[proofplan] The equivalence is exactly the cancellation property for multiplication by a nonzero element. In an [integral domain](/page/Integral%20Domain), the absence of zero divisors turns an equality $cx=cy$ into $x=y$ by subtracting and factoring. Conversely, if every nonzero multiplication map is injective, then a product $ab=0_R$ with $a \ne 0_R$ forces $b=0_R$ by comparing $m_a(b)$ with $m_a(0_R)$, so the ring has no zero divisors. [/proofplan] [step:Use the absence of zero divisors to prove every nonzero multiplication map is injective] Assume that $R$ is an integral domain. Let $c \in R$ satisfy $c \ne 0_R$. We prove that \begin{align*} m_c: R &\to R \end{align*} is injective. Let $x,y \in R$ and suppose $m_c(x)=m_c(y)$. By the definition of $m_c$, this means $cx=cy$. Subtracting $cy$ from both sides and using distributivity gives \begin{align*} c(x-y)=0_R. \end{align*} Since $R$ is an integral domain, it has no zero divisors. Because $c \ne 0_R$, the equality $c(x-y)=0_R$ implies $x-y=0_R$. Hence $x=y$. Therefore $m_c$ is injective. [guided] Assume that $R$ is an integral domain, and fix an element $c \in R$ with $c \ne 0_R$. To prove that the multiplication map \begin{align*} m_c: R &\to R \end{align*} is injective, we must prove that whenever two inputs have the same image, the inputs are equal. Let $x,y \in R$ and assume $m_c(x)=m_c(y)$. By the definition of the multiplication map, this equality is \begin{align*} cx=cy. \end{align*} We want to isolate the difference between $x$ and $y$. Since $R$ is a ring, its additive structure is an abelian group, so subtraction is available. Subtracting $cy$ from both sides gives \begin{align*} cx-cy=0_R. \end{align*} By distributivity in the ring, the left-hand side factors as \begin{align*} c(x-y)=0_R. \end{align*} Now the defining property of an integral domain is used: a nonzero product cannot be equal to $0_R$. Since $c \ne 0_R$ and $c(x-y)=0_R$, the other factor must be zero, so \begin{align*} x-y=0_R. \end{align*} Adding $y$ to both sides gives $x=y$. Thus equal images under $m_c$ imply equal inputs, and therefore $m_c$ is injective. [/guided] [/step] [step:Use injectivity of nonzero multiplication maps to rule out zero divisors] Conversely, assume that for every $c \in R$ with $c \ne 0_R$, the map $m_c: R \to R$ is injective. We prove that $R$ has no zero divisors. Let $a,b \in R$ satisfy $ab=0_R$, and assume $a \ne 0_R$. Since $a$ is nonzero, the map \begin{align*} m_a: R &\to R \end{align*} is injective by hypothesis. By the definition of $m_a$, \begin{align*} m_a(b)=ab=0_R. \end{align*} Also, \begin{align*} m_a(0_R)=a0_R=0_R. \end{align*} Thus $m_a(b)=m_a(0_R)$. Injectivity of $m_a$ gives $b=0_R$. Therefore every product $ab=0_R$ in $R$ has $a=0_R$ or $b=0_R$. Since $R$ is already assumed to be nonzero and commutative, $R$ is an integral domain. [/step]