Unity of Subring is not necessarily Unity of Ring
Theorem
Let $\struct {S, +, \circ}$ be a ring with unity whose unity is $1_S$.
Let $\struct {T, + \circ}$ be a subring of $\struct {S, + \circ}$ whose unity is $1_T$.
Then it is not necessarily the case that $1_T = 1_S$.
Proof
Let $\struct {S, + \times}$ be the ring formed by the set of order $2$ square matrices over the real numbers $R$ under (conventional) matrix addition and (conventional) matrix multiplication.
Let $T$ be the subset of $S$ consisting of the matrices of the form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ for $x \in \R$.
We have that:
- $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} -y & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} x - y & 0 \\ 0 & 0 \end{bmatrix}$
and so $T$ is closed under subtraction.
From Matrices of the Form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ we have that $\struct {T, \times}$ is a subsemigroup of $\struct {S, \times}$.
It follows from the Subring Test that $\struct {T, + \circ}$ be a subbring of $\struct {S, + \circ}$.
From (some result somewhere) $\struct {S, \times}$ has an identity $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which is not in $\struct {T, \times}$.
However, note that:
\(\ds \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\) | \(=\) | \(\ds \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}\) |
demonstrating that $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ is the identity of $\struct {T, \times}$.
The result follows.
$\blacksquare$
Also see
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 56.3$ Subrings and subfields