Left ideal in matrix ring

hbghlyj

Let \(R = M_2(\mathbb{C})\), the ring of \(2 \times 2\) matrices over the complex numbers.

Consider the subset \(I \subseteq R\) consisting of all matrices of the form
\[
\begin{pmatrix}
x & 0 \\
y & 0
\end{pmatrix}
\]
where \(x, y \in \mathbb{C}\).

To verify that \(I\) is a left ideal, first note that it is an additive subgroup of \(R\) (closed under addition and additive inverses). Now take any \(A = \begin{pmatrix} p & q \\ r & s \end{pmatrix} \in R\) and any \(M = \begin{pmatrix} x & 0 \\ y & 0 \end{pmatrix} \in I\). Then
\[
A M = \begin{pmatrix} p & q \\ r & s \end{pmatrix} \begin{pmatrix} x & 0 \\ y & 0 \end{pmatrix} = \begin{pmatrix} p x + q y & 0 \\ r x + s y & 0 \end{pmatrix} \in I,
\]
since the second column is zero. Thus, \(R I \subseteq I\).

To show that \(I\) is not a right ideal, take \(M = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \in I\) and \(B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \in R\). Then
\[
M B = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \notin I,
\]
since the second column is not zero. Thus, \(I R \not\subseteq I\).

hbghlyj

There are two two-sided ideals in \(M_2(\mathbb{C})\): the zero ideal \(\{0\}\) and the entire ring \(M_2(\mathbb{C})\).

To prove this, suppose \(I\) is a nonzero two-sided ideal of \(R = M_2(\mathbb{C})\). Let \(A = (a_{pq}) \in I\) be nonzero, so some entry \(a_{ij} = \alpha \neq 0\). Denote by \(E_{kl}\) the matrix unit with \(1\) in position \((k, l)\) and zeros elsewhere. Then \(E_{ki} A E_{jl}\) is the matrix with \(\alpha\) in position \((k, l)\) and zeros elsewhere. Hence, \(\frac{1}{\alpha} E_{ki} A E_{jl} = E_{kl}\). Since \(I\) is a two-sided ideal, \(E_{ki} A \in I\) (left multiplication) and then \((E_{ki} A) E_{jl} \in I\) (right multiplication), so a scalar multiple of \(E_{kl}\) is in \(I\).

Since ideals in \(M_n(F)\) for a field \(F\) are \(F\)-vector spaces (as scalar multiplication by \(\lambda \in F\) corresponds to left or right multiplication by \(\lambda I\), where \(I\) is the identity matrix), it follows that \(E_{kl} \in I\).

Now, since \(I\) contains \(E_{kl}\), right multiplication by \(E_{lm}\) yields \(E_{kl} E_{lm} = E_{km} \in I\) (using the Kronecker delta property \(E_{kl} E_{lm} = \delta_{l m'} E_{k m}\) with \(m' = l\)). Similarly, left multiplication by \(E_{nk}\) yields \(E_{nk} E_{kl} = E_{nl} \in I\). Repeating this process shows that \(I\) contains all matrix units \(E_{rs}\) for \(1 \leq r, s \leq 2\).

The matrix units span \(R\) as a \(\mathbb{C}\)-vector space, so \(I = R\). Thus, the only two-sided ideals are \(\{0\}\) and \(R\).

hbghlyj

The set of all matrices in \(M_2(\mathbb{Z})\) whose entries are even integers is a proper nonzero two-sided ideal. To see that it is a two-sided ideal, note that it is closed under addition and that left or right multiplication of any such matrix by an arbitrary matrix in \(M_2(\mathbb{Z})\) yields a matrix whose entries are even integers (as each entry is an integer linear combination of even integers). It is proper because the identity matrix is not in it, and it is nonzero because, for example, \(2I\) is in it.