Consider the ring \(\rm R=\left\{\Sigma_{n \in Z} a_n X^n \left|a_n \in Z; \ and \ a_n \ne 0\ only\ for\ finitely\ many \ n \in Z\right.\right\}\) where addition and multiplication are given by \(\rm \Sigma_{n \in Z}a_n X^n+\rm \Sigma_{n \in Z}b_n X^n=\rm \Sigma_{n \in Z}(a_n+b_n)X^n\)

\(\rm \left(\rm \Sigma_{n \in Z}a_n X^n\right)\rm (\Sigma_{n \in Z}b_m X^m)=\rm \Sigma_{k \in Z}(\Sigma_{n+m=k}a_nb_m) X^k\)

Which of the following statements is true? 

Concept:  

Maximal Ideal: A maximal ideal I in a ring R is an ideal such that the quotient ring R/I is a field. 
 

Prime Ideal: A prime ideal P in a ring R is an ideal such that if the product of two elements is in P ,

then at least one of the elements must be in P .

Explanation:

\(R = \left\{ \sum_{n \in \mathbb{Z}} a_n X^n \mid a_n \in \mathbb{Z}, \text{ and } a_n \neq 0 \text{ for finitely many } n \in \mathbb{Z} \right\}\)

The addition and multiplication in the ring are defined as

\(\left( \sum_{n \in \mathbb{Z}} a_n X^n \right) + \left( \sum_{n \in \mathbb{Z}} b_n X^n \right) = \sum_{n \in \mathbb{Z}} (a_n + b_n) X^n \)

\(\left( \sum_{n \in \mathbb{Z}} a_n X^n \right) \left( \sum_{n \in \mathbb{Z}} b_m X^m \right) = \sum_{k \in \mathbb{Z}} \left( \sum_{n+m=k} a_n b_m \right) X^k\)

Option 1:

The addition in this ring is clearly commutative since the sum of polynomials in any ring is commutative.

Now consider the multiplication. In standard polynomial rings, multiplication is commutative as long as the

coefficients come from a commutative ring (in this case, integers \( \mathbb{Z} \) ).

Since \( \mathbb{Z} \) is commutative under multiplication, and the exponents \(X^n\) follow commutative multiplication rules

(i.e., \(X^n X^m = X^{n+m} \)), the ring R is also commutative under multiplication.

Therefore, the statement R is not commutative is false.

Option 2:

 The ideal \(X-1 \) would be maximal if R/\(X-1 \)  is a field.

However, this is not necessarily true in the ring R as described, since R/\(X-1 \) is unlikely

to be a field (it may reduce to a simpler ring, but not a field).
 

Option 3:

(X - 1, 2) is a standard type of ideal in certain polynomial rings, particularly over integers. For a prime ideal,

the condition that multiplication of elements should remain within the ideal must hold. 

Option 4:

\( (X, 5)\) as a Maximal Ideal: If \( (X, 5)\) is maximal, the quotient R/\( (X, 5)\) should be a field.

While in certain rings this could lead to a field, this needs further validation.

Therefore, option 3) is correct.