UFDs .... Counterexample - I_8 = Z/8Z

In summary, Joseph J. Rotman's book: A First Course in Abstract Algebra with Applications (Third Edition) discusses the concept of degree functions in linear polynomials, and how they can be used to prove that the linear factors are irreducible. It is possible for a ring $R$ not to be an integral domain, and for linear polynomials in $R[x]$ to not be a unique factorization domain. However, if $R$ is a UFD, then so is $R[x]$.
  • #1
Math Amateur
Gold Member
MHB
3,990
48
I am reading Joseph J. Rotman's book: A First Course in Abstract Algebra with Applications (Third Edition) ...

I am currently focused on Section 3.6 Unique Factorization ...

I need help with an aspect of Example 3.70 ...

The relevant text from Rotman's book is as follows:https://www.physicsforums.com/attachments/4651

[NOTE: In the above text, Rotman uses \(\displaystyle \mathbb{I}_8\) for the integers \(\displaystyle \text{ mod } m\) instead of the more usual \(\displaystyle \mathbb{Z} / \mathbb{Z}_8\) ... ... ]In the above text, Rotman mentions that it is not "obvious" that the linear factors are irreducible ... ...Now, on page 263 he proves that a linear polynomial \(\displaystyle f(x) \in k[x]\) where \(\displaystyle k\) is a field is always irreducible ... ... as follows:View attachment 4652Now Rotman's proof that linear polynomials seems to rest on \(\displaystyle k\) being a field ... but why ... that is my problem ...My thinking on the issue is that \(\displaystyle k\) being a field means \(\displaystyle k[x]\) is a domain ... and, I think a Euclidean ring/domain ... and that \(\displaystyle k[x]\) needs to be a Euclidean ring/domain for a degree function to exist in \(\displaystyle k[x]\) ... ... and Rotman's argument that linear factors are irreducible rests on an argument related to the degrees of polynomials ...Can someone please confirm that my thinking is correct ... or point out deficiencies or errors in my thinking ...

Another question I have is the following:

How in Example 3.85 do we, in fact, prove that each of the linear factors in the example are actually irreducible ...

Hope someone can help ...

Peter
 
Physics news on Phys.org
  • #2
Peter said:
Now Rotman's proof that linear polynomials [are irreducible] seems to rest on \(\displaystyle k\) being a field ... but why ... that is my problem ...
The main thing that goes wrong if $k$ is not a field is that the property $\deg(gh) = \deg(g) + \deg(h)$ no longer holds. For example, in Rotman's ring $\mathbb{I}_8[x]$, $(4x+1)(2x+1) = 6x+1$, so that the linear polynomial $6x+1$ is not irreducible.

The presence of zero-divisors ($4\times2=0$ in $\mathbb{I}_8$) undermines any theory of unique factorisation.
 
  • #3
Opalg said:
The main thing that goes wrong if $k$ is not a field is that the property $\deg(gh) = \deg(g) + \deg(h)$ no longer holds. For example, in Rotman's ring $\mathbb{I}_8[x]$, $(4x+1)(2x+1) = 6x+1$, so that the linear polynomial $6x+1$ is not irreducible.

The presence of zero-divisors ($4\times2=0$ in $\mathbb{I}_8$) undermines any theory of unique factorisation.
Thanks Opalg ... yes, see the point regarding zero divisors ...

So, then, worrying about degree functions is misguided? Can you comment on that issue ...?

Peter
 
  • #4
Rotman no doubt uses the notation $\Bbb I_n$ instead of $\Bbb Z_n$, to avoid confusion in the case of $n = p$, a prime, because $\Bbb Z_p$ might mean either: the integers modulo $p$, or the $p$-adic integers, which are an entirely different thing (instead of measuring distances between numbers "linearly", we can measure "how close to powers of $p$" they are, which yields an entirely different notion of "metric").

A ring $R$ need not be a field for $R[x]$ to be a unique factorization domain, however, if it is not even an integral domain, $R[x]$ CANNOT be a UFD, as all UFD's are NECESSARILY integral domains.

In fact, if $R$ is a UFD, then so is $R[x]$. So for example, we have unique factorization in $\Bbb Z[x]$, even though $\Bbb Z$ is definitely not a field.

In my opinion, a more "spectacular" demonstration of non-unique factorization occurs when $R = \Bbb Z[\sqrt{-5}]$: for example:

$6 = 2(3) = (1+\sqrt{-5})(1 - \sqrt{-5})$

Here, the problem is *not* that we have zero divisors, it is something more subtle. It was thought that such extensions of the integers would share their properties-indeed, it was quite a shock when it was discovered this was not so.
 
  • #5
Deveno said:
Rotman no doubt uses the notation $\Bbb I_n$ instead of $\Bbb Z_n$, to avoid confusion in the case of $n = p$, a prime, because $\Bbb Z_p$ might mean either: the integers modulo $p$, or the $p$-adic integers, which are an entirely different thing (instead of measuring distances between numbers "linearly", we can measure "how close to powers of $p$" they are, which yields an entirely different notion of "metric").

A ring $R$ need not be a field for $R[x]$ to be a unique factorization domain, however, if it is not even an integral domain, $R[x]$ CANNOT be a UFD, as all UFD's are NECESSARILY integral domains.

In fact, if $R$ is a UFD, then so is $R[x]$. So for example, we have unique factorization in $\Bbb Z[x]$, even though $\Bbb Z$ is definitely not a field.

In my opinion, a more "spectacular" demonstration of non-unique factorization occurs when $R = \Bbb Z[\sqrt{-5}]$: for example:

$6 = 2(3) = (1+\sqrt{-5})(1 - \sqrt{-5})$

Here, the problem is *not* that we have zero divisors, it is something more subtle. It was thought that such extensions of the integers would share their properties-indeed, it was quite a shock when it was discovered this was not so.
Thanks for a most interesting post, Deveno ...

BUT ... given that the problem with $R = \Bbb Z[\sqrt{-5}]$ is not zero divisors ... then what is the source or cause of the failure of unique factorization in this case ... ?

Peter
 
  • #6
Peter said:
BUT ... given that the problem with $R = \Bbb Z[\sqrt{-5}]$ is not zero divisors ... then what is the source or cause of the failure of unique factorization in this case ... ?
Nothing special. It's in the nature of things. About the example provided by Deveno, I quote from Lectures on Modern Algebra (Paul Dubreil, M.L. Dubreil-Jacotin)

This example, which is due to Dedekind has played an important part in the foundation of theory of ideals. This theory was created to supply a kind of unique factorisation in rings which are not unique factorisation domains.
 
  • #7
Fernando Revilla said:
Nothing special. It's in the nature of things. About the example provided by Deveno, I quote from Lectures on Modern Algebra (Paul Dubreil, M.L. Dubreil-Jacotin)

This example, which is due to Dedekind has played an important part in the foundation of theory of ideals. This theory was created to supply a kind of unique factorisation in rings which are not unique factorisation domains.
Thanks Fernando ... thanks for the help ...

You write: " ... Nothing special. It's in the nature of things. ... " ... that serves me right for looking for a reason for everything ... :)

Thanks again, Fernando ... definitely appreciate the clarity and help ...

Peter
 
  • #8
Motivated in part by an investigation of Fermat's Last Theorem (the infamous:

$x^n + y^n = z^n$)

Ernst Kummer (1810-1893) began looking into what we would call "integral algebraic extensions". In particular he was interested in the cyclotomic extensions of the rationals obtaining by adjoining $n$-th roots of unity (these are complex numbers for which $z^n = 1$. Geometrically, they lie on the vertices of a regular $n$-gon inscibed on the unit circle, with a vertex at $(1,0)$. They form a cyclic group of order $n$ under complex multiplication).

He almost succeeded in proving FLT, but his proof had a fatal flaw: it assumed that $\Bbb Q(\zeta_p)$ (here $\zeta_p$ is a primitive $p$-th root of unity-this notation is "more or less" standard) was a unique factorization domain.

It turns out this although this is indeed true for some of the smaller primes, it is not true for (for example) $p = 23$. So Kummer began looking for a kind of "substitute" for primes. He found what he was looking for in the form of what he called "ideal numbers". See: https://en.wikipedia.org/wiki/Ideal_number

Richard Dedekind (1831-1916) recognized that ideal numbers formed what we would now call an ideal, and made a first-rate discovery- that trying to factor elements into unique factors was the wrong approach, rather factoring the RING itself into prime ideals yielded more satisfactory results.

Richard Dedekind was a mathematical giant: not only did he lay the foundations for ring theory (a task ably and astonishingly extended by Emmy Noether), he was one of the first mathematicians to realize the importance of groups, his construction of the real numbers is astonishingly simple, beautiful and deep (and often the "gold standard" in many "rigorous" constructions presented at math departments across the world), but also was one of the first to re-cast Cantor's bewildering hierarchy of infinities into a more manageable language of mappings between sets.

You can find his "Essays on the Theory of Numbers" here. It's a good read (warning: the material is a bit dated, and it's not always easy going-Dedekind was writing for mathematicians, not students). In much of modern mathematics, the properties of the natural numbers (and their big brother-the integers) are "taken for granted".

Indeed, in the category of rings with unity (rings with unity, with the only mappings between them restricted to ring-homomophisms that preserve unity) the integers are an initial object, which is merely to say there is a unique ring-homomorphism:

$\phi: \Bbb Z \to R$ given by $\phi(n) = n\cdot 1_R$

(here, $2\cdot r$ means $r+r$, and so on).

Another way of saying this is: every ring is an extension ring of a quotient ring of the integers. Which quotient we have is determined by the characteristic of a ring. Often, we are interested in rings of characteristic 0, which are extension rings of the integers: $\Bbb Z \cong \Bbb Z_0 = \Bbb Z/0\Bbb Z = \Bbb Z/\{0\}$, or rings of prime characteristic $p$ which are extensions of $\Bbb Z_p = \Bbb Z/p\Bbb Z$.

Why are some rings "good", and some rings "bad" (hard to predict/calculate in)?

I cannot give a *comprehensive* answer, but the short answer is this: monoids are not as "nice" as groups. Invertibility, in a monoid, introduces a certain regularity: bijective functions are more manageable than non-bijective functions. With non-bijective functions (every monoid can be realized as a set of such functions on itself) we "lose information" we flow TO, but "going back" is uncertain.

Divisibility, is an attempt to "go back". This is why theorems involving divisibility are *so* important. As a simple example, if I want to know that a certain positive integer is a multiple of $6$: I need only check that it is even, and a multiple of $3$. I can take two completely different approaches to each one of these tasks. This is sufficient (and not merely necessary) because $\Bbb Z$ is a UFD, and $2$ and $3$ are prime (and thus co-prime).

The same principles apply if we want to know if $p(x) \in \Bbb Q[x]$ is a multiple of $x^3 - 1$, we check if $1$ is a root, and if $x^2 + x + 1$ leaves no remainder upon (synthetic) division.
 
  • #9
Deveno said:
Motivated in part by an investigation of Fermat's Last Theorem (the infamous:

$x^n + y^n = z^n$)

Ernst Kummer (1810-1893) began looking into what we would call "integral algebraic extensions". In particular he was interested in the cyclotomic extensions of the rationals obtaining by adjoining $n$-th roots of unity (these are complex numbers for which $z^n = 1$. Geometrically, they lie on the vertices of a regular $n$-gon inscibed on the unit circle, with a vertex at $(1,0)$. They form a cyclic group of order $n$ under complex multiplication).

He almost succeeded in proving FLT, but his proof had a fatal flaw: it assumed that $\Bbb Q(\zeta_p)$ (here $\zeta_p$ is a primitive $p$-th root of unity-this notation is "more or less" standard) was a unique factorization domain.

It turns out this although this is indeed true for some of the smaller primes, it is not true for (for example) $p = 23$. So Kummer began looking for a kind of "substitute" for primes. He found what he was looking for in the form of what he called "ideal numbers". See: https://en.wikipedia.org/wiki/Ideal_number

Richard Dedekind (1831-1916) recognized that ideal numbers formed what we would now call an ideal, and made a first-rate discovery- that trying to factor elements into unique factors was the wrong approach, rather factoring the RING itself into prime ideals yielded more satisfactory results.

Richard Dedekind was a mathematical giant: not only did he lay the foundations for ring theory (a task ably and astonishingly extended by Emmy Noether), he was one of the first mathematicians to realize the importance of groups, his construction of the real numbers is astonishingly simple, beautiful and deep (and often the "gold standard" in many "rigorous" constructions presented at math departments across the world), but also was one of the first to re-cast Cantor's bewildering hierarchy of infinities into a more manageable language of mappings between sets.

You can find his "Essays on the Theory of Numbers" here. It's a good read (warning: the material is a bit dated, and it's not always easy going-Dedekind was writing for mathematicians, not students). In much of modern mathematics, the properties of the natural numbers (and their big brother-the integers) are "taken for granted".

Indeed, in the category of rings with unity (rings with unity, with the only mappings between them restricted to ring-homomophisms that preserve unity) the integers are an initial object, which is merely to say there is a unique ring-homomorphism:

$\phi: \Bbb Z \to R$ given by $\phi(n) = n\cdot 1_R$

(here, $2\cdot r$ means $r+r$, and so on).

Another way of saying this is: every ring is an extension ring of a quotient ring of the integers. Which quotient we have is determined by the characteristic of a ring. Often, we are interested in rings of characteristic 0, which are extension rings of the integers: $\Bbb Z \cong \Bbb Z_0 = \Bbb Z/0\Bbb Z = \Bbb Z/\{0\}$, or rings of prime characteristic $p$ which are extensions of $\Bbb Z_p = \Bbb Z/p\Bbb Z$.

Why are some rings "good", and some rings "bad" (hard to predict/calculate in)?

I cannot give a *comprehensive* answer, but the short answer is this: monoids are not as "nice" as groups. Invertibility, in a monoid, introduces a certain regularity: bijective functions are more manageable than non-bijective functions. With non-bijective functions (every monoid can be realized as a set of such functions on itself) we "lose information" we flow TO, but "going back" is uncertain.

Divisibility, is an attempt to "go back". This is why theorems involving divisibility are *so* important. As a simple example, if I want to know that a certain positive integer is a multiple of $6$: I need only check that it is even, and a multiple of $3$. I can take two completely different approaches to each one of these tasks. This is sufficient (and not merely necessary) because $\Bbb Z$ is a UFD, and $2$ and $3$ are prime (and thus co-prime).

The same principles apply if we want to know if $p(x) \in \Bbb Q[x]$ is a multiple of $x^3 - 1$, we check if $1$ is a root, and if $x^2 + x + 1$ leaves no remainder upon (synthetic) division.



Hi Deveno ...

Thanks for a very interesting and informative post ...

You write:

" ... ... I cannot give a *comprehensive* answer ... "

Well your answer is, I think, as comprehensive as it gets ...

Thanks again,

Peter
 

1. What is a UFD?

A UFD, or Unique Factorization Domain, is a type of commutative ring in mathematics that has the property that every non-zero element can be uniquely factored into a product of irreducible elements.

2. What are some examples of UFDs?

Some examples of UFDs include the integers (Z), polynomials over a field, and Gaussian integers.

3. What is a counterexample?

A counterexample is an example that disproves a statement or claim. In this case, I_8 = Z/8Z is a counterexample to the idea that all integers are UFDs.

4. What is I_8?

I_8 refers to the ideal generated by the integer 8 in the ring Z/8Z. This ideal consists of all elements in Z/8Z that are multiples of 8.

5. How does I_8 serve as a counterexample?

I_8 serves as a counterexample because it shows that not all integers can be uniquely factored in Z/8Z. In fact, in this ring, 8 can be factored as 2 * 2 * 2, but it can also be factored as 4 * 2, demonstrating that the factorization is not unique.

Similar threads

  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
4
Views
1K
  • Linear and Abstract Algebra
Replies
5
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
10
Views
2K
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
8
Views
2K
Back
Top