\(\DeclarePairedDelimiterX{\Set}[2]{\{}{\}}{#1 \nonscript\;\delimsize\vert\nonscript\; #2}\) \( \DeclarePairedDelimiter{\set}{\{}{\}}\) \( \DeclarePairedDelimiter{\parens}{\left(}{\right)}\) \(\DeclarePairedDelimiterX{\innerproduct}[1]{\langle}{\rangle}{#1}\) \(\newcommand{\ip}[1]{\innerproduct{#1}}\) \(\newcommand{\bmat}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\barray}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\mat}[1]{\begin{matrix}#1\end{matrix}}\) \(\newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}}\) \(\newcommand{\mathword}[1]{\mathop{\textup{#1}}}\)
Needs:
Rings
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Subrings

Definition

Let $(R, +, \cdot )$ be a ring. A ring $(S, +, \cdot )$ is a subring of $(R, +, \cdot )$ if $S \subset R$.

Verification

If $(R, +, \cdot )$ and $S \subset R$, then $+$ is associative and commutative on $S$ because it is on $R$. Likewise $\cdot $ is associative on $S$ and $+$ and $\cdot $ distribute over each on $S$ because they do on $R$. So we have restricted the number of conditions to check, and arrive at our first statement of sufficient conditions on $S$ that ensure $(S, +, \cdot )$ is a ring.

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