ring -> field

**petter** · January 3rd 2009, 08:56 AM

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 ?

**NonCommAlg** · January 3rd 2009, 06:38 PM

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 $0 \neq a \in A$ and $a^k=0, \ k > 1.$ then obviously $a$ is not invertible. thus $a^2 = \pm a,$ which gives us: $0=a^k=\pm a.$ so $a=0,$ i.e. $A$ has no non-zero nilpotent element.

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

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

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

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

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

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