Irreducible Elements and Maximal Ideals in Integral Domains ....

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.3: Modules Over Principal Ideal Domains ... and I need some help in order to formulate a proof of Proposition 4.3.5 Part (iii)... ...

Proposition 4.3.5 reads as follows:https://www.physicsforums.com/attachments/8308
https://www.physicsforums.com/attachments/8309
Bland does not prove part (iii) of the above proof ...

Can someone please help me to formulate a rigorous proof ...

Peter

 

[Opalg]

https://www.physicsforums.com/attachments/8310

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.3: Modules Over Principal Ideal Domains ... and I need some help in order to formulate a proof of Lemma 4.3.5 Part (iii)... ...

Lemma 4.3.5 reads as follows:

View attachment 8311

Bland does not prove part (iii) of the above proof ...

Can someone please help me to formulate a rigorous proof ...

Peter

I believe that (3) is incorrect as it stands. What is true is that an element $a\in R$ is irreducible if and only if $(a)$ is maximal among the principal ideals of $R$. So (3) would be correct if $R$ is a PID. Since Section 4.3 is entitled Modules Over PIDs, maybe Paul Bland had PIDs in mind when stating the result. But in general an integral domain need not be a PID.

For a counterexample, let $R = \Bbb{Z}[x]$, which is an integral domain but not a PID. The element $x$ is irreducible in $R$, but $(x)$ is not a maximal ideal because it is contained in $\langle 2,x\rangle$, which is a larger, proper (but not principal) ideal.

 

[Math Amateur]

I believe that (3) is incorrect as it stands. What is true is that an element $a\in R$ is irreducible if and only if $(a)$ is maximal among the principal ideals of $R$. So (3) would be correct if $R$ is a PID. Since Section 4.3 is entitled Modules Over PIDs, maybe Paul Bland had PIDs in mind when stating the result. But in general an integral domain need not be a PID.

For a counterexample, let $R = \Bbb{Z}[x]$, which is an integral domain but not a PID. The element $x$ is irreducible in $R$, but $(x)$ is not a maximal ideal because it is contained in $\langle 2,x\rangle$, which is a larger, proper (but not principal) ideal.

Thanks Opalg ...

Hmm ... given what you have pointed out, I'm beginning to wonder about the other parts of Proposition 4.3.5 ... ...

Peter

 

[Olinguito]

I think we also have to exclude any ring with two or three elements (e.g. $\mathbb Z/2\mathbb Z$, $\mathbb Z/3\mathbb Z$). Such a ring is an integral domain (in fact a field) and a PID, but $\langle 0\rangle$ is a maximal ideal and $0$ is not an irreducible element.

Let $R$ be an integral domain that is also a PID and has more than three elements.

Suppose $a\in R$ is irreducible, and consider an ideal $\langle b\rangle$ such that $\langle a\rangle\subseteq\langle b\rangle\subseteq R$. Then $a=cb$ for some $c\in R$. As $a$ is irreducible, either $b$ or $c$ is a unit. If $b$ is a unit, then $\langle b\rangle=R$ ($\because$ for any $r\in R$, $r=b(b^{-1}r)\in\langle b\rangle$). If $c$ is a unit, then $a$ and $b$ are associates and so generate the same ideal, i.e. $\langle a\rangle=\langle b\rangle$. This shows that the ideal $\langle a\rangle$ is maximal.

Now suppose the ideal $\langle a\rangle$ is maximal. Since $R$ has more than three elements, $a\ne0$; also, as $\langle a\rangle\ne R$ by the definition of a maximal ideal, $a$ is not a unit. Let $a=bc$ for some $b,c\in R$. Then $\langle a\rangle\subseteq\langle c\rangle$; as $\langle a\rangle$ is maximal, either $\langle a\rangle=\langle c\rangle$ or $\langle c\rangle=R$. If $\langle a\rangle=\langle c\rangle$ then $c\in\langle a\rangle$ $\implies$ $c=da$ for some $d\in R$ $\implies$ $a=bc=bda$ $\implies$ $bd=1$ since $a\ne0$ and $R$ is an integral domain. Thus $b$ is a unit. If $\langle c\rangle=R$, then $1\in\langle c\rangle$ and so $c$ is a unit. This shows that $a$ is irreducible.

In fact, for any integral domain $R$ that is a PID and has more than three elements, the following are equivalent: for any $a\in R$,

• $a$ is irreducible,
• $\langle a\rangle$ is a maximal ideal,
• the quotient ring $R/\langle a\rangle$ is a field.

 

[Math Amateur]

I think we also have to exclude any ring with two or three elements (e.g. $\mathbb Z/2\mathbb Z$, $\mathbb Z/3\mathbb Z$). Such a ring is an integral domain (in fact a field) and a PID, but $\langle 0\rangle$ is a maximal ideal and $0$ is not an irreducible element.

Let $R$ be an integral domain that is also a PID and has more than three elements.

Suppose $a\in R$ is irreducible, and consider an ideal $\langle b\rangle$ such that $\langle a\rangle\subseteq\langle b\rangle\subseteq R$. Then $a=cb$ for some $c\in R$. As $a$ is irreducible, either $b$ or $c$ is a unit. If $b$ is a unit, then $\langle b\rangle=R$ ($\because$ for any $r\in R$, $r=b(b^{-1}r)\in\langle b\rangle$). If $c$ is a unit, then $a$ and $b$ are associates and so generate the same ideal, i.e. $\langle a\rangle=\langle b\rangle$. This shows that the ideal $\langle a\rangle$ is maximal.

Now suppose the ideal $\langle a\rangle$ is maximal. Since $R$ has more than three elements, $a\ne0$; also, as $\langle a\rangle\ne R$ by the definition of a maximal ideal, $a$ is not a unit. Let $a=bc$ for some $b,c\in R$. Then $\langle a\rangle\subseteq\langle c\rangle$; as $\langle a\rangle$ is maximal, either $\langle a\rangle=\langle c\rangle$ or $\langle c\rangle=R$. If $\langle a\rangle=\langle c\rangle$ then $c\in\langle a\rangle$ $\implies$ $c=da$ for some $d\in R$ $\implies$ $a=bc=bda$ $\implies$ $bd=1$ since $a\ne0$ and $R$ is an integral domain. Thus $b$ is a unit. If $\langle c\rangle=R$, then $1\in\langle c\rangle$ and so $c$ is a unit. This shows that $a$ is irreducible.

In fact, for any integral domain $R$ that is a PID and has more than two elements, the following are equivalent: for any $a\in R$,

• $a$ is irreducible,
• $\langle a\rangle$ is a maximal ideal,
• the quotient ring $R/\langle a\rangle$ is a field.

Thanks for a most interesting and VERY helpful post Olinguito ...

I very much appreciate your help ...

Thanks again ...

Peter

 

[Math Amateur]

Thanks for a most interesting and VERY helpful post Olinguito ...

I very much appreciate your help ...

Thanks again ...

Peter

In his very interesting post above, Olinguito indirectly implies (I think) that (up to an isomorphism) there is only one ring of two elements and one ring of three elements ... but ... how would we prove this ... ?

Can someone please demonstrate ow we prove this fact ...

... then it would follow that all rings of two elements are isomorphic to \(\displaystyle \mathbb{Z} / 2 \mathbb{Z}\) and all rings of three elements are isomorphic to \(\displaystyle \mathbb{Z} / 3 \mathbb{Z}\) and thus have \(\displaystyle \langle 0 \rangle\) as a maximal ideal where 0 is NOT an irreducible element ...Thanks again to Olinguito for a very interesting and helpful post ...

Peter

 

[Olinguito]

Here is an beautiful result about finite fields.

Given a prime $p$ and a positive integer $n$, there exists an up-to-isomorphism unique field with exactly $p^n$ elements. Conversely, the number of elements in any finite field is always of the form $p^n$ for some prime $p$ and positive integer $n$.​

I truly find this one of the most beautiful results in mathematics, and abstract algebra in particular. We can denote the isomorphically unique field with $q=p^n$ elements $\mathbb F_q$. This is the splitting field of the polynomial $x^q-x$ over a field with $p$ elements (e.g. $\mathbb Z/p\mathbb Z$); it is also the quotient ring of the ring of polynomials over a field with $p$ elements by the principal ideal generated by an irreducible polynomial of degree $n$.

 

[Olinguito]

I think we also have to exclude any ring with two or three elements (e.g. $\mathbb Z/2\mathbb Z$, $\mathbb Z/3\mathbb Z$). Such a ring is an integral domain (in fact a field) and a PID, but $\langle 0\rangle$ is a maximal ideal and $0$ is not an irreducible element.

Argh, I wasn’t thinking things through. What we’re interested in is any integral domains that is also a PID but not a field. $\mathbb Z/2\mathbb Z$ and $\mathbb Z/3\mathbb Z$ are fields, but so is any finite integral domain. In any case, a field has no maximal ideals (the only ideals are $\langle 0\rangle$ and the whole field itself) or irreducible elements (every element is either zero or a unit) so the result is vacuously true for fields.

 

[Math Amateur]

I think we also have to exclude any ring with two or three elements (e.g. $\mathbb Z/2\mathbb Z$, $\mathbb Z/3\mathbb Z$). Such a ring is an integral domain (in fact a field) and a PID, but $\langle 0\rangle$ is a maximal ideal and $0$ is not an irreducible element.

Let $R$ be an integral domain that is also a PID and has more than three elements.

Suppose $a\in R$ is irreducible, and consider an ideal $\langle b\rangle$ such that $\langle a\rangle\subseteq\langle b\rangle\subseteq R$. Then $a=cb$ for some $c\in R$. As $a$ is irreducible, either $b$ or $c$ is a unit. If $b$ is a unit, then $\langle b\rangle=R$ ($\because$ for any $r\in R$, $r=b(b^{-1}r)\in\langle b\rangle$). If $c$ is a unit, then $a$ and $b$ are associates and so generate the same ideal, i.e. $\langle a\rangle=\langle b\rangle$. This shows that the ideal $\langle a\rangle$ is maximal.

Now suppose the ideal $\langle a\rangle$ is maximal. Since $R$ has more than three elements, $a\ne0$; also, as $\langle a\rangle\ne R$ by the definition of a maximal ideal, $a$ is not a unit. Let $a=bc$ for some $b,c\in R$. Then $\langle a\rangle\subseteq\langle c\rangle$; as $\langle a\rangle$ is maximal, either $\langle a\rangle=\langle c\rangle$ or $\langle c\rangle=R$. If $\langle a\rangle=\langle c\rangle$ then $c\in\langle a\rangle$ $\implies$ $c=da$ for some $d\in R$ $\implies$ $a=bc=bda$ $\implies$ $bd=1$ since $a\ne0$ and $R$ is an integral domain. Thus $b$ is a unit. If $\langle c\rangle=R$, then $1\in\langle c\rangle$ and so $c$ is a unit. This shows that $a$ is irreducible.

In fact, for any integral domain $R$ that is a PID and has more than three elements, the following are equivalent: for any $a\in R$,

• $a$ is irreducible,
• $\langle a\rangle$ is a maximal ideal,
• the quotient ring $R/\langle a\rangle$ is a field.

Hi Olinguito ... thanks again for your posts and help ...

Now ... just a clarification ...

In the post above, you write the following ...


" ... ... If $\langle a\rangle=\langle c\rangle$ then $c\in\langle a\rangle$ $\implies$ $c=da$ for some $d\in R$ $\implies$ $a=bc=bda$ $\implies$ $bd=1$ since $a\ne0$ and $R$ is an integral domain. ... ... "

My question is as follows:

How/why does \(\displaystyle bd = 1\) depend on R being an integral domain ... ?

It seems to me that \(\displaystyle bd = 1 \) would follow from \(\displaystyle a \neq 0\) in a ring with identity that was not an integral domain ... ... but I have to say ... I suspect I may be missing something ... can you help ...

Peter

 

[Olinguito]

$a=1\cdot a=bda$ $\implies$ $0=bda-1\cdot a=(bd-1)\cdot a$ and since $R$ is an integral domain and $a\ne0$, $bd-1=0$, i.e. $bd=1$.

In general, the law of cancellation holds in any integral domain, provided the element being canceled is not zero. That is to say, if $x,y,z$ are elements of an integral domain and $xy=xz$, then either $x=0$ or $y=z$.

 

[Math Amateur]

$a=1\cdot a=bda$ $\implies$ $0=bda-1\cdot a=(bd-1)\cdot a$ and since $R$ is an integral domain and $a\ne0$, $bd-1=0$, i.e. $bd=1$.

In general, the law of cancellation holds in any integral domain, provided the element being canceled is not zero. That is to say, if $x,y,z$ are elements of an integral domain and $xy=xz$, then either $x=0$ or $y=z$.

Thanks for the help Olinguito ...

Your assistance is very much appreciated ...

Peter

 

[steenis]

Thanks Opalg ...

Hmm ... given what you have pointed out, I'm beginning to wonder about the other parts of Proposition 4.3.5 ... ...

Peter

Parts (1), (2), (5), and (6) of Proposition 4.3.5 are correct if R is an (integral) domain.
Part (3) needs the extra condition that R is a PID, thanks to Opalg
Part (4) is a consequence of part (3), so in this part, R must be a PID too.

Rotman - Advanced Modern Algebra (2003), page 322-323 confirms the statement of Opalg.

 

[Opalg]

Part (4) is a consequence of part (3), so in this part, R must be a PID too.

In fact, (4) holds in any integral domain, because it relies on the correct part of (3).

To be more precise, (3) states "$a\in R$ is irreducible if and only if $(a)$ is a maximal ideal of $R$". That is an "if and only if" statement, in which the "if" part is correct, but the "only if" part is not. But (4) follows from the "if" part and is therefore true.

 

[Math Amateur]

In fact, (4) holds in any integral domain, because it relies on the correct part of (3).

To be more precise, (3) states "$a\in R$ is irreducible if and only if $(a)$ is a maximal ideal of $R$". That is an "if and only if" statement, in which the "if" part is correct, but the "only if" part is not. But (4) follows from the "if" part and is therefore true.

Thanks to Steenis and Opalg for clarifying Bland Proposition 4.3.5 ...

Hmm... ... seems that Bland made a bit of a mess of that Proposition ...

Peter

 

[steenis]

In fact, (4) holds in any integral domain, because it relies on the correct part of (3).

To be more precise, (3) states "$a\in R$ is irreducible if and only if $(a)$ is a maximal ideal of $R$". That is an "if and only if" statement, in which the "if" part is correct, but the "only if" part is not. But (4) follows from the "if" part and is therefore true.

You are right. I have a proof that uses PID, but it can be done without PID, independent of (3). I do not see a proof that (4) follows from the "if" part.

This is not my favourite part of algebra, never liked it.

Also, I never can remember what is the "if" part and what is the "only if " part of "if and only if".
I think:
"if and only if" = $\Longleftrightarrow$
"if" = $\Longleftarrow$
"only if "= $\Longrightarrow$
Right ?

 

[Opalg]

You are right. I have a proof that uses PID, but it can be done without PID, independent of (3). I do not see a proof that (4) follows from the "if" part.

This is not my favourite part of algebra, never liked it.

Also, I never can remember what is the "if" part and what is the "only if " part of "if and only if".
I think:
"if and only if" = $\Longleftrightarrow$
"if" = $\Longleftarrow$
"only if "= $\Longrightarrow$
Right ?

I have the same trouble with "if" and "only if", and I'm not 100% sure that I had them the right way round in my previous comment. (Thinking)

I agree that the best way to prove that prime implies irreducible in an integral domain is to argue directly from the definitions (not using (3) at all). Suppose that $a\in R$ is prime, and that $a = bc$. We must show that either $b$ or $c$ is a unit. Since $a$ is prime and divides $bc$, $a$ must divide $b$ or $c$. Suppose that $a$ divides $b$, say $b=ad$. Then $a = bc = adc$, and so $a(1-dc) = 0$. But $R$ has no zero-divisors, and since $a\ne0$ it follows that $dc=1$. Therefore $c$ is a unit.

 

[steenis]

Thank you Opalg.

My proof, almost the same:
Suppose p is a prime element, 0≠p=ab. Then a|p and b|p. But also p|ab implying p|a or p|b. So a or b is an associate of p. Suppose a and p are associates, then p=au for some invertible u in R. R is an integral domain, so b=u, b is a unit and p is irreducible.