is (y-c^2). But both (x-c) and (x+c) have preimage (y-c^2) under the ring map k[y]-->k[x], y-->x^2. I.e. the two preimages of c^2 are obtained by considering y-c^2 , substituting x^2 for y, then factoring x^2-c^2 in the larger ring k[x], an extension of the subring k[x^2] ≈ k[y].

The degree of the map, the number of inverese images of a general point, is equal to the degree of the field extension of quotient fields of these rings, k(x) over k(y), which is two. But there is one point of the y axis, namely zero which only has one preimage. But if we proceed as above and find its preimages by factoring x^2-0^2, we get (x-0)(x+0), or two copies of the point x=0. Thus the map is ramified over y=0, since although there is only one preimage, that preimage counts twice.

In th same way we can regard the ring inclusion Z in Z or in any ring R of integers of a number field, as corresponding to a geometric map. We make the sets of prime ideals, into topological spaces, spec(Z) and spec(R), with finite closed sets. Then we use the ring inclusion from Z to R to pull back prime ideals and hence to define a continuous map from spec(R)-->spec(Z). E.g. the inclusion Z in Z pulls back the prime ideals (2+i) and (2-i) to the prime ideal (5). I.e. the inverse image of (5) in spec(R) is the pair of prime ideals (2+i) and (2-i). The field extension Q in Q(i) has degree two and this is the degree of the geometric map spec(Z)-->spec(Z). Thus every prime ideal should have two preimages, counted properly, i.e. by morphism's formulas.

What about the prime (7) in spec(Z)? It generates a prime ideal (7) also in spec(Z), but the residue field here is (Z/7), which has degree 2 over the residue field Z/7. In the prior cases, the residue fields Z/5 and Z/(2-i) seem to be isomorphic (they both have 5 elements). Thus morphism's formulas always give 2.

So the primes of form 4k+3 in Z are ramification points of the map spec(Z)-->spec(Z),

while those of form 4k+1 are the “good guys”, i.e. unramified, and having the right number of preimages.

Morphism’s theorem is that all primes have the same number of preimages under the map spec(R)-->spec(Z) and that if a prime is ramified, it is ramified symmetrically, i.e. every preimage point has the same ramification. Geometrically this is plausible, since presumably the Galois group (or inertia group?) acts on the space upstairs like a covering map, transitively, so every preimage point has to look like every other. I.e. the ramification points upstairs in the ring of integers should be the points that are fixed by a non trivial subgroup of the Galois action, and these isotropy subgroups should all be conjugates, hence have the same order.

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morphism
Homework Helper
Morphism’s theorem is that all primes have the same number of preimages under the map spec(R)-->spec(Z) and that if a prime is ramified, it is ramified symmetrically, i.e. every preimage point has the same ramification. Geometrically this is plausible, since presumably the Galois group (or inertia group?) acts on the space upstairs like a covering map, transitively, so every preimage point has to look like every other. I.e. the ramification points upstairs in the ring of integers should be the points that are fixed by a non trivial subgroup of the Galois action, and these isotropy subgroups should all be conjugates, hence have the same order.
That's spot on. (The galois group does indeed act transitively on the primes upstairs lying above any given prime downstairs. The isotropy subgroup above a prime p is called the decomposition group at p, and is a very important subgroup of Gal(K/Q); for one thing, it surjects onto the galois group of the residue field at p.)

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mathwonk
Homework Helper
Thank you morphism. Now what about the non galois case? If we take any finite extension of Q, and normalize the integers in there we get a ring of integers R, such that R is a module of finite type over Z, and the map spec(R)-->spec(Z) should again have degree equal to the degree of the field extension, but not necessarily of the "Galois" group.

But every prime in Z should again have the same number of preimages in some sense.

e.g. although now the exponents in the factorization of p in R need not be equal, they should all add up to the same number, or rather give the same number after multiplication by the degree of the residue field extension.

just guessing......???

OOps, am I assuming unique factorization? there seems always to be a notion of preimages, but they may not correspond to a unique factorization in some cases? do theyn give in some sense a substitute for that notion?

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morphism
Homework Helper
Yup, that's right. If there are r distinct primes lying over p, with exponents e_i and inertial degrees f_i, then ##\sum_{i=1}^r e_i f_i = [K : \mathbb Q]##.

re: unique factorization: The map specR -> specZ is explicitly given by ##P \mapsto P \cap \mathbb Z##. It's not obvious that this map is surjective -- this is the "going up" lemma in this setting. Its counterpart, the going down lemma, says (amongst other things) that the primes in the preimage of p are in fact the primes that occur in the unique factorization of pR. This is easy to see: it follows from the fact that P contains pR iff it divides it. (And of course ##P \cap \mathbb Z = p\mathbb Z \iff P \supset pR##.)

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mathwonk
Homework Helper
apparently i am using the terminology wrong. i was calling any prime ramified if it had fewer than the maximal number of distinct prime factors upstairs. the traditional terminology seems to restrict that word to the case where this is caused by exponents higher than one, not where, as above, it is caused by degree of residue field extension greater than one. thus it seems the only traditionally "ramified" prime is 2. those of form 4k+3 are perhaps called "inertial". But again I am a beginner who has never studied this topic.

morphism
Homework Helper
Right.

There are many good reasons for distinguishing between primes that don't "split completely" and those which are ramified. Let me give you one reason that's motivated by geometric considerations. The context is the analogy between extensions of number fields and extensions of function fields (i.e., maps of algebraic curves). To be concrete (and perhaps a little dishonest - forgive me), the relevant part of this analogy states that corresponding to the notion of a finite extension K/Q of number fields is the notion of a finite extension C(z,w)/C(z) of function fields, "i.e." a finite branched cover f:X->P^1 of the Riemann sphere by a compact Riemann surface X. The primes of (the ring of integers of) K correspond to points of X (this is a lie, but not a big one). The splitting of primes specR->specZ corresponds to the pulling back of divisors f*:Div(P^1)->Div(X) (another minor lie). The exponents in the prime factorization correspond to the ramification indices in pullback.

Now here's the punchline: the primes in K that ramify (in the sense that they have e_i > 1 for some i) are completely detected by an ideal in the ring of integers (called the different). Under the number field <-> function field analogy, the different corresponds to the ramification divisor of the covering map X->P^1.

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Hurkyl
Staff Emeritus
Gold Member
For another description, look at what happens if you don't assume
(Assume k is algebraically closed if need be.)​

Consider the real polynomial x²+1. Being degree 2, the zero set of this polynomial "ought" to consist of two points on the real line, counting multiplicity. But, alas, its factorization only includes a single polynomial; a single point of Spec R[x]! But we 'know' that the zero set is really two points: i and -i.

This is one of the intuitions you want for higher degree ideals, I think: a degree n prime ideal corresponds in some sense to n individually indistinguishable points.

Going back to your Gaussian integer example, the ideal (7) corresponds to two "geometric" points. One way to manifest them is as the two different residue maps to F49 (or to $\overline{\mathbf{F}_7}$, if you prefer). e.g. the "function" i has the value $\sqrt{-1}\in \mathbf{F}_{49}$ at one point, and $-\sqrt{-1}\in \mathbf{F}_{49}$ at the other point (after choosing a square root of -1 in F49, of course). The "function" 7, of course, vanishes at both points.

The ideal (2), however, corresponds to a double point.

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mathwonk
Homework Helper
Hurkyl, So these geometric points sound like the geometric "k valued points" where k is the algebraic closure of Z/(7), in the language of Mumford's redbook.

so although spec(k) only has one point, there are two maps of spec(k) to spec(Z) taking that point to the point (7) in spec(Z). hence there are two ways to pullback "functions" in Z from spec(Z) to spec(k). I.e. two ring maps from Z to k. Under both maps, 7 goes to zero, but the two maps correspond exactly to the two choices for the image of i in k.

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mathwonk
Homework Helper
morphism, does the "different" ideal allow us to define a canonical divisor in spec(Z) by analogy with Hurwitz' formula for the ramification divisor of a map?

but i guess first we would need a canonical divisor in spec(Z).

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mathwonk