Chapter 2.5, Problem 2E

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A.

1. Claim. If $\left(R,+,\cdot \right)$ is a ring, $a,b\in R,$ and $a\ne 0,$ then the equation $ax=b$ has a unique solution.
   "Proof." Suppose that p and q are two solutions to $ax=b.$ Then $ap=b$ and $aq=b.$ Therefore, $ap=aq.$ Therefore, $p=q.$
2. Claim. If $\left(R,+,\cdot \right)$ is a finite integral domain, then $\left(R,+,\cdot \right)$ is a field.
   "Proof." Suppose that R has n elements. Let $x\in R$ . Then the $n+1$ powers of $x:e=x^0,x,x^2,x^3,...,x^n$ are not all distinct. Therefore, $x^t=x^r$ for integers t and r, where we may assume that $t<r.$ Then $x^{-t}{x}^t=x^{-t}{x}^r,$ Therefore, $e=x^{r-t}.$ and therefore $e=x\cdot x^{r-t-1}.$ Thus, $e=x\cdot x^{r-t-1}.$ has an inverse. Hence R is a field.
3. Claim. Let $\left(R,+,\cdot \right)$ be an integral domain with a, b, $c\in R$ and unity element 1. If $ab=ac$ and $a\ne 0,$ than $b=c.$
   "Proof." Suppose that $ab=ac$ and $a\ne 0.$ Then $b=1b=\left(a^{-1}a\right)b=a^{-1}\left(ab\right)=a^{-1}\left(ac\right)=\left(a^{-1}a\right)c=1c=c.$ Therefore $b=c.$