Proposition 30.1.4.label Let $n \in \natp$, then

  1. (1)

    $M_{n}(\complex)$ admits no nontrivial two-sided ideals.

  2. (2)

    $M_{n}(\complex)$ admits no multiplicative functionals.

Proof. Let $x = (x_{ij}) \in M_{n}(\complex) \setminus \bracs{0}$, then there exists $1 \le i, j \le n$ such that $x_{ij}\ne 0$. In which case, for any $1 \le k, l \le n$ and $\lambda \in \complex$, there exists $y_{k}, z_{l} \in M_{n}(\complex)$ such that $y_{k}xz_{l}$ is the matrix with $\lambda$ on its $(k, l)$ entry and $0$ everywhere else. Therefore every non-trivial two-sided ideal of $M_{n}(\complex)$ is $M_{n}(\complex)$.$\square$

Post a Comment

Name:Email:
Please enter the tag of the current page (WN) to post the comment.
Tag: