A ring \(R\) is called simple if its only two-sided ideals are \(\{0\}\) and \(R\).
Suppose \(R\) is a commutative ring with identity \(1 \ne 0\). Prove that \(R\) is simple if and only if \(R\) is a field.
Let \(k\) be a field and \(n \ge 1\). Show that the ring \(M_n(k)\) of \(n \times n\) matrices with entries in \(k\) is a simple ring.
Solution
Goal: Prove that commutative unital simple rings are fields in (a), and that matrix rings \(M_n(k)\) over a field are simple in (b) using matrix units.
1. Part (a): \(R\) is simple if and only if \(R\) is a field.
1.1. Forward direction (\(\implies\)):- Assume \(R\) is simple, commutative, with \(1 \ne 0\).- Let \(x \in R \setminus \{0\}\) be a non-zero element.- The principal ideal \((x) = R x = \{r x \mid r \in R\}\) is an ideal of \(R\).- Since \(x = 1 \cdot x \in (x)\) and \(x \ne 0\), the ideal \((x) \ne \{0\}\).- Because \(R\) is simple, the only non-zero ideal is \(R\), so \((x) = R\).- Since \(1 \in R = (x)\), there exists \(y \in R\) such that \(y x = 1\).- By commutativity, \(x y = y x = 1\), so \(x\) is a unit in \(R\).- Since every non-zero element of \(R\) is invertible, \(R\) is a field.
1.2. Reverse direction (\(\impliedby\)):- Assume \(R\) is a field.- Let \(I \subseteq R\) be an ideal such that \(I \ne \{0\}\).- Choose a non-zero element \(x \in I \setminus \{0\}\).- Since \(R\) is a field, \(x^{-1} \in R\).- Since \(I\) is an ideal, \(1 = x^{-1} x \in I\).- For every \(r \in R\), \(r = r \cdot 1 \in I\), so \(I = R\).- Thus the only ideals of \(R\) are \(\{0\}\) and \(R\), proving \(R\) is simple.
2. Part (b): \(M_n(k)\) is a simple ring.
2.1. Let \(I \subseteq M_n(k)\) be a non-zero two-sided ideal.
2.2. Matrix units: For \(1 \le i, j \le n\), let \(E_{i j} \in M_n(k)\) denote the standard matrix unit with 1 in the \((i, j)\)-entry and 0 elsewhere.
2.3. Multiplication rule for matrix units:\[E_{p q} E_{r s} = \delta_{q r} E_{p s} = \begin{cases} E_{p s} & \text{if } q = r, \\ 0 & \text{if } q \ne r. \end{cases}\]
2.4. Non-zero entry extraction:- Since \(I \ne \{0\}\), choose a non-zero matrix \(A = (a_{p q}) \in I \setminus \{0\}\).- There exist indices \(r, c \in \{1, \dots, n\}\) such that \(a_{r c} \ne 0\).- Since \(k\) is a field, the scalar \(a_{r c} \in k^\times\) has an inverse \(a_{r c}^{-1} \in k\).
2.5. Generation of all matrix units:- For any desired indices \(i, j \in \{1, \dots, n\}\), compute the product:\[E_{i r} A E_{c j} = E_{i r} \left( \sum_{p=1}^n \sum_{q=1}^n a_{p q} E_{p q} \right) E_{c j} = \sum_{p, q} a_{p q} (E_{i r} E_{p q}) E_{c j} = \sum_{q} a_{r q} E_{i q} E_{c j} = a_{r c} E_{i j}.\]- Since \(A \in I\) and \(I\) is a two-sided ideal, \(E_{i r} A E_{c j} \in I\), so \(a_{r c} E_{i j} \in I\).- Multiplying by the central element \(a_{r c}^{-1} I_n \in M_n(k)\) gives\[E_{i j} = (a_{r c}^{-1} I_n)(a_{r c} E_{i j}) \in I \quad \text{for all } 1 \le i, j \le n.\]
2.6. Inclusion of the identity:- The \(n \times n\) identity matrix \(I_n\) is the sum of diagonal matrix units:\[I_n = \sum_{i=1}^n E_{i i} \in I.\]
2.7. Full ring equality:- For any matrix \(M \in M_n(k)\), \(M = M I_n \in I\), so \(I = M_n(k)\).- Thus the only two-sided ideals of \(M_n(k)\) are \(\{0\}\) and \(M_n(k)\), so \(M_n(k)\) is simple.
3. Conclusion:
Commutative unital simple rings are fields, and $M_n(k)$ is a simple ring.