@srestha ma'am

is ring , field in syllabus ?

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+14 votes

Consider the set $S = \{1, ω, ω^2\}$, where $ω$ and $ω^2$ are cube roots of unity. If $*$ denotes the multiplication operation, the structure $(S, *)$ forms

- A Group
- A Ring
- An integral domain
- A field

+29 votes

Best answer

Answer: A

$$\underset{\begin{array}{|c|c|c|c|} \hline \text{} & \text{1}& \omega & \omega^2 \\\hline \text{1} & \text{1}& \omega & \omega^2 \\\hline \omega & \omega & \omega^2 & \text{1}\\\hline \omega^2 & \omega^2 & \text{1} & \omega\\\hline \end{array}}{\text{Cayley Table}}$$

The structure $(S,*)$ satisfies closure property, associativity and commutativity. The structure also has an identity element $(=1)$ and an inverse for each element. So, the structure is an Abelian group.

$$\underset{\begin{array}{|c|c|c|c|} \hline \text{} & \text{1}& \omega & \omega^2 \\\hline \text{1} & \text{1}& \omega & \omega^2 \\\hline \omega & \omega & \omega^2 & \text{1}\\\hline \omega^2 & \omega^2 & \text{1} & \omega\\\hline \end{array}}{\text{Cayley Table}}$$

The structure $(S,*)$ satisfies closure property, associativity and commutativity. The structure also has an identity element $(=1)$ and an inverse for each element. So, the structure is an Abelian group.

0

Can anyone explain what are ring, field, and integral domain and why the given structure is not one of them?

+7

In the question, it is only asked for multiplication so why even do we need to check for ring or field because for both of them two binary operations should be defined. right? Thus it clearly not a ring or a field.

+10 votes

In this question, only one binary operation is given so it cannot be a Ring or Integral domain or field.For Ring or Integral domain or for field there must be Two binary operations are required and the algebraic structure looks like (S ,+ ,⨉) .

So ,option b ,c and d are False.

Now check the properties of groups.It satisfies all the conditions of groups.

Hence, it is a Group.

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