Group isomorphism and Polynomial ring modulo ideal

In summary, when trying to show isomorphism between groups, it is not enough to show that the order of each element within the group is the same. You also have to show that the binary operator is the same.
  • #1
X-il3
3
0
Hi everyone.

I have two questions that I hope you can help me with.

First when trying to show isomorphism between groups is it enough to show that the order of each element within the group is the same in the other group? For example the groups (Z/14Z)* and (Z/9Z)*. They are both of order 6 and both have
1 element of order 1
1 element of order 2
2 elements of order 3
2 elements of order 6
And does the binary operator have to be the same in both groups when doing group isomorphism?

And the second question(actually the third one:smile:). I am trying to write the multiplication table for the following ring
Z_2[X]/(x^2 + x +). I know that this ring has 4 elements. Is it correct that the elements are the following
1
x + 1
x^2 + 1
x^2 + x + 1
?

Hope you can help me with this,

X-il3
 
Physics news on Phys.org
  • #2
X-il3 said:
First when trying to show isomorphism between groups is it enough to show that the order of each element within the group is the same in the other group?

No.

And does the binary operator have to be the same in both groups when doing group isomorphism?

What do you mean by 'same'? If f is an isomorphism then f(xy)=f(x)f(y), if that's what you mean.


Z_2[X]/(x^2 + x +).

Is there a missing 1 there?

I know that this ring has 4 elements. Is it correct that the elements are the following
1
x + 1
x^2 + 1
x^2 + x + 1

Where is 0? Why haven't you written down x or x^x+x? (You have, by the way, but you have made a strange choice of representatives of the elements, which is why I ask, since it implies you've not really understood what you're doing.
 
  • #3
matt grime said:
Is there a missing 1 there?
Yes there was 1 missing there.


matt grime said:
Where is 0? Why haven't you written down x or x^x+x? (You have, by the way, but you have made a strange choice of representatives of the elements, which is why I ask, since it implies you've not really understood what you're doing.
I thought the identity element in multiplication is 1 and therefore no 0. How would you represent the elements in this ring? More like this
0
x
x^2
x^2 + x
? Leaving the 1 out from all the elements?

X-il3
 
  • #4
Now where's 1 gone? (Again, it is still there in disguise which makes more confused as toy what you're trying to do).
 
  • #5
matt grime said:
Now where's 1 gone? (Again, it is still there in disguise which makes more confused as toy what you're trying to do).

Ok here is one more try. Hope I get it right this time :smile:.
Since we are working in Z_2 then a = -a where a is an element in Z_2.

We know that X^2 + x + 1 = 0 = [0]. Moving numbers around now we then get the remaining 3 elements in the ring.
x^2 + x = 1 = [1]
x^2 + 1 = x = [x]
x + 1 = x^2 = [x^2]

X-il3
 

1. What is a group isomorphism?

A group isomorphism is a type of mapping between two groups that preserves their algebraic structure. This means that the operation between elements in the first group will be the same as the operation between their corresponding elements in the second group. In other words, a group isomorphism is a bijective homomorphism between two groups.

2. How is a group isomorphism related to polynomial rings?

A group isomorphism can also be extended to polynomial rings. In this case, the isomorphism maps the elements of one polynomial ring to the elements of another polynomial ring, preserving the algebraic structure. This is useful in studying the properties of polynomial rings and their subrings.

3. What is a polynomial ring modulo ideal?

A polynomial ring modulo ideal is a type of quotient ring, where the elements are polynomials and the ideal is a subset of the polynomial ring that satisfies certain conditions. This ring is formed by taking the remainders of polynomial division by the ideal. It is denoted as R/I, where R is the polynomial ring and I is the ideal.

4. How is group isomorphism related to polynomial ring modulo ideal?

Group isomorphism and polynomial ring modulo ideal are related in the sense that the isomorphism between two groups can be extended to an isomorphism between their corresponding polynomial rings modulo ideal. This means that if two groups are isomorphic, their corresponding polynomial rings modulo ideal will also be isomorphic.

5. What are some applications of group isomorphism and polynomial ring modulo ideal?

Group isomorphism and polynomial ring modulo ideal have various applications in mathematics and other fields. In algebra, they are used to study the properties of groups and polynomial rings. In number theory, they are used to study the structure of number fields. In physics, they are used to study symmetries and transformations. They also have applications in cryptography, coding theory, and computer science.

Similar threads

  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
17
Views
4K
Replies
2
Views
965
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
33
Views
3K
  • Linear and Abstract Algebra
Replies
1
Views
642
Replies
3
Views
2K
  • Linear and Abstract Algebra
Replies
3
Views
1K
  • Linear and Abstract Algebra
Replies
5
Views
1K
Back
Top