We are given the set $S = \left\{\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} : a, b \in \mathbb{Z} \right\}$ and the ring of 2x2 matrices over integers, $M_2$. We need to determine if S is a left ideal, a right ideal, both, or neither.
Checking for Left Ideal Property
For S to be a left ideal in $M_2$, for any matrix $r = \begin{pmatrix} p & q \\ s & t \end{pmatrix} \in M_2$ and any matrix $x = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \in S$, the product $rx$ must belong to S.
Let's compute the product $rx$: $ rx = \begin{pmatrix} p & q \\ s & t \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} = \begin{pmatrix} pa & qb \\ sa & tb \end{pmatrix} $ For $rx$ to be in S, the off-diagonal elements must be zero. That is, $qb = 0$ and $sa = 0$ must hold for all integers $a, b, q, s$.
Consider $r = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \in M_2$ and $x = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \in S$. Then, $rx = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. The resulting matrix $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has a non-zero off-diagonal element (top-right). Therefore, $rx \notin S$. Thus, S is not a left ideal.
Checking for Right Ideal Property
For S to be a right ideal in $M_2$, for any matrix $r = \begin{pmatrix} p & q \\ s & t \end{pmatrix} \in M_2$ and any matrix $x = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \in S$, the product $xr$ must belong to S.
Let's compute the product $xr$: $ xr = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} p & q \\ s & t \end{pmatrix} = \begin{pmatrix} ap & aq \\ bs & bt \end{pmatrix} $ For $xr$ to be in S, the off-diagonal elements must be zero. That is, $aq = 0$ and $bs = 0$ must hold for all integers $a, b, q, s$.
Consider $r = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \in M_2$ and $x = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \in S$. Then, $xr = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. The resulting matrix $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has a non-zero off-diagonal element (top-right). Therefore, $xr \notin S$. Thus, S is not a right ideal.
Conclusion
Since the set S fails both the left ideal and right ideal conditions, it is neither a left ideal nor a right ideal.
Important Questions from Rings & Ideals
If the ring R is a commutative ring with unity, then the polynomial ring R[X] is-
Let R = (Z2 × Z2, +,.) forms a ring of module 2 such that (a, b) + (c, d) = (a + c, d + d) and (a, b) (c. d) = (a.c, b.d) for (a, b), (c, d) ∈ Z2 × Z2 then-
The set of all units in a ring R with unity forms ______.
Let C[0, 1] be the ring of all real valued continuous function on [0, 1].
Let A = {f ∈ C[0, 1] ∶ \(f\left( \frac{1}{4}\right)=f\left( \frac{3}{4}\right)\) = 0}. Then which of the following statements are true?
Which of the following statements is NOT true?
