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I Definition of an irreducible element in an integral domain

  1. Feb 18, 2017 #1
    Joseph A. Gallian, in his book, "Contemporary Abstract Algebra" (Fifth Edition) defines an irreducible element in a domain as follows ... (he also defines associates and primes but I'm focused on irreducibles) ...



    ?temp_hash=d48504d3474d6b78ff7865ba0298a62c.png



    I am trying to get a good sense of this definition ...

    My questions are as follows:

    (1) Why are we dealing with a definition restricted to an integral domain ... why can't we deal with a general ring ... presumably we don't want zero divisors ... but why ...

    (2) What is the logic or rationale for excluding a unit ...that is why is a unit not allowed to be an irreducible element ..

    (3) We read that for an irreducible element ##a##, if ##a = bc## then ##b## or ##c## is a unit ... ... why is this ... ... ? ... ... ... presumably for an irreducible we want to avoid a situation where ##a## has a "genuine" factorisation ... but how does ##b## or ##c## being a unit achieve this ...


    Hope someone can help ...

    Peter
     

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  3. Feb 18, 2017 #2

    fresh_42

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    Hi Peter,
    I don't see the reason for this restriction either. It is not needed. I have a book on number theory, which defines irreducibility on commutative rings with ##1##. Commutative, for otherwise one has to distinguish left and right, which only complicates it, so commutativity is for the lazy ones (and number theory deals with ##\mathbb{Z}## which is Abelian). The ##1## is needed, for otherwise the term unit wouldn't make sense. However, I assume, that Gallian wants to show that the two definitions prime and irreducible are the same and this holds true only in integral domains (if I remember correctly. To find out whether I'm correct on this, one might try to find a counterexample in ##\mathbb{Z}_{12}## or a matrix ring).
    A unit ##u## is an element for which there is another element ##v##, such that ##uv=1##. Next you could also write ##u=u^2v=uvu=(uv)^nu## and so on. The concept simply wouldn't make much sense to investigate on units. And again have the coincidence of the two definitions in mind. Why primes shouldn't be units is a similar question and I think, we've already discussed this in an earlier thread. In the end it's for the same reason: it would water down the definition close to uselessness.
    Same as above: units can be added arbitrarily so we want to distinguish between "proper" decompositions (reducible elements) and "improper" decompositions, i.e. those which only allow units as factors (irreducible elements).
     
  4. Feb 18, 2017 #3
    Thanks for those thoughts, fresh_42 ... ...

    Just now reflecting on and thinking about what you have said ...

    Thanks again ...

    Peter
     
  5. Feb 18, 2017 #4

    lavinia

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    The concept of prime and non-factorizable do not require an integral domain, only a commutative ring with identity.

    However the study of divisibility theory naturally leads to integral domains. Integral domains have the crucial property that if an element ##a## is not zero and ##ab=ac## then ##b=c## (since ##a(b-c)=0## and an integral domain has no zero divisors). This in turn implies that a prime factorization of any element of the ring is essentially unique. By "essentially" is meant that if ##a## has two factorizations ##a = up_1...p_{n} = vq_1...q_{m}## where ##u## and ##v## are units and the ##p_{i}##'s and ##q_{j}##'s are primes then ##n=m## and there is a permutation of the ##q_{j}##'s so that ##p_{i}|q_{i}## and ##q_{i}|p_{i}##. (the symbol "##|##" means "divides".)

    The integers are an integral domain and have the further property that primes and non-factorizables are the same. But in an arbitrary integral domain this may not be true. It is simple to show that a prime must be a non-factorizable but a non-factorizable may not be prime. In this case it is possible for there to be distinct factorizations of an element into non-factorizables. These factorizations may not have the same length and no two may divide each other as in a prime factorization. A ring in which every element has an essentially unique factorization into non-factorizables is called a Unique Factorization Domain. One can show that in a UFD that non-factorizables and primes are the same.

    - Two good exercises:

    1) Prove that in an integral domain prime factorizations are essentially unique.

    2) Find an integral domain in which there are non-factorizables that are not prime.
     
  6. Feb 18, 2017 #5
    Thanks Lavinia... most illuminating and most helpful!!

    Peter


    NOTE: Just now puzzling over finding an integral domain where irreducibles are not prime ...???
     
  7. Feb 18, 2017 #6

    fresh_42

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    Be sure not to chose a UFD (unique factorization domain) and consider the implication of ring properties listed on the Wiki-link.
     
  8. Feb 19, 2017 #7

    lavinia

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    For the unique factorization proof , try an induction on the lengths of the factorizations.

    Good examples of rings are the integers with algebraic irrational numbers adjoined, for instance the ring generated by the integers and the square root of 3 or of -5.
     
    Last edited: Feb 20, 2017
  9. Feb 21, 2017 #8

    lavinia

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    If you like I can walk you through an example with a series of small exercises.
     
  10. Feb 21, 2017 #9
    Very kind of you, Lavinia ...

    I suspect that would be extremely helpful ...

    Peter
     
  11. Feb 21, 2017 #10

    lavinia

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    The ring ##R## will be the integers with the square root of ##-5## adjoined. This is the set of all complex numbers of the form ##a + b\sqrt{5}i## where ##a## and ##b## are integers.

    Note that the square of the complex norm ##|x|^2 = a^2+5b^2## is an integer and ##|xy|^2=|x|^2|y|^2##.

    Use the square of the norm to show that the units of ##R## are ##±1## that ##3## is non-factorizable in ##R## and that ##2±\sqrt{5}i## are also both non-factorizable in ##R##.

    From the equation ##9 = 3⋅3 = (2+\sqrt{5}i)(2-\sqrt{5}i)## conclude that ##3## is not prime in ##R##.
     
  12. Feb 22, 2017 #11

    Thanks for the exercise Lavinia ...

    =================================================================================

    First, show that the units of ##R = \mathbb{Z} [ \sqrt{ -5 } = \{ a + b \sqrt{ 5 } i \ | \ a, b \in \mathbb{Z} \}## are ##\pm 1##

    In order to show that the units of ##R## are ##\pm 1##, we first define a function (norm) ##N \ : \ R \ \longrightarrow \mathbb{Z}## such that for ##r \in R## :

    ##N(r) = (a + b \sqrt{ 5 } i ) ( a - b \sqrt{ 5 } i) = a^2 + 5 b^2##

    Then we have that if ##r \in R## is a unit ... ... then ... ... ##rs = 1## for some ##s \in R## and so ... ...

    ##1 = N(rs) = N(r)N(s)## ... ...

    ... and so it follows that ##r \in R## is a unit if and only if ##N(r) = 1## ... ...



    Now, consider ##r = a + b \sqrt{ 5 } i## ... ...

    If r is a unit ... ... we have that ##N(r) = a^2 + 5 b^2 = 1## ...

    But ... ##r = \pm 1## are the only values satisfying ##N(r) = 1## ... ...

    Hence the only units of ##R## are ##\pm 1## ... ...


    Is that correct?


    Rest of the answers will follow shortly ...

    Peter
     
  13. Feb 22, 2017 #12

    fresh_42

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    Yes, this is correct.

    Although I personally think you should have proven the equations ##N(rs)=N(r)N(s)## and ##N(1)=1## at the beginning because they are the essential part of this proof. You also use ##N(r) \geq 0## whithout mentioning. Do you see where?

    And to be peeky: You have shown that a unit has to be either ##1## or ##-1##. You haven't shown, that they are actually units, so your "hence" is not 100% true. A remark would do, but I want you to see the difference between a necessary condition, which you have shown, and a sufficient condition, which you have not. I admit, this is a bit of a too simple example for this, but nevertheless a good habit to get used to.
     
  14. Feb 23, 2017 #13
    Thanks fresh_42 ... yes ... accept your points ...

    Thanks for the critique ... most helpful, as usual ...

    Peter
     
  15. Feb 23, 2017 #14
    Hi Lavinia, fresh_42

    In this post I will attempt to show that 3 is irreducible (non-factorizable) in ##R = \mathbb{Z} [ \sqrt{ -5 } ] = \{ a + b \sqrt{ 5 } i \ | \ a, b \in \mathbb{Z} \}##


    Let ##3 = rs## where ##r,s \in R## ...

    Then ##N(3) = 9 = N(rs) = N(r) N(s)##

    Now, ##N(r) N(s) = 9 \ \Longrightarrow \ N(r) = 1,3## or ##9##

    But note that ##N(r) = 3## is impossible as there are no integers ##a, b## for which ##a^2 + 5b^2 = 3##


    Now, if ##N(r) = 1## then r is a unit

    ... and if ##N(r) = 9##, then ##N(s) = 1## and ##s## is a unit ... ...


    So .. ... ##3## is not a unit and whenever ##3 = rs## we have that one of ##r## or ##s## is a unit ...

    ... thus ##3## is irreducible ... that is it does not have a 'genuine' factorization in R ... that is, 3 is non-factorizable in R


    Is that correct?

    Peter


    ===================================================================


    *** EDIT ***

    Just realised that for ##2 + \sqrt{ -5 } i## and ##2 - \sqrt{ -5 }i## we have

    ##N(2 + \sqrt{ -5 } i) = N(2 + \sqrt{ -5 } i) = 9 = N(3)## ... ...

    and so the argument for the irreducibility or non-factorizability of ##2 + \sqrt{ -5 } i## and ##2 - \sqrt{ -5 }i## in ##R## is similar to the argument for the irreducibility of ##3## in ##R## ...

    Peter
     
    Last edited: Feb 23, 2017
  16. Feb 24, 2017 #15

    lavinia

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    Correct. Well done.

    - Some things to think about:

    This exercise illustrates the importance of distinguishing primes and irreducibles from units. The definitions that exclude units are not arbitrary.

    Using the norm makes otherwise difficult questions easy by translating questions about the ring ##R## into questions about the integers.This exercise is only one example. The translation involves two things: the norm is non-zero except for zero; the norm factors over multiplication. These two properties make everything work.

    The norm can be used to study any subring of the complex numbers - for instance the integers with the seventh roots of one adjoined or the rational numbers with the square root of 13 adjoined.

    - Here are a couple more exercises about the ring ##R## that use the norm.

    ##R## is an integral domain

    If the norm squared of a ring element is a prime number then the ring element is non-factorizable.
     
    Last edited: Feb 24, 2017
  17. Feb 25, 2017 #16
    Hi Lavinia, fresh_42 ... ...

    Thanks for the exercises, Lavinia ... and also ... thanks to you both for your thoughts/critique regarding previous exercises ... ...


    In this post I will attempt to show that ##R = \mathbb{Z} [ \sqrt{ -5 } ] = \{ a + b \sqrt{ 5 } i \ | \ a, b \in \mathbb{Z} \}## is an integral domain ... ...

    Since ##R## is clearly a commutative ring with a unity ... ... we have to show that ##R## has no zero divisors ...

    In other words. we have to show that there exist no elements of ##R## such that

    ##(a + b \sqrt{ 5 } i) ( c + d \sqrt{ 5 } i) = 0##

    where ##(a + b \sqrt{ 5 } i) \neq 0## and ##( c + d \sqrt{ 5 } i) \neq 0## ... ...


    Now, ##(a + b \sqrt{ 5 } i) ( c + d \sqrt{ 5 } i) = 0##

    ##\Longrightarrow N( (a + b \sqrt{ 5 } i) ( c + d \sqrt{ 5 } i) ) = N(0) = 0 ##

    ##\Longrightarrow N( (a + b \sqrt{ 5 } i) ) N( ( c + d \sqrt{ 5 } i) ) = 0##

    ##\Longrightarrow (a^2 + 5b^2) (c^2 + 5d^2) = 0##

    BUT ... there is no way that this equation can be satisfied with at least one (and possibly both) of ##a, b## non-zero and also at least one (and possibly both) of ##c, d## non-zero ... since all terms are squares of integers and thus positive with at least one of ##a,b## non-zero and at least one of ##c,d## non-zero ...

    Thus there are no zero divisors in R ... so R is an integral domain ...


    Is that correct?

    Peter
     
  18. Feb 25, 2017 #17

    lavinia

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    Correct.
     
  19. Feb 25, 2017 #18

    fresh_42

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    Yes, it is correct. You could have formalized your last step by:

    ##(a^2+5b^2)(c^2+5d^2) = 0##
    ##\Longrightarrow a^2+5b^2= 0 \; \vee \; c^2+5d^2=0 ## since ##\mathbb{Z}## is an integral domain
    ##\Longrightarrow \textrm{(w.l.o.g.) } \;a^2+5b^2 = 0##
    ##\Longrightarrow \; a= 0 \wedge b=0## since ##\mathbb{R}## is an ordered field (the reason why squares are positive)

    but this is only to give the entire proof a homogeneous look and to use "without loss of generality", not that anything is wrong with what you wrote. Of course w.l.o.g. isn't needed as one could as well do both cases successively as it is often the case when this terminology is used.
     
  20. Feb 25, 2017 #19



    Hi lavinia, fresh_42 ...

    Lavinia ... you asked me to carry out the following exercise about the ring ##R = \mathbb{Z} [ \sqrt{ -5 } ] = \{ a + b \sqrt{ 5 } i \ | \ a, b \in \mathbb{Z} \}## ... ... :

    "If the norm squared of a ring element is a prime number then the ring element is non-factorizable."


    I am having some trouble with this exercise and must be misunderstanding something ... let me explain ...

    It appears to me that no element of R has the property that its norm squared equals a prime ...

    because ..


    Let the ring element with this property be ##r = a + b \sqrt{ 5 } i##

    Now the norm squared of ##r## is a prime, say ##p## ...

    ... ... that is ##N(a + b \sqrt{ 5 } i)^2 = p## ... where ##p## is a prime ... ...

    Thus ##(a^2 + 5b^2)^2 = p## ... ...

    So ... ##a^2 + 5b^2 = \sqrt{p}## ... ... ... ... (1)

    BUT ...

    ... there is a problem since ##a,b## are integers, and hence ##a^2, b^2## are positive integers ...

    ... and so ##a^2 + 5b^2## is a positive integer ...

    BUT ... no positive integer can equal the square root of a prime since the square root of a prime number is an irrational number ...



    So I have a problem ... but i suspect I have misunderstood something ...

    Can you help ...?

    Peter
     
  21. Feb 26, 2017 #20

    lavinia

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    It is just a confusion of language that I created. ##a^2+5b^2## is the norm squared. The norm is ## (a^2+5b^2)^{1/2}##

    All of your proofs used the norm squared. Sorry for the confusion.

    - If ##x+iy## is a complex number then its norm is its distance to the origin. From the Pythagorean Theorem this is ##(x^2+y^2)^{1/2}##. The square of the norm is the square of the distance to the origin and this is ##(x^2+y^2)##. If the number is ##a^2+\sqrt5bi## then the square of its norm is ##a^2+5b^2##.
     
    Last edited: Feb 26, 2017
  22. Feb 27, 2017 #21

    mathwonk

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    This sort of subdivision of elements of a ring confused me for a long time, until it sort of became clear at one point. I agree Lavinia's comments are helpful in particular. The whole point of understanding a ring seems to be understanding how its elements factor. As she points out this is harder to make sense of in the presence of zero divisors, since you can't cancel them from a factorization.

    But for the same reason you can cancel non zero divisors, even if the ring has other zero divisors. Units are uninteresting for factorization because they divide everything, or maybe I should have said they are easy to understand with regard to factorization since they do divide everything.

    So the interesting and feasible problem is to understand the factorization of those elements that are not zero divisors but also are not units.

    I.e. one separates ring elements into zero divisors and non zero divisors, and within the non zero divisors one also separates out the units. So what is left to understand is the factorization of non unit, non zero divisors. This has at least a chance of being possible essentially uniquely, into irreducibles. As Lavinia makes clear, factorization (of non zero divisors) into primes is always unique, but it is not clear that a lot of primes exist. It is easier to show irreducibles exist in many cases, so the best situation is that irreducibles be also prime.

    Niote that if a non zero divisor factors at all, the factors cannot be zero divisors, so you never introduce zero divisors when exploring factorization of a non zero divisor.

    I.e. if ab = ac with b ≠c, then a(b-c) = 0 so a is a zero divisor. This says that sticking with non zero divisors will preserve cancellation. Then if a = bc, and c is a zero divisor, say cd = 0, with d ≠ 0, then also ad = bcd = 0 and a is also a zero divisor, so factoring non zero divisors does not introduce zero divisors.

    So you could study factorization in any commutative ring, if you stick to the non zero divisors, but most books don't do this. In fact this is unnecessary, since apparently in any ring, any ideal consisting entirely of zero divisors and maximal with this property, is prime. Then after modding out by such an ideal, one is reduced to studying essentially the same factorization problem, but in an integral domain.

    Life would be simpler if all the zero divisors formed an ideal, so you could just mod it out, but this can be false.

    http://math.stackexchange.com/questions/100941/do-the-zero-divisors-form-an-ideal
     
  23. Feb 27, 2017 #22

    fresh_42

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    It is. ##2\cdot 3=0## in ##\mathbb{Z}_6## so both are zero divisors, whereas ##2+3=5## is a unit.
    The process of factoring out is a double-edged sword in algebra. E.g. ##\mathfrak{Z}(G/\mathfrak{Z}(G)) \neq_{i.g.} \{1\}##, i.e. a factored out center doesn't necessarily lead to a center-less group.
     
  24. Feb 27, 2017 #23

    mathwonk

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    I was in error to assert that factoring out a maximal ideal of zero divisors yields essentially the same factorization problem in the quotient ring. Rather, in the quotient ring, elements that were zero divisors before may become non zero divisors, say if the element that previously multiplied it to zero has now become zero. E.g. in the product ring ZxZ, factoring out by the zero divisor of form (0,n) yields just Z, which is isomorphic to the subring Zx{0} of the original ring, which consists entirely of zero divisors there.

    fresh42's example seems typical. If you are familiar with "partitions of identity" in topology, you can see that the function identically equal to 1, is a sum of functions each of which has small support, hence is a zero divisor. In the ring ZxZ, also we have zero divisors (1,0) and (0,1) whose sum is the unit element of the ring. Indeed the identity element is frequently a sum of such "orthogonal idempotents".

    In fact notice that fresh's example Z/6Z is isomorphic to the product ring Z/2Z x Z/3Z, by the "chinese remainder theorem". where (1,1) is the unit element, so the elements 2 and 3 in Z/6 correspond to the elements (0,2) and (1,0), which do add up to the unit (1,2) whose inverse is itself.


    In geometry the phenomenon is associated to rings of function on spaces having more than one irreducible component. I.e. a function that vanishes on one but not both components, is annihilated by multiplying by a function that vanishes only on the other component. E.g. the quotient ring R[X,Y]/(XY) is the ring of function on the union of the X and Y axes. both X and Y are zero divisors, since they each vanish on one axis. Modding out by one of them say Y, yields the quotient ring R[X,Y](Y) ≈ R[X], the polynomial ring on the X axis, an integral domain.

    So it is seems in this case that one can reasonably study factorization in these separate quotient rings that correspond to the rings of functions on a single component at a time, but it is not clear to me how this relates to the problem of factorization of non zero divisors in the original ring, or whether that latter question is even particularly interesting. It is beginning to look to me as if factorization is just more natural and interesting in a domain.

    In geometry it is also nice when the zero locus of a single p[olynomial has dimension one less than the whole space. If a space has two pieces of the same dimension, the zero locus of a function vanishing on one piece has dimension the same as the whole space. So dimension theory is more complicated in such rings. I just do not know much about more general rings than domains.

    I guess one could look at a simple example like ZxZ, where say (4,3) is divisible by (2,1), so irreducible elements would have to have both entries irreducible in Z. But in this ring even factoring zero divisors like (n,0) would seem interesting, at least factoring them using only things of form (r,0), just unneccessary, since it agrees with that in Z. Maybe someone can find a reference where someone has looked at this.

    I guess in my rather simple example of ZxZ, it might make sense just to factor every element of form (n,m) into (1,m) and (n,1) and then factor further as usual in Z. So this does not seem too interesting?

    indeed what would we say about the factors? i.e. (3,3) = (3,1).(1,3), so i guess an irreducible would have to be a unit in one entry and an irreducible integer in the other?

    so if a ring had a finite number of maximal ideals of zero divisors, maybe an irreducible would be an element that is irreducible modulo one of them, and a unit modulo the others?

    in my example of R[X,Y]/(XY) maybe like (X+1)? I.e. this is X+1, mod Y, and 1 mod X. I.e. notice this function equals X+1 on the X axis, and 1 on the Y axis.

    So maybe my original guess was correct in this sense, i.e. this way perhaps one does reduce to factorizability in all of the quotients of the ring by its maximal ideals of zero dvisors. So one really does need to work essentially only in domains.

    then what is the irreducible factorization of X+Y?
     
    Last edited: Feb 27, 2017
  25. Feb 28, 2017 #24

    mathwonk

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    no i was too optimistic. I.e. in a product ring the components of the space are disjoint, whereas they intersect in the case of the X and y axes. So in the latter case although elements like X+Y+1 do factor as (X+1)(Y+1), elements like X+Y do not. I.e. this element seems to be irreducible. ???divi

    more generally, if f is a function of x divisible by x, and g is a polynomial in y divisible by y, then f+g+1 equals (f+1)(g+1). but what about f+g?

    interesting question by the OP!!
     
    Last edited: Feb 28, 2017
  26. Feb 28, 2017 #25
    Thanks lavinia ... understand now ...

    Peter
     
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