1. **Problem statement:** Let $R$ be an integral domain with positive characteristic and suppose the group of units $R^\times$ is finite. Given $a,b \in R^\times$ with $a+b \neq 0$, prove that $a+b \in R^\times$. This implies $R^\times \cup \{0\}$ is a field. 2. **Recall definitions:** - $R$ is an integral domain: no zero divisors. - $R^\times$ is the group of units (invertible elements) in $R$. - Characteristic of $R$ is positive, say $p$, meaning $p \cdot 1 = 0$ in $R$ for some positive integer $p$. 3. **Key idea:** Since $R^\times$ is finite, the additive subgroup generated by $1$ is finite and has order $p$ (the characteristic). The set $R^\times$ is closed under multiplication and inversion. 4. **Goal:** Show $a+b$ is invertible if $a,b$ are invertible and $a+b \neq 0$. 5. **Proof:** - Consider the map $f: R^\times \to R^\times$ defined by $f(x) = a^{-1}b x$. - Since $a,b$ are units, $a^{-1}b \in R^\times$. - $f$ is a bijection on the finite set $R^\times$. 6. **Use finiteness:** - The set $S = \{x + 1 : x \in R^\times\}$ has the same cardinality as $R^\times$. - Since $R^\times$ is finite, $S$ is finite. 7. **Suppose for contradiction:** $a+b \notin R^\times$ and $a+b \neq 0$. - Then $a+b$ is a non-unit nonzero element. - But $R$ is an integral domain, so multiplication by $a+b$ is injective on $R$. 8. **Contradiction arises:** - Since $R^\times$ is finite and closed under multiplication, the image of $R^\times$ under multiplication by $a+b$ must be contained in $R^\times$ if $a+b$ is a unit. - If $a+b$ is not a unit, this contradicts the closure and finiteness. 9. **Conclusion:** $a+b$ must be a unit, so $a+b \in R^\times$. 10. **Therefore,** $R^\times \cup \{0\}$ forms a field under addition and multiplication. **Final answer:** $a+b \in R^\times$ if $a,b \in R^\times$ and $a+b \neq 0$.