1. ## ring -> field

Let be a commutative ring with elements , . Suppose that if is not inversable, then . Prove that is a field.

I know the following result:
A finite ring with unit and with no non-zero nilpotent elements is a finite direct product of fields.How can this result help me ?

2. Originally Posted by petter
Let be a commutative ring with elements , . Suppose that if is not inversable, then . Prove that is a field.

I know the following result:
A finite ring with unit and with no non-zero nilpotent elements is a finite direct product of fields.How can this result help me ?
suppose $\displaystyle 0 \neq a \in A$ and $\displaystyle a^k=0, \ k > 1.$ then obviously $\displaystyle a$ is not invertible. thus $\displaystyle a^2 = \pm a,$ which gives us: $\displaystyle 0=a^k=\pm a.$ so $\displaystyle a=0,$ i.e. $\displaystyle A$ has no non-zero nilpotent element.

hence $\displaystyle A=\bigoplus_{i=1}^m Q_i,$ where $\displaystyle Q_1, \cdots , Q_m$ are finite fields. if $\displaystyle m=1,$ we're done. so suppose that $\displaystyle m \geq 2.$ since $\displaystyle |A|$ is odd, $\displaystyle |Q_i|$ must be odd for all $\displaystyle 1 \leq i \leq m.$ consider two cases:

case 1: $\displaystyle |Q_i| > 3,$ for some $\displaystyle i.$ without loss of generality we assume that $\displaystyle i=1.$ choose $\displaystyle x \in Q_1 - \{0,1,-1 \}$ and let $\displaystyle a=(x, 0, \cdots , 0).$ then $\displaystyle a$ is clearly a non-invertible element

of $\displaystyle A,$ but $\displaystyle a^2 = (x^2 , 0, \cdots , 0) \neq (\pm x, 0 , \cdots , 0)= \pm a,$ because $\displaystyle x \notin \{0,1,-1 \}.$ contradiction!

case 2: $\displaystyle |Q_i|=3, \ \forall i.$ if $\displaystyle m=2,$ then $\displaystyle |A|=9,$ which is impossible because the problem assumes that $\displaystyle |A| \geq 11.$ therefore $\displaystyle m \geq 3.$ now let $\displaystyle a=(1,-1, 0, \cdots , 0) \in A.$ then $\displaystyle a$ is a

non-invertible element of $\displaystyle A$ and $\displaystyle a^2=(1,1, 0, \cdots , 0) \neq \pm a,$ which is again a contradiction! Q.E.D.