# I Definition of an irreducible element in an integral domain

1. Feb 18, 2017

### Math Amateur

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) ...

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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2. Feb 18, 2017

### Staff: Mentor

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).

3. Feb 18, 2017

### Math Amateur

Thanks for those thoughts, fresh_42 ... ...

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

Thanks again ...

Peter

4. Feb 18, 2017

### lavinia

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.

5. Feb 18, 2017

### Math Amateur

Thanks Lavinia... most illuminating and most helpful!!

Peter

NOTE: Just now puzzling over finding an integral domain where irreducibles are not prime ...???

6. Feb 18, 2017

### Staff: Mentor

Be sure not to chose a UFD (unique factorization domain) and consider the implication of ring properties listed on the Wiki-link.

7. Feb 19, 2017

### lavinia

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
8. Feb 21, 2017

### lavinia

If you like I can walk you through an example with a series of small exercises.

9. Feb 21, 2017

### Math Amateur

Very kind of you, Lavinia ...

I suspect that would be extremely helpful ...

Peter

10. Feb 21, 2017

### lavinia

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 $R$.

11. Feb 22, 2017

### Math Amateur

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?

Peter

12. Feb 22, 2017

### Staff: Mentor

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.

13. Feb 23, 2017

### Math Amateur

Thanks fresh_42 ... yes ... accept your points ...

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

Peter

14. Feb 23, 2017

### Math Amateur

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
15. Feb 24, 2017

### lavinia

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
16. Feb 25, 2017

### Math Amateur

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

17. Feb 25, 2017

### lavinia

Correct.

18. Feb 25, 2017

### Staff: Mentor

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.

19. Feb 25, 2017

### Math Amateur

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

20. Feb 26, 2017

### lavinia

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