Understanding zero divisors & ##\mathbb{Z_m}## in Ring Theory

[chwala]

Actually, no, these are cyclic subgroups of the group of all rotations. This group of all rotations is uncountable and uncountable groups cannot be cyclic.

Let me check on this...

 

[WWGD]

Let me check on this...

And rotation group is a nice example of infinite groups having finite subgroups and having " torsion": Nonzero elements A with $A^n =0$, which seems counterintuitive. Notice the ring of countable integers being ( very) finitely generated by $$\{1,-1\}$$

 

[SammyS]

⋅⋅Ok, i see your point we cannot pick any item in $\left [a_0\right]$, in that case we ought to have;

$[6]⋅[3]= 0 mod 9$
$[3]⋅[6]= 0 mod 9$
$[6]⋅[6]= 0 mod 9$

I can also see that,
$[2]⋅[5]= 1 mod 9$ implying that $[2]$ is an inverse of $[5]$, same to $7$ and $4$...

Your idea is correct, but your notation is incorrect.

You can state that in $\mathbb{Z_9}$ , the Ring of Integers Modulo 9: $~ [6]⋅[3]=[18]=[0]$ . This is written in terms of the residue classes mod 9., which are the actual elements of the ring.

Added in Edit:​

I should have also explicitly enumerated the elements (members) of this ring.
They are: [0], [1], [2], [3], [4], [5], [6], [7], [8] .
Thus $\displaystyle \mathbb{Z_9}=\left\{[0],\,[1],\,[2],\,[3],\,[4],\,[5],\,[6],\,[7],\,[8]\right\}$ .

end of Edit.​

Or as a congruence, you can state that $~ 6⋅3\equiv 0 \mod 9$ .

Or with modular arithmetic: $~ 6⋅3=0$ modulo $9$ .

Using the mod operation (a function) you could write: $~~ (6⋅3) \!\!\!\mod9=0$.

For LaTeX: when writing mod, precede it with a back-slash as in \mod .

##2⋅5=1\mod9## gives $2⋅5=1\mod9$

##2⋅5=1mod9## gives $2⋅5=1mod9$

 

[chwala]

And rotation group is a nice example of infinite groups having finite subgroups and having " torsion": Nonzero elements A with $A^n =0$, which seems counterintuitive. Notice the ring of countable integers being ( very) finitely generated by $$\{1,-1\}$$

I am familiarising with literature to respond amicably to the question, i would like clarification on the highlighted and i also hope that i am looking at the 'correct' literature;

Link
http://campbell.mcs.st-and.ac.uk/~john/MT4521/Lectures/L3.html

Now on the complex number $e^{iθ}$ this to my understanding is another representation of $cos θ + i sin θ$. They basically used the conversion from the cartesian coordinates to polar form. We know that $e^{iθ}=cos θ + i sin θ$. (Euler's equation)
Now on the second part where they indicate that the rotation in a unit circle has determinant of $1$...this is dependant (in my thinking) on the direction of rotation;
if the direction is say $90^0$ anticlockwise then our matrix of rotation would be;

$\begin{pmatrix} 1& 0\\ 0 &1\end {pmatrix}$

if the rotation is say $270^0$ anticlockwise then our rotation matrix would be;
$\begin{pmatrix} 1& 0\\ 0 &-1\end {pmatrix}$
giving us determinant values of $1$ and $-1$ respectively. I hope i am getting this right.

 

[mathwonk]

To me, typical examples of zero divisors occur in product rings, like ZxZ, where (1,0) x (0,1) = (0,0).

Note also that although in Z/6Z, we have 2.3 = 6 = 0, in fact Z/6Z ≈ Z/2Z x Z/3Z, via the map sending 1 to (1,1), and that 2 and 3 in Z/6Z correspond under this isomorphism to (0,2) and (1,0).

 

[chwala]

To me, typical examples of zero divisors occur in product rings, like ZxZ, where (1,0) x (0,1) = (0,0).

Note also that although in Z/6Z, we have 2.3 = 6 = 0, in fact Z/6Z ≈ Z/2Z x Z/3Z, via the map sending 1 to (1,1), and that 2 and 3 in Z/6Z correspond under this isomorphism to (0,2) and (1,0).

Nice to read this...i think i will try and go through the whole link then get more insight so as to deal with post $27$.

 

[mathwonk]

The insight for me here comes from geometry. If a space S is composed of two pieces, X and Y, then in the ring of functions on S, two functions f and g, such that f vanishes on X, and g vanishes on Y, will have a product that vanishes on all of S. The connection with the previous remark is that this ring of functions on S, is a subring of the product of the ring of functions on X and the ring of functions on Y. Namely a function on S is identified with a pair of functions (f,g), where f is a function on X, g is a function on Y, and f and g agree on the intersection of X and Y. So in algebraic geometry, an algebraic set which is "irreducible", i.e. consists of only one piece, like the x axis, has a ring of functions with no zero divisors, namely just polynomials in X. But a space composed of more than one piece, like the union of the x and y axes, has a ring of functions which does have zero divisors, namely the polynomials in both X and Y, but where the product XY, and any monomial containing XY, is set equal to zero.

 

[mathwonk]

I guess it is even simpler when viewed algebraically, as quotient rings. I.e. in a quotient ring, something is set equal to zero, and it just depends on whether or not the thing you set equal to zero is or is not a prime. The ring of functions on the x-axis in the plane, can be viewed as the polynomial ring in X and Y where Y is set equal to zero, and Y is prime. The ring of functions on the union of x and y axes, is the polynomial ring in X and Y, but with the non prime product XY set equal to zero. The ring Z/6Z is the ring Z but with the non prime product 6 = 2.3 set equal to zero, while Z/7Z is Z with the prime 7 set equal to zero. When a product AB is set equal to zero, of course the elements A and B become zero divisors in the quotient ring.

 

[SammyS]

If i understand your question we are looking for $xy≡0 mod 9$ with the condition that $x≠0 mod 9$and $y≠0 mod 9$ .

Notice that if you use a back-slash ( \ ) with 'mod', the above sentence is rendered as follows:

If I understand your question we are looking for $xy=0\mod 9$ with the condition that $x≠0\mod 9$ and $y\ne0 \mod 9$ .

Now onto my primary reasons for making this reply ...

Now we have,
$[a_0]$=$\{\ldots, -9,0,9,18,27 \ldots\}$
$[a_1]$=$\{\ldots, -8,1,10,19,28 \ldots\}$
if we pick $[a_0]$=$9$ and $[a_1]$=$1$, then $[a_0]⋅[a_1]=9⋅1=0 mod 9$

just a minute i check $1$ cannot be a zero divisor in any ring!

Second attempt.

$[a_0]$=$\{\ldots, -9,0,9,18,27 \ldots\}$
$[a_2]$=$\{\ldots, -7,2,11,20,29 \ldots\}$
if we pick $[a_0]$=$9$ and $[a_2]$=$2$, then $[a_0]⋅[a_2]=9⋅2=0 mod 9$ therefore $9$ and $2$ are zero divisors.

It is improper to state things such as $[a_0]$=$9$ , $[a_2]$=$2$ , etc.
Presumably, by what you have stated early in the quoted post (Post #17), $a_0$ is some integer multiple of $9$ .
Similarly $a_2$ is one of the integers in the set $\{\dots\, -7,\,2,\,11,\,20,\, \dots \}$ . In fact you can say that $a_2 \in[2]$ meaning that
$a_2=9k+2$ for some integer, $k$ .

You can pick $a_0$ to be $9$ . That makes $[a_0]=[9]$ . This refers to a set, not an individual integer.

That leads me to mention another misconception you seem to have. There is nothing special about $a_0,\,a_1,\,a_2,\,a_3,\$ etc. That is not any sort of general or widely accepted notation. The author of one of the attachments in Post #2, was simply picking a member from one of the residue classes, and using an index (in a subscript) to indicate from which class the pick was made. The point being made was that any particular class can be labeled using any of its members.

One last caution:
Be careful when referring to the $1$ or the $0$ of a ring. Those are, generically speaking, the multiplicative identity and the additive identity, respectively.

For the ring $\mathbb{Z}_9$ , the "1" is $[1]$, This is not the integer $1$.

Similarly, the zero of this ring is $[0]$. This is not the integer $0$.

 

[chwala]

Notice that if you use a back-slash ( \ ) with 'mod', the above sentence is rendered as follows:

If I understand your question we are looking for $xy=0\mod 9$ with the condition that $x≠0\mod 9$ and $y\ne0 \mod 9$ .

Now onto my primary reasons for making this reply ...

It is improper to state things such as $[a_0]$=$9$ , $[a_2]$=$2$ , etc.
Presumably, by what you have stated early in the quoted post (Post #17), $a_0$ is some integer multiple of $9$ .
Similarly $a_2$ is one of the integers in the set $\{\dots\, -7,\,2,\,11,\,20,\, \dots \}$ . In fact you can say that $a_2 \in[2]$ meaning that
$a_2=9k+2$ for some integer, $k$ .

You can pick $a_0$ to be $9$ . That makes $[a_0]=[9]$ . This refers to a set, not an individual integer.

That leads me to mention another misconception you seem to have. There is nothing special about $a_0,\,a_1,\,a_2,\,a_3,\$ etc. That is not any sort of general or widely accepted notation. The author of one of the attachments in Post #2, was simply picking a member from one of the residue classes, and using an index (in a subscript) to indicate from which class the pick was made. The point being made was that any particular class can be labeled using any of its members.

One last caution:
Be careful when referring to the $1$ or the $0$ of a ring. Those are, generically speaking, the multiplicative identity and the additive identity, respectively.

For the ring $\mathbb{Z}_9$ , the "1" is $[1]$, This is not the integer $1$.

Similarly, the zero of this ring is $[0]$. This is not the integer $0$.

Thanks @SammyS noted...

 

[chwala]

Notice that if you use a back-slash ( \ ) with 'mod', the above sentence is rendered as follows:

If I understand your question we are looking for $xy=0\mod 9$ with the condition that $x≠0\mod 9$ and $y\ne0 \mod 9$ .

Now onto my primary reasons for making this reply ...

It is improper to state things such as $[a_0]$=$9$ , $[a_2]$=$2$ , etc.
Presumably, by what you have stated early in the quoted post (Post #17), $a_0$ is some integer multiple of $9$ .
Similarly $a_2$ is one of the integers in the set $\{\dots\, -7,\,2,\,11,\,20,\, \dots \}$ . In fact you can say that $a_2 \in[2]$ meaning that
$a_2=9k+2$ for some integer, $k$ .

You can pick $a_0$ to be $9$ . That makes $[a_0]=[9]$ . This refers to a set, not an individual integer.

That leads me to mention another misconception you seem to have. There is nothing special about $a_0,\,a_1,\,a_2,\,a_3,\$ etc. That is not any sort of general or widely accepted notation. The author of one of the attachments in Post #2, was simply picking a member from one of the residue classes, and using an index (in a subscript) to indicate from which class the pick was made. The point being made was that any particular class can be labeled using any of its members.

One last caution:
Be careful when referring to the $1$ or the $0$ of a ring. Those are, generically speaking, the multiplicative identity and the additive identity, respectively.

For the ring $\mathbb{Z}_9$ , the "1" is $[1]$, This is not the integer $1$.

Similarly, the zero of this ring is $[0]$. This is not the integer $0$.

This part is getting clear to me, in general,
if $a=b\mod n$ then it follows that $a=b+kn$ for some $k∈ \mathbb{Z}$
For example, we know that
$1≡5\mod 2$ then using $1=5+2k$ then the other equivalent values would be given as follows;
$5\mod 2≡7$ when $k=1$.
$5\mod 2≡9$ when $k=2$.
$5\mod 2≡11$ when $k=3$...

 

[SammyS]

This part is getting clear to me, in general,
if $a=b\mod n$ then it follows that $a=b+kn$ for some $k∈ \mathbb{Z}$
For example, we know that
$1≡5\mod 2$ then using $1=5+2k$ then the other equivalent values would be given as follows;

Technically , for the above $k=-2$, although that's fine.
Also, usually the mod 2 is enclosed in parentheses as in: $\quad 1≡5\ \ (\mod 2)$

$5\mod 2≡7$ when $k=1$.
$5\mod 2≡9$ when $k=2$.
$5\mod 2≡11$ when $k=3$...

As mathematical function, $5\mod2$ gives the remainder for 5 divided by 2, in the sense of Euclidean Division. In other words, $5\mod2=1$ because $5=2\cdot2+1$ , 2 being the quotient, 1 being the remainder. A rather poor example.

Better: $7\mod2=1$ because $7=2\cdot3+1$ , 3 being the quotient, 1 being the remainder.

Now, if you want to write any of these as congruence mod 2 , then the $\mod2$ usually appears in parentheses at the tail end of the statement.

$5\equiv7\ \ (\mod 2 \ )$ means $7=5+2k\text{, where }k=1$ .

A bit cleaner way to consider such a congruence is to see that the following is true.

$5\equiv7\ \ (\mod 2 \ )$ means $7-5=2k\$ for some integer, $k$.

 

[chwala]

Technically , for the above $k=-2$, although that's fine.
Also, usually the mod 2 is enclosed in parentheses as in: $\quad 1≡5\ \ (\mod 2)$As mathematical function, $5\mod2$ gives the remainder for 5 divided by 2, in the sense of Euclidean Division. In other words, $5\mod2=1$ because $5=2\cdot2+1$ , 2 being the quotient, 1 being the remainder. A rather poor example.

Better: $7\mod2=1$ because $7=2\cdot3+1$ , 3 being the quotient, 1 being the remainder.

Now, if you want to write any of these as congruence mod 2 , then the $\mod2$ usually appears in parentheses at the tail end of the statement.

$5\equiv7\ \ (\mod 2 \ )$ means $7=5+2k\text{, where }k=1$ .

A bit cleaner way to consider such a congruence is to see that the following is true.

$5\equiv7\ \ (\mod 2 \ )$ means $7-5=2k\$ for some integer, $k$.

This ought to be pretty obvious to me i.e The Euclidean algorithm...sometimes my thinking is not straight,
anyway,
i note that if;
$7=2\cdot3+1$ then we may either have;
$7\mod2=1$ or
$7\mod3=1$

In general,
$a=b⋅c+d$ then it follows that;
$a\mod c=d$
$a\mod b=d$ dependant on either $b$ or $c$ being the quotient and divisor respectively.

 

[SammyS]

This ought to be pretty obvious to me i.e. The Euclidean algorithm...sometimes my thinking is not straight ,
anyway,
i note that if;
$7=2\cdot3+1$ then we may either have;
$7\mod2=1$ or
$7\mod3=1$

True, but in the present discussion, the context was $\mod 2$ .

Thus $2$ is the divisor, $3$ is the quotient, and the remainder is $1$ .

 

[chwala]

I guess it is even simpler when viewed algebraically, as quotient rings. I.e. in a quotient ring, something is set equal to zero, and it just depends on whether or not the thing you set equal to zero is or is not a prime. The ring of functions on the x-axis in the plane, can be viewed as the polynomial ring in X and Y where Y is set equal to zero, and Y is prime. The ring of functions on the union of x and y axes, is the polynomial ring in X and Y, but with the non prime product XY set equal to zero. The ring Z/6Z is the ring Z but with the non prime product 6 = 2.3 set equal to zero, while Z/7Z is Z with the prime 7 set equal to zero. When a product AB is set equal to zero, of course the elements A and B become zero divisors in the quotient ring.

Hey, just confirm that i am getting you right, for Z/7Z we do not have zero divisors! For Z/6Z yes we do have zero divisors.

I can see that under Z/7Z; we have inverses that are commutative under multiplication i.e

$2⋅4=1$
$3⋅5=1$
$1⋅1=1$
$6⋅6=1$ only

 

[chwala]

Am still looking at the literature, i need confirmation on $6$;

Now we know that;

My understanding- In reference to 6;
Let $r=3$ and $p=2$, then it follows that,
$3^2+3^2=18\mod2=0$
$(3+3)^2=36\mod2=0$

satisfies definition 2.3.1 (1)$3^2⋅ 3^2=81\mod2=1$
$(3⋅3)^2=81\mod2=1$

satisfies definition 2.3.1 (2)using another example say, $r=2$ and $p=3$, then it follows that,
$2^3+2^3=16\mod3=1$
$(2+2)^3=64\mod3=1$

satisfies definition 2.3.1 (1)
$2^3⋅ 2^3=64\mod3=1$
$(2⋅2)^3=64\mod3=1$

satisfies definition 2.3.1 (2)thus the pth power map is a ring homomorphism.

 

[chwala]

Also here; i need insight on the inverse of a matrix...

Are they assuming that in the given symmetric matrix; both $n$ values are units? What if we have non- unit $n$ values? ...then the inverse would not be equal to $n$.

 

[Maarten Havinga]

The p-power is a homomorphism indeed based on your observations. The $1^p=1$ part is trivial but when handing something in do add it.

As to the matrix inverse, there's a slight confusion. They mean the inverse of the homomorphism as a map, not the inverse of the matrix. It's a rather trivial statement.

 

[chwala]

The p-power is a homomorphism indeed based on your observations. The $1^p=1$ part is trivial but when handing something in do add it.

As to the matrix inverse, there's a slight confusion. They mean the inverse of the homomorphism as a map, not the inverse of the matrix. It's a rather trivial statement.

Thanks... we can't have $1^p=1$ because $p$ has to be prime ... unless I am missing something here.

 

[SammyS]

Also here; i need insight on the inverse of a matrix...

View attachment 301452Are they assuming that in the given symmetric matrix; both $n$ values are units? What if we have non- unit $n$ values? ...then the inverse would not be equal to $n$.

What do you mean by "both $n$ values" being "units" ?

As @Maarten Havinga states, they are referring to the inverse map .

Show that the map given as $\psi$ is the inverse (as in function inverse) of the map given as $\phi$ .

 

[chwala]

What do you mean by "both $n$ values" being "units" ?

As @Maarten Havinga states, they are referring to the inverse map .

Show that the map given as $\psi$ is the inverse (as in function inverse) of the map given as $\phi$ .

What do you mean by "both $n$ values" being "units" ? ...$n=1$.

Now understood.