# Units and prime elements in euclidean rings

1. Jun 7, 2008

### SiddharthM

A general question.

A unit element is one that has it's multiplicative inverse in the ring.

An element p is prime if whenever p=ab then either a or b is a unit element.

Can a prime be a unit element?

The answer is, i think, no but thus far I've been unable to find a contradiction.

2. Jun 7, 2008

### matt grime

Usually one specifies a prime cannot be a unit. You will not reach a contradiction with your definition, since a 'unit' is a 'prime' according to that: if you write u=ab for any unit u, then necessarily both a and b are units (I'm assuming all your rings are commutative).

3. Jun 7, 2008

### mathwonk

one divides elements of a ring into three or four categories. zero, units, zero divisors, and others. i guess primes are elements x which are none of the above, and when x divides yz, then x divides either y or z.

usually one assumes that zero is the only zero divisor, i.e. one usually defines primes only in a domain.

tHE DEFINITION That when x = ab either a or b is uniT, is usually called an irreducible elemEnt x.

irreducible only implies prime in a unique factorization domain.

Last edited: Jun 7, 2008
4. Jun 22, 2008

### DeaconJohn

SiddhearthM,

There is more behind mathwonk's suggestion than he is letting on. Your definition of of "prime" is the more intuitive one, and it was most people would be likely to understand a prime as being before Kummer and Dedekind's theory of ideal elements was on the scene. Before Kummer, people kind of assumed that these two definitions were equivalent in an normal algebraic number field. Many people think it was likely that that was Fermat's mistake in his original proof of FLT.

It turns out that the "non-intuitive" definitions that mathwonk puts forward are not only standard today but also they clear up a lot of stuff that is pretty hard to understand even after people started using mathwonk's definition a couple 100 years ago or so.

Nothing that I have written here is original. Most of it is a paraphrase from the first page of one of the chapters in the middle of Ian Stewart's book "Introduction to Algebraic Number Theory and Fermat's Last theorem." If anybody notieces that I got the author's name or the book title slightly wrong, please correct me.

Deacon John

5. Jun 23, 2008

### gel

Really?
That's essentially Lamé's error, because any number field where prime==irreducible is a unique factorization domain. That was >200 years after Fermat though.

6. Jun 24, 2008

### DeaconJohn

Gel,

Thanks for your comment. Yes, you are right, I spoke too quickly. I have forgotten exactrly what the proof is that most people who have considereded the matter deeply think that Fermat probably wrote down.

It was in a first year graduate class at NYU that I saw it. A professor Shapiro, his first name might have been Shapiro put it on the blackboard. If I remember correctly, the proof works for the Gaussian integers put not for the rational integers. Of course, the Professor challenged us to tell him why it did not work for the integers. None of us could. When he showed us why, none of could believe we didn't see it right away. There were between 50 and 100 budding young mathematicians in the class.

Maybe it was the way he presented it, but, he said that most people who were familiar with what he had just put on the board "felt" that that must have been what Fermat was thinking.

I had read many timnes later essentialy the same remark. The last time was recently in the context of a discussion of Lame's error that I tried to paraphrase in my previous note. I must have misread that last note because, aren't the Gaussin integers an UFD?

I'm not trained in arithmetic geometry. I'm essentially trying to learn it on my own. Have been from for the last 15 years. But, I get a lot of help from my friends.

It's great to be able to write stuff here and have people like you point out where you have misunderstood something.

Oh, to summarize --

1) Aren't the Gaussina integers a UFD?

2) Is not Q(squareroot(-23) the first imaginary quadratic number field that is not a UFD?

Thanks again.

DJ

P.S., I was trained in differential and algebraic topology, partial differential equations, differential geometry, C* and W* algebras and quantum field theory - that is - elementary particle theory - especially the problem of quantizing the gravitational field. That was in 1967-1972 - the years I was in graduate school. Precious years.

DJ

7. Jun 24, 2008

### DeaconJohn

Gel, I'm trying to understand ramification in imaginary quadratic number fields. I know it has something to do with prime ideals and rational prime numbers and

3 = (1-squareroot(5))(1+squareroot(-5))/2

but I just can't quite put it all together in a way that I can remember.

I don't have Cassels and Frolich or Neukrich, two of the classics. They are out of print. I have not been able to get copies. The books I have (and that I can understand) explain different bits and pieces in different books. My favorite, Ian Steward (and David Tall?) "Introduction to ALgebraic Number Theory and Fermat's last theorem," doesn't discuss ramification at all. It goes right up to the front door and stops. Oh, I have one book that does a good job but the discussion is spread out over 75 pages with lots of rabbit trails and "exercises for the reader." That's Marcus, "number fields."

8. Jun 24, 2008

### gel

I don't specialise in number theory either -- probability and stochastic processes is more my field -- but it can be fun.
I think that what you are probably getting at is that 6 decomposes in two distict ways in $Z[\sqrt{-5}]$,
$$6 = 2.3 = (1+\sqrt{-5})(1-\sqrt{-5}).$$
Taking norms, you can see that $2,3,1\pm\sqrt{-5}$ are irreducible, as they have norms 4,9,6 respectively and no element has norm 2 or 3 (the norm of $a+b\sqrt{-5}$ is $a^2+5b^2$).
It follows that this is not a UFD.
So, you should have said -5, not -23 in your previous post (but, http://en.wikipedia.org/wiki/23_%28number%29" [Broken]).
btw, the proper prime decomposition using ideals is
\begin{align*} &(2)=(2,1+\sqrt{-5})^2\\ &(3)=(3,1+\sqrt{-5})(3,1-\sqrt{-5})\\ &(6)=(2,1+\sqrt{-5})^2(3,1+\sqrt{-5})(3,1-\sqrt{-5}). \end{align*}

Ramification is different. 2, and $5 = -(\sqrt{-5})^2$ have repeated prime factors in $Z[\sqrt{-5}]$, so they ramify. More generally, the set of rational primes which ramify in any given number field is finite, and equals the set of prime factors of the discriminant of the field.

For your other question, the Gaussian integers, Z, is a UFD.
Given any two a,b in Z with b non-zero you can write a=pb+q for Gaussian integers p,q with |q|<|b| (just let p be the closest Gaussian integer to a/b). This means that the Euclidean algorithm works for finding the gcf of Gaussian integers -- i.e. Z is Euclidean -- and is therefore a UFD.

Last edited by a moderator: May 3, 2017
9. Jun 24, 2008

### gel

Interestingly, unique factorization and ramification can also be related to ideas in algebraic topology and differential geometry. This might sound all rather complicated, but it is fascinating.

A number field k whose ring of integers is R can be understood as a geometric object (actually, a scheme).
- points are the prime ideals (forming the spectrum of R, spec(R)).
- closed sets are the finite sets and the whole of spec(R). This is the Zariski topology.
- if S is a closed subset of spec(R), the rational (i.e. smooth) functions on the complement S are those elements of k which can be written as x=a/b for a,b in R and b coprime to the primes in S. The 'value' of x at a point P is its image in the quotient field R/P.

Given a point P on a differentiable manifold M, you have the tangent space -- or better, the cotangent space. If m(P) is the space of smooth functions vanishing at P then the quotient m(P)/m(P)2 can be identified with the cotangent space.

Similarly, a point P of spec(R) has (co)tangent space P/P^2.
An extension of number fields, k1 subset of k2, gives an onto map Spec(R2)->Spec(R1).
The primes which ramify are, in a sense, degenerate points of this map - they are the points at which the induced map on the tangent space is singular.
In particular, for quadratic extensions of Q, one of 3 things can happen for each point of Spec(Z) (i.e. for each rational prime). Either they remain prime, so the map is 1-1 at that point, or they split into 2 primes so the map is 2 to 1, or they ramify so the map is 1-1 but singular.

I'm getting a bit out of my depth now, but here goes...the class group looks a lot like a first cohomology group (Spec(R) is one dimensional, so no higher cohomology groups). UFDs are then a bit like manifolds with trivial cohomology.

10. Jun 25, 2008

### DeaconJohn

Thanks, I'm thinking about it. DJ

11. Jun 25, 2008

### DeaconJohn

Gel,

For most of your post, you were talking about the things I am trying to learn regarding the geometric aspects of number theory and then you leap-frogged way ahead of me when you brought up the analogy between the class group and the cohomology group. That's a suprising connection, suprising to me, that is.

You certainly have a gift for a coherent summary of a large body of material.

Actually, my problem with ramification and splitting was that I did not have any good examples of the basic definitions. Last night I found an example that clarified the basic definitions for me. Namely, the Wikipedia example of splitting and ramification in the Gaussian integers:

http://en.wikipedia.org/wiki/Splitt...sions#Example_.E2.80.94_the_Gaussian_integers

What would you recommend as a reference for the circle of ideas that you are talking about in your post referenced above? Especially for the part about the class group looking a bit like the first cohomology group. Is that etale cohomology you have in mind? that would be my guess as the best candidate for such an object.

My reason for that guess is that it is the etale cohomology that links most closely to motives and algebraic k-theory and to the Weil conjectures (polished off by Pierre Deligne in .... 1974?).

So, I would suspect that etale would normally be most closely related to number theory. Although crystalline cohomology seems to "pop up" for a few "special" primes from time to time, it is etale that is used for "almost all" primes, in what I've seen so far.

Yours,

DJ

Last edited: Jun 25, 2008
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