Basic Field Theory Question

**rockinmoroccan** · February 8th 2010, 11:07 PM

Let F be a field of which alpha, beta are components. Is it possible for (alpha)x(beta)=0, if alpha and beta are both nonzero?

Any ideas? Thanks for your help.

**Drexel28** · February 8th 2010, 11:19 PM

Originally Posted by rockinmoroccan

Let F be a field of which alpha, beta are components. Is it possible for (alpha)x(beta)=0, if alpha and beta are both nonzero?

Any ideas? Thanks for your help.

Isn't every field an integral domain?

**rockinmoroccan** · February 8th 2010, 11:26 PM

yeah...

**Drexel28** · February 8th 2010, 11:27 PM

Originally Posted by rockinmoroccan

yeah...

What's the definition of an integral domain??

**rockinmoroccan** · February 8th 2010, 11:58 PM

well i think one facet of an integral domain was that it has no divisors of zero, meaning there is no solution to the original problem?

**Swlabr** · February 9th 2010, 12:49 AM

Originally Posted by Drexel28

Isn't every field an integral domain?

Surely the point of this question is to prove that it is an integral domain.

To prove this result you should use the fact that every non-zero element has an inverse combined with the fact that zero does not have an inverse.

**Drexel28** · February 9th 2010, 01:57 PM

Originally Posted by rockinmoroccan

Let F be a field of which alpha, beta are components. Is it possible for (alpha)x(beta)=0, if alpha and beta are both nonzero?

Any ideas? Thanks for your help.

Originally Posted by Swlabr

Surely the point of this question is to prove that it is an integral domain.

To prove this result you should use the fact that every non-zero element has an inverse combined with the fact that zero does not have an inverse.

Really? Ok. First note that every field, by definition, is a commutative division ring. And thus, $\alpha\ne 0\implies \alpha^{-1}$ exists. So $\alpha\cdot \beta=0\implies \beta=0\cdot\alpha^{-1}$ . But, it is relatively easy to see that $\alpha^{-1}\cdot 0=\alpha^{-1}\left(0+0\right)=\alpha^{-1}\cdot 0+\alpha^{-1}\cdot 0\implies \alpha^{-1}\cdot 0=0$ and thus $\beta=0$ .

**Swlabr** · February 9th 2010, 11:56 PM

Originally Posted by Drexel28

Really? Ok. First note that every field, by definition, is a commutative division ring. And thus, $\alpha\ne 0\implies \alpha^{-1}$ exists. So $\alpha\cdot \beta=0\implies \beta=0\cdot\alpha^{-1}$ . But, it is relatively easy to see that $\alpha^{-1}\cdot 0=\alpha^{-1}\left(0+0\right)=\alpha^{-1}\cdot 0+\alpha^{-1}\cdot 0\implies \alpha^{-1}\cdot 0=0$ and thus $\beta=0$ .

The hint wasn't directed at you...

**rockinmoroccan** · February 10th 2010, 10:16 PM

thanks for all the help guys

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