Verifying Answers to "Zero Divisors & Isomorphism Theorem"

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SUMMARY

The discussion centers on verifying the properties of the rings Z x Z and Z x Z_5 regarding zero divisors and isomorphism. It is established that Z x Z contains zero divisors, specifically the matrix $\begin{pmatrix}0&1\\0&0\end{pmatrix}$, which confirms it is not isomorphic to Z x Z_5, which has five elements. Both rings are infinite, and the units for Z x Z are (1,1), (1,-1), (-1,1), and (-1,-1), while the units for Z x Z_5 include (1,1), (1,2), (1,3), (1,4), (-1,1), (-1,2), (-1,3), and (-1,4).

PREREQUISITES
  • Understanding of ring theory, specifically the concepts of zero divisors and units.
  • Familiarity with matrix representation of rings.
  • Knowledge of the structure of Z (integers) and Z_5 (integers modulo 5).
  • Basic understanding of isomorphism in algebraic structures.
NEXT STEPS
  • Study the properties of zero divisors in various ring structures.
  • Learn about the concept of isomorphism in ring theory.
  • Explore the units of other finite and infinite rings.
  • Investigate the implications of commutativity in ring operations.
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Mathematicians, algebra students, and educators focusing on abstract algebra, particularly those interested in ring theory and its applications.

Joe20
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I have gotten the following answer to (a) and (b) which require verification on them. I have also attached the theorem for reference.

(a) Z x Z => have zero divisors
The matrix has no zero divisors (no nonzero matrix when multiplied to the matrix gives zero element)
Hence not isomorphic. (b) Z x Z => have 2 elements
Z x Z subscript 5 => have 5 elements ( [0,0] [0,1] [0,2] [0,3] [0,4] )
Hence not isomorphic.
 

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Your reasoning for part (a) is incorrect since $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ is a zero divisor: $\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}1&1\\0&0\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}$ Also, your reasoning for (b) is incorrect: both $\Bbb Z\times \Bbb Z$ and $\Bbb Z \times \Bbb Z_5$ are infinite rings.
 
Euge said:
Your reasoning for part (a) is incorrect since $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ is a zero divisor: $\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}1&1\\0&0\end{pmatrix} = \begin{pmatrix}0&0\\0&0\end{pmatrix}$ Also, your reasoning for (b) is incorrect: both $\Bbb Z\times \Bbb Z$ and $\Bbb Z \times \Bbb Z_5$ are infinite rings.
So how do I go about approaching these two questions? I have no clue.
 
Alexis87 said:
So how do I go about approaching these two questions? I have no clue.

For (a), let's start with the first property.
Are they both commutative?
What do we get for (a,b)x(c,d) and (c,d)x(a,b) in both cases?

For (b), which units do they have?
 
I have done up part (a), hence need verification on that as attached.

Next for part (b), I am not sure how to identify the unit(s) from Z x Z and Z x Z subscript 5. May need someone's help.

Thanks.
 

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Alexis87 said:
I have done up part (a), hence need verification on that as attached.

Next for part (b), I am not sure how to identify the unit(s) from Z x Z and Z x Z subscript 5. May need someone's help.

Thanks.
First consider $\Bbb Z$ and $\Bbb Z_5$. The units in $\Bbb Z$ are $\pm 1$ and the units of $\Bbb Z_5$ are the nonzero elements of $\Bbb Z_5$. Next, consider the fact that if $R$ and $S$ are rings with unity, then the units of $R\times S$ are of the form $(u,v)$ where $u$ is a unit in $R$ and $v$ is a unit in $S$.
 
for part b, i have gotten the units for Z x Z as (1,1) , (1,-1) , (-1,1) , (-1, -1) and units for Z x Z subscript 5 as (1, 1), (1,2), (1,3) , (1,4), (-1,1), (-1,2), (-1,3), (-1,4).

hence would require someone to verify the results stated. Thanks
 
Alexis87 said:
for part b, i have gotten the units for Z x Z as (1,1) , (1,-1) , (-1,1) , (-1, -1) and units for Z x Z subscript 5 as (1, 1), (1,2), (1,3) , (1,4), (-1,1), (-1,2), (-1,3), (-1,4).

hence would require someone to verify the results stated. Thanks
It's correct.
 

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