Exercise 1.3

Question

Exercise 3: Show
    \begin{itemize}
        \item[(a)] If $x 
eq 0$ and $xy = xz$, then $y = z$.
        \item[(b)] If $x 
eq 0$ and $xy = x$ then $y = 1$. 
        \item[(c)] If $x 
eq 0$ and $xy = 1$, then $y = 1/x$.
        \item[(d)] If $x 
eq 0$ then $1/(1/x) = x$.
    \end{itemize}

ghostofgarborg · Accepted Answer

(a): Solution 1: $x \neq 0$ implies that $x$ has a multiplicative inverse. Multiply by $1/x$.

However, we can prove this without resorting to the existence of inverses, and get a solution that carries over to a larger class of rings, i.e. integral domains like $\mathbb Z$. Solution 2: Note that proposition 1.16b implies that $xy=0$ implies that $x=0$ or $y=0$. If $xy = xz$, then
    $x(y-z)=0$. If $ x \neq 0$, then by prop. 1.16b, $(y-z) = 0$. The claim follows.

(b): Follows from (a) by letting $z = 1$.

(c): Follows from (a) by letting $z = x^{-1}$.

(d): Follows by applying (a) to  
    $$ \left( \frac 1 x \right) \frac 1 {\left( \frac 1 x \right)} = \left( \frac 1 x \right) x $$

Amit Awasthi · Answer

\begin{proof}
      If $xy = xz$, the axioms (M) from Definition 1.12 give
      \begin{equation*}
        y = 1y = \frac{x}{x} y = \frac{xy}{x} = \frac{xz}{x} = \frac{x}{x} z = 1z = z.
      \end{equation*}
      This proves $(a)$. Take $z = 1$ in $(a)$ to obtain $(b)$. Take $z = 1/x$ in $(a)$ to obtain $(c)$. Since $x(1/x) = 1$, $(c)$ with $1/x$ in place of $x$ gives $(d)$.
\end{proof}

Exercise 1.3

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Exercise 1.3

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