Non-Commutative Ring with Unity and 2 Ideals not necessarily Division Ring

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Theorem

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_F$ and whose unity is $1_F$.

Let $\struct {R, +, \circ}$ specifically not be commutative.

Let $\struct {R, +, \circ}$ be such that the only ideals of $\struct {R, +, \circ}$ are $\set {0_R}$ and $R$ itself.


Then it is not necessarily the case that $\struct {R, +, \circ}$ is a division ring.


Proof

Let $S$ be the set of square matrices of order $2$ over the real numbers $\R$.

$S$ is not a division ring, as for example:

$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$

and so both $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ are proper zero divisors of $S$.


Let $J$ be an ideal of $S$ containing a non-zero matrix $A$ whose $\tuple {r, s}$th entry is $\lambda \ne 0$

Let $E_{i j}$ be the matrix of order $2$ defined as:

$e_{a b} = \begin{cases} 1 & : a = i, b = j \\ 0 & : \text {otherwise} \end{cases}$

Thus for example:

$E_{1 2} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$

Then for all $i, j \in \set {1, 2}$:

\(\displaystyle \dfrac 1 \lambda E_{i r} A E_{s j}\) \(=\) \(\displaystyle \dfrac 1 \lambda E_{i r} A E_{s j}\)
\(\displaystyle \) \(=\) \(\displaystyle E_{i j}\)



It follows that:

$\forall i, j \in \set {1, 2}: E_{i j} \in J$

It remains to be shown that $J$ is the whole of $S$.


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