Verifying Answers to "Zero Divisors & Isomorphism Theorem"

In summary, the conversation discusses a problem where the matrices do not have zero divisors and are not isomorphic. The conversation also mentions the incorrect reasoning for parts (a) and (b) and talks about identifying units for Z x Z and Z x Z subscript 5. The units for Z x Z are determined to be (1,1), (1,-1), (-1,1), and (-1,-1), while the units for Z x Z subscript 5 are (1,1), (1,2), (1,3), (1,4), (-1,1), (-1,2), (-1,3), and (-1,4). These results are then verified.
  • #1
Joe20
53
1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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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  • #6
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 $S$.
 
  • #7
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
 
  • #8
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.
 

FAQ: Verifying Answers to "Zero Divisors & Isomorphism Theorem"

What are zero divisors in mathematics?

Zero divisors are elements in a mathematical structure (such as a ring or a group) that when multiplied by another element, result in zero. In other words, they are non-zero elements that produce a product of zero.

How do you verify if an element is a zero divisor?

To verify if an element is a zero divisor, you can simply multiply it with every other element in the structure and see if any of the resulting products is equal to zero. If at least one product is zero, then the element is a zero divisor.

What is the Isomorphism Theorem?

The Isomorphism Theorem is a fundamental concept in abstract algebra that states that two mathematical structures are isomorphic if they have the same algebraic properties, despite potentially having different names and symbols for their elements. In simpler terms, it means that two structures are equivalent in terms of their mathematical operations and relationships.

How do you prove that two structures are isomorphic?

To prove that two structures are isomorphic, you need to show that there exists a bijective homomorphism (a function that preserves the structure and operations of the elements) between the two structures. This means that the elements in one structure can be mapped to the elements in the other structure in a way that maintains their algebraic properties.

Can a zero divisor ever be an invertible element in a mathematical structure?

No, a zero divisor cannot be an invertible element. This is because an invertible element's product with any other element in the structure must result in a non-zero element, which is not possible for a zero divisor. In other words, a zero divisor does not have an inverse element.

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