Integral Domains and Characteristics - Char(D)

[Math Amateur]

I am reading Beachy and Blair's book: Abstract Algebra (3rd Edition) and am currently studying Proposition 5.211.

I need help with the proof of the proposition.

Proposition 5.211 and its proof read as follows:View attachment 2837
View attachment 2838

In the proof we read the following:

" ... ... ... Since \(\displaystyle \phi ( \mathbb{Z} ) \) inherits the property that D has no nontrivial divisors of zero, this shows that \(\displaystyle \mathbb{Z} / ker( \phi ) \) must be an integral domain. Thus either \(\displaystyle ker( \phi ) = 0 \), in which case \(\displaystyle \text{char}(D) = 0 \), or \(\displaystyle ker( \phi ) = n \mathbb{Z} \) for some positive number n. ... ... "

Can someone please explain exactly why \(\displaystyle \mathbb{Z} / ker( \phi ) \) being an integral domain implies that either \(\displaystyle ker( \phi ) = 0 \) or \(\displaystyle ker( \phi ) = n \mathbb{Z} \) for some positive number n.

Further, can someone please clarify why \(\displaystyle ker( \phi ) = 0 \) implies that \(\displaystyle \text{char}(D) = 0 \).

Peter

 

[Deveno]

$D$ has no zero-divisors, so $\phi(\Bbb Z)$, which is a SUBSET of $D$, cannot have any zero-divisors.

Since $\Bbb Z/\text{ker }\phi$ is isomorphic to an integral domain, it is an integral domain.

Now $\text{ker }\phi$ is an ideal of the integers, and $\Bbb Z$ is a principal ideal domain.

Hence $\text{ker }\phi = (n) = n\Bbb Z$ for some integer $n$ (remember, ideals are first and foremost additive subgroups, and $\Bbb Z$ under addition is cyclic, and a subgroup of a cyclic group is itself cyclic).

This shows that $\phi(\Bbb Z) \cong \Bbb Z/(n) = \Bbb Z/n\Bbb Z = \Bbb Z_n$.

There are just two cases, here: $n = 0$, or $n \neq 0$. If $n = 0$, the ideal generated by 0 is just the trivial additive subgroup $\{0\}$ of the integers, and:

$\Bbb Z/(0) = \Bbb Z$

( the coset $k + 0\Bbb Z$ has just a single element, $k$).

In this case, we have a copy of the integers "inside" $D$.

Otherwise, we have a copy of $\Bbb Z_n$ inside $D$. I will digress for a moment to consider the "special case" $n = 1$.

In this case, we get $\phi(\Bbb Z) \cong \Bbb Z/\Bbb Z \cong \{0\}$. In particular, we have:

$1 = 1\cdot 1 = \phi(1) = \phi(0) = 0 \cdot 1 = 0$, so that $D$ must be a trivial ring. This is uninteresting.

So we proceed to the more fruitful case where $n \geq 2$.

If $n = km$, for $1 < k,m < n$ (that is, $n$ is composite), it is clear that:

$\phi(k),\phi(m)$ are zero-divisors of $D$, since:

$\phi(k)\phi(m) = \phi(km) = \phi(n) = 0$ (since $n \in \text{ker }\phi$).

Since $D$ is a domain, and HAS no zero-divisors, $n$ cannot be composite, and since it is greater than $1$, must be prime.

****************

By definition char($D$) is the additive order of $1$ in $D$, if this is finite. If $1$ is of infinite order, we say $D$ is of characteristic 0. Why the confusing use of 0 here rather than $\infty$ I cannot say, perhaps someone else knows.

 

[Euge]

I am reading Beachy and Blair's book: Abstract Algebra (3rd Edition) and am currently studying Proposition 5.211.

I need help with the proof of the proposition.

Proposition 5.211 and its proof read as follows:View attachment 2837
View attachment 2838

In the proof we read the following:

" ... ... ... Since \(\displaystyle \phi ( \mathbb{Z} ) \) inherits the property that D has no nontrivial divisors of zero, this shows that \(\displaystyle \mathbb{Z} / ker( \phi ) \) must be an integral domain. Thus either \(\displaystyle ker( \phi ) = 0 \), in which case \(\displaystyle \text{char}(D) = 0 \), or \(\displaystyle ker( \phi ) = n \mathbb{Z} \) for some positive number n. ... ... "

Can someone please explain exactly why \(\displaystyle \mathbb{Z} / ker( \phi ) \) being an integral domain implies that either \(\displaystyle ker( \phi ) = 0 \) or \(\displaystyle ker( \phi ) = n \mathbb{Z} \) for some positive number n.

Further, can someone please clarify why \(\displaystyle ker( \phi ) = 0 \) implies that \(\displaystyle \text{char}(D) = 0 \).

Peter

Having $\ker(\phi) = 0$ or $\ker(\phi) = n\Bbb Z$ for some positive integer $n$ is not a consequence of the fact that $Z/\ker(\phi)$ is an integral domain, but rather of the fact that the subgroups of $\Bbb Z$ are of the form $nZ$, where $n$ is a nonnegative integer. The part of the proof where the structure of an integral domain comes to play is after mentioning $Z/\text{ker}(\phi) \approx \Bbb Z_n$. This is because $\Bbb Z/\text{ker}(\phi)$ is an integral domain if and only if $n$ is prime. Then $n$ will be the least positive integer such that $n \cdot 1 = 0$ and consequently $\text{char}(D) = n$, a prime.

If $\text{ker}(\phi) = 0$, then $n \cdot 1\neq 0$ for any positive integer $n$. In other words, there is no smallest positive integer $n$ such that $n \cdot 1 = 0$. So, $\text{char}(D) = 0$ by definition.

 

[Euge]

By definition char($D$) is the additive order of $1$ in $D$, if this is finite. If $1$ is of infinite order, we say $D$ is of characteristic 0. Why the confusing use of 0 here rather than $\infty$ I cannot say, perhaps someone else knows.

From the perspective of viewing char($D$) as the order of 1, I can see why you might define char($D$) to be $\infty$ if 1 has infinite order. But allowing char($D$) to be 0 instead of $\infty$ makes it consistent with the order of ker($\phi$): char($D$) = $n$ if and only if $|\text{ker}(\phi)| = n$.

 

[Deveno]

From the perspective of viewing char($D$) as the order of 1, I can see why you might define char($D$) to be $\infty$ if 1 has infinite order. But allowing char($D$) to be 0 instead of $\infty$ makes it consistent with the order of ker($\phi$): char($D$) = $n$ if and only if $|\text{ker}(\phi)| = n$.

While that has a certain appeal to it, it almost certainly isn't true, for example, if $D = \Bbb Z_p$ and $\phi$ is the canonical projection $\phi: \Bbb Z \to \Bbb Z_p$:

$\phi(k) = [k]_p$

It is clear that the order of $\text{ker }\phi$ is infinite, consisting all all integers $\{n \in \Bbb Z: n = tp, t\in \Bbb Z\} = p\Bbb Z$.

 

[Euge]

While that has a certain appeal to it, it almost certainly isn't true, for example, if $D = \Bbb Z_p$ and $\phi$ is the canonical projection $\phi: \Bbb Z \to \Bbb Z_p$:

$\phi(k) = [k]_p$

It is clear that the order of $\text{ker }\phi$ is infinite, consisting all all integers $\{n \in \Bbb Z: n = tp, t\in \Bbb Z\} = p\Bbb Z$.

Sorry, I had a typo. In the statement "char($D$) = $n$ if and only if |ker($\phi$)| = $n$", the letter $n$ is supposed to be 0. Although I can't say for sure, I don't think char($D$) is always defined to be 0 if it's not positive.

 

[Math Amateur]

$D$ has no zero-divisors, so $\phi(\Bbb Z)$, which is a SUBSET of $D$, cannot have any zero-divisors.

Since $\Bbb Z/\text{ker }\phi$ is isomorphic to an integral domain, it is an integral domain.

Now $\text{ker }\phi$ is an ideal of the integers, and $\Bbb Z$ is a principal ideal domain.

Hence $\text{ker }\phi = (n) = n\Bbb Z$ for some integer $n$ (remember, ideals are first and foremost additive subgroups, and $\Bbb Z$ under addition is cyclic, and a subgroup of a cyclic group is itself cyclic).

This shows that $\phi(\Bbb Z) \cong \Bbb Z/(n) = \Bbb Z/n\Bbb Z = \Bbb Z_n$.

There are just two cases, here: $n = 0$, or $n \neq 0$. If $n = 0$, the ideal generated by 0 is just the trivial additive subgroup $\{0\}$ of the integers, and:

$\Bbb Z/(0) = \Bbb Z$

( the coset $k + 0\Bbb Z$ has just a single element, $k$).

In this case, we have a copy of the integers "inside" $D$.

Otherwise, we have a copy of $\Bbb Z_n$ inside $D$. I will digress for a moment to consider the "special case" $n = 1$.

In this case, we get $\phi(\Bbb Z) \cong \Bbb Z/\Bbb Z \cong \{0\}$. In particular, we have:

$1 = 1\cdot 1 = \phi(1) = \phi(0) = 0 \cdot 1 = 0$, so that $D$ must be a trivial ring. This is uninteresting.

So we proceed to the more fruitful case where $n \geq 2$.

If $n = km$, for $1 < k,m < n$ (that is, $n$ is composite), it is clear that:

$\phi(k),\phi(m)$ are zero-divisors of $D$, since:

$\phi(k)\phi(m) = \phi(km) = \phi(n) = 0$ (since $n \in \text{ker }\phi$).

Since $D$ is a domain, and HAS no zero-divisors, $n$ cannot be composite, and since it is greater than $1$, must be prime.

****************

By definition char($D$) is the additive order of $1$ in $D$, if this is finite. If $1$ is of infinite order, we say $D$ is of characteristic 0. Why the confusing use of 0 here rather than $\infty$ I cannot say, perhaps someone else knows.

Thanks for the considerable help Deveno ... Appreciate the help ...

just working through your post in detail now

Peter

- - - Updated - - -

Sorry, I had a typo. In the statement "char($D$) = $n$ if and only if |ker($\phi$)| = $n$", the letter $n$ is supposed to be 0. Although I can't say for sure, I don't think char($D$) is always defined to be 0 if it's not positive.

Thanks Euge ... appreciate your contribution to this issue ...

Peter

 

[Math Amateur]

Having $\ker(\phi) = 0$ or $\ker(\phi) = n\Bbb Z$ for some positive integer $n$ is not a consequence of the fact that $Z/\ker(\phi)$ is an integral domain, but rather of the fact that the subgroups of $\Bbb Z$ are of the form $nZ$, where $n$ is a nonnegative integer. The part of the proof where the structure of an integral domain comes to play is after mentioning $Z/\text{ker}(\phi) \approx \Bbb Z_n$. This is because $\Bbb Z/\text{ker}(\phi)$ is an integral domain if and only if $n$ is prime. Then $n$ will be the least positive integer such that $n \cdot 1 = 0$ and consequently $\text{char}(D) = n$, a prime.

If $\text{ker}(\phi) = 0$, then $n \cdot 1\neq 0$ for any positive integer $n$. In other words, there is no smallest positive integer $n$ such that $n \cdot 1 = 0$. So, $\text{char}(D) = 0$ by definition.

Thanks for the help Euge ... Just working through this now ... appreciate your support in helping me understand this proof fully ...

Peter

 

[Math Amateur]

$D$ has no zero-divisors, so $\phi(\Bbb Z)$, which is a SUBSET of $D$, cannot have any zero-divisors.

Since $\Bbb Z/\text{ker }\phi$ is isomorphic to an integral domain, it is an integral domain.

Now $\text{ker }\phi$ is an ideal of the integers, and $\Bbb Z$ is a principal ideal domain.

Hence $\text{ker }\phi = (n) = n\Bbb Z$ for some integer $n$ (remember, ideals are first and foremost additive subgroups, and $\Bbb Z$ under addition is cyclic, and a subgroup of a cyclic group is itself cyclic).

This shows that $\phi(\Bbb Z) \cong \Bbb Z/(n) = \Bbb Z/n\Bbb Z = \Bbb Z_n$.

There are just two cases, here: $n = 0$, or $n \neq 0$. If $n = 0$, the ideal generated by 0 is just the trivial additive subgroup $\{0\}$ of the integers, and:

$\Bbb Z/(0) = \Bbb Z$

( the coset $k + 0\Bbb Z$ has just a single element, $k$).

In this case, we have a copy of the integers "inside" $D$.

Otherwise, we have a copy of $\Bbb Z_n$ inside $D$. I will digress for a moment to consider the "special case" $n = 1$.

In this case, we get $\phi(\Bbb Z) \cong \Bbb Z/\Bbb Z \cong \{0\}$. In particular, we have:

$1 = 1\cdot 1 = \phi(1) = \phi(0) = 0 \cdot 1 = 0$, so that $D$ must be a trivial ring. This is uninteresting.

So we proceed to the more fruitful case where $n \geq 2$.

If $n = km$, for $1 < k,m < n$ (that is, $n$ is composite), it is clear that:

$\phi(k),\phi(m)$ are zero-divisors of $D$, since:

$\phi(k)\phi(m) = \phi(km) = \phi(n) = 0$ (since $n \in \text{ker }\phi$).

Since $D$ is a domain, and HAS no zero-divisors, $n$ cannot be composite, and since it is greater than $1$, must be prime.

****************

By definition char($D$) is the additive order of $1$ in $D$, if this is finite. If $1$ is of infinite order, we say $D$ is of characteristic 0. Why the confusing use of 0 here rather than $\infty$ I cannot say, perhaps someone else knows.

Hi Deveno ... thanks again for the help ...

Just checking something that has been bothering me for some time ...

You write:

" ... ... This shows that $\phi(\Bbb Z) \cong \Bbb Z/(n) = \Bbb Z/n\Bbb Z = \Bbb Z_n$. ... ... ... "

What is the (exact) difference between \(\displaystyle \Bbb Z/n\Bbb Z\) and \(\displaystyle \Bbb Z_n\)?

Are \(\displaystyle \Bbb Z/n\Bbb Z\) and \(\displaystyle \Bbb Z_n\) identical i.e is this just different notations for the same thing .. ... or are they isomorphic?

(I am assuming it makes sense to ask such a question ... )

Peter

 

[Deveno]

Yes (to both questions- the identity mapping is an isomorphism).

As groups, it is usual to see the notation $\Bbb Z/n\Bbb Z$, as these rings provide an ready source of finite cyclic groups (their additive group).

As rings, $\Bbb Z/(n)$ is not uncommon, as this is typical notation of a quotient ring of a principal ideal domain (compare $F[x]/(f(x))$).

I prefer the notation $\Bbb Z_n$, for its compactness. However, $\Bbb Z_p$ is also used to denote the $p$-adic integers, so careful attention to context is required.

Generally, in algebra "one doesn't care", the structure of two isomorphic groups/rings/etc. is the same, only the names have been changed. Therefore, while:

$\{\overline{0},\overline{1},\overline{2},\overline{3}\}$

$\{e,(1\ 2\ 3\ 4), (1\ 3)(2\ 4), (1\ 4\ 3\ 2)\}$

$\{1,i,-1,-i\}$

$\left\{\begin{bmatrix}1&0\\0&1 \end{bmatrix}, \begin{bmatrix}0&-1\\1&0 \end{bmatrix}, \begin{bmatrix}-1&0\\0&-1 \end{bmatrix}, \begin{bmatrix}0&1\\-1&0 \end{bmatrix}\right\}$

$\{e,a,a^2,a^3\}$

$\{[\ \ ],X,XX,X^{\ast}\}$

are all different SETS (indeed, different kinds of "objects"), they all, in some sense, represent the same "group", a cyclic group of order 4.

"Group" is a "structure-template", one could make the argument that there are no such things as "groups" at all, only "other things" that have "the group properties". One need not, in point of fact, know any examples of groups at all, in order to study group theory. It turns out that enough "real things" have these properties (frieze patterns, plane tilings, molecular crystals, forces applied to an object, etc.) to make the study of groups not entirely divorced from reality.

A similar observation holds for any category, and the SAME observation holds for the notion of "category" itself. Algebraists (in particular) are the "behaviorists" of mathematics, what things ontologically are, is a matter for the logicians.

It is, moreover, common in mathematics to blur the line between "isomorphism" and "equality". The field of fractions is a perfect case in point: The original integral domain $D$ is not the same thing as the embedding of $D$ in $D\times D^{\ast}/\sim$, clearly the integer $1$ isn't the same thing as $4/4$ (a whole pie is not the same thing as a pie cut into 4 quarter-pieces), but we treat it "as if it were". We do the same thing with monetary exchanges: a dollar bill isn't "the exact same thing" as 4 25-cent pieces. That is "isomorphism" at work.

Computer programming does a similar thing: values are assigned to a variable TYPE. Comparing variables of different types requires conversion, the conversion routine itself is like a homomorphism. If the conversion is reversible, it represents an isomorphism (as you can see, the reversibility of such a routine is highly dependent on the allowable data input/output ranges). For example if a REAL variable is turned into an INTEGER variable (usually by truncation), some information is lost, and the mapping is not bijective.

I myself prefer to think of $\Bbb Z_n$ as "the finite group (ring) of integers mod $n$" and $\Bbb Z/n\Bbb Z$ as: "the group (ring) of integers" (mod $n$), that is: the latter is the quotient group, and the former is the image of the map that sends:

$k \mapsto \overline{k}$.

that is: I prefer to see "quotient objects" as "homomorphic images", because the quotient construction (via cosets) seems to me to be unnecessarily notationally difficult, writing complex numbers as:

$a + bx + \langle x^2 + 1\rangle$ seems burdensome, and I prefer to write them as $a + bi$, where $i$ is a root of $x^2 + 1$.

Since these two ways are "parallel", I am in no danger of making an egregious error in doing so.

 

[Math Amateur]

Yes (to both questions- the identity mapping is an isomorphism).

As groups, it is usual to see the notation $\Bbb Z/n\Bbb Z$, as these rings provide an ready source of finite cyclic groups (their additive group).

As rings, $\Bbb Z/(n)$ is not uncommon, as this is typical notation of a quotient ring of a principal ideal domain (compare $F[x]/(f(x))$).

I prefer the notation $\Bbb Z_n$, for its compactness. However, $\Bbb Z_p$ is also used to denote the $p$-adic integers, so careful attention to context is required.

Generally, in algebra "one doesn't care", the structure of two isomorphic groups/rings/etc. is the same, only the names have been changed. Therefore, while:

$\{\overline{0},\overline{1},\overline{2},\overline{3}\}$

$\{e,(1\ 2\ 3\ 4), (1\ 3)(2\ 4), (1\ 4\ 3\ 2)\}$

$\{1,i,-1,-i\}$

$\left\{\begin{bmatrix}1&0\\0&1 \end{bmatrix}, \begin{bmatrix}0&-1\\1&0 \end{bmatrix}, \begin{bmatrix}-1&0\\0&-1 \end{bmatrix}, \begin{bmatrix}0&1\\-1&0 \end{bmatrix}\right\}$

$\{e,a,a^2,a^3\}$

$\{[\ \ ],X,XX,X^{\ast}\}$

are all different SETS (indeed, different kinds of "objects"), they all, in some sense, represent the same "group", a cyclic group of order 4.

"Group" is a "structure-template", one could make the argument that there are no such things as "groups" at all, only "other things" that have "the group properties". One need not, in point of fact, know any examples of groups at all, in order to study group theory. It turns out that enough "real things" have these properties (frieze patterns, plane tilings, molecular crystals, forces applied to an object, etc.) to make the study of groups not entirely divorced from reality.

A similar observation holds for any category, and the SAME observation holds for the notion of "category" itself. Algebraists (in particular) are the "behaviorists" of mathematics, what things ontologically are, is a matter for the logicians.

It is, moreover, common in mathematics to blur the line between "isomorphism" and "equality". The field of fractions is a perfect case in point: The original integral domain $D$ is not the same thing as the embedding of $D$ in $D\times D^{\ast}/\sim$, clearly the integer $1$ isn't the same thing as $4/4$ (a whole pie is not the same thing as a pie cut into 4 quarter-pieces), but we treat it "as if it were". We do the same thing with monetary exchanges: a dollar bill isn't "the exact same thing" as 4 25-cent pieces. That is "isomorphism" at work.

Computer programming does a similar thing: values are assigned to a variable TYPE. Comparing variables of different types requires conversion, the conversion routine itself is like a homomorphism. If the conversion is reversible, it represents an isomorphism (as you can see, the reversibility of such a routine is highly dependent on the allowable data input/output ranges). For example if a REAL variable is turned into an INTEGER variable (usually by truncation), some information is lost, and the mapping is not bijective.

I myself prefer to think of $\Bbb Z_n$ as "the finite group (ring) of integers mod $n$" and $\Bbb Z/n\Bbb Z$ as: "the group (ring) of integers" (mod $n$), that is: the latter is the quotient group, and the former is the image of the map that sends:

$k \mapsto \overline{k}$.

that is: I prefer to see "quotient objects" as "homomorphic images", because the quotient construction (via cosets) seems to me to be unnecessarily notationally difficult, writing complex numbers as:

$a + bx + \langle x^2 + 1\rangle$ seems burdensome, and I prefer to write them as $a + bi$, where $i$ is a root of $x^2 + 1$.

Since these two ways are "parallel", I am in no danger of making an egregious error in doing so.

Thanks Deveno ... Very clear and most helpful!

Peter