# Thread: Prove it is a field

1. ## Prove it is a field

Prove that a commutative ring F is a field iff each equation ax+b=c
(a,b,c is a member of F and a does not equal 0) has a unique solution in F

2. Originally Posted by mandy123
Prove that a commutative ring F is a field iff each equation ax+b=c
(a,b,c is a member of F and a does not equal 0) has a unique solution in F

well, the condition you have is clearly equaivalent to say that $ax=b$ has a unique solution for any $a \neq 0.$ if $F$ is a field and $0 \neq a \in F,$ then from $ax=b$ you'll get $x=a^{-1}b$ and you're done.

but the converse is less trivial: so we have that any equation $ax=b, \ a \neq 0,$ has a unique solution in $F.$ first we show that $F$ has no zero divisor: suppose that $uv=0$ and $u \neq 0.$ now the

equation $ux=0$ has two solutions $x=0,v.$ thus $v=0.$ this proves that $F$ has no zero divisor. next we show that $F$ is unitary: choose any $0 \neq a \in F.$ then $ax_0=a$ for some $x_0 \in F.$ now

let $0 \neq b \in F.$ then there exists $x_1 \in F$ such that $bx_1=b.$ but then $ab(x_1-x_0)=0.$ since $F$ has no zero divisor, we must have $x_1=x_0.$ clearly $0x_0=0.$ thus: $\forall b \in F: bx_0=b.$ therefore

$x_0=1_F.$ finally for any $0 \neq a \in F$ the equation $ax=1$ has a (unique) solution, which means any non-zero element of $F$ is invertible. $\Box$