# Thread: Basic Field Theory Question

1. ## Basic Field Theory Question

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.

2. 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?

3. yeah...

4. Originally Posted by rockinmoroccan
yeah...
What's the definition of an integral domain??

5. 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?

6. 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.

7. 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, $\displaystyle \alpha\ne 0\implies \alpha^{-1}$ exists. So $\displaystyle \alpha\cdot \beta=0\implies \beta=0\cdot\alpha^{-1}$. But, it is relatively easy to see that $\displaystyle \alpha^{-1}\cdot 0=\alpha^{-1}\left(0+0\right)=\alpha^{-1}\cdot 0+\alpha^{-1}\cdot 0\implies \alpha^{-1}\cdot 0=0$ and thus $\displaystyle \beta=0$.

8. Originally Posted by Drexel28
Really? Ok. First note that every field, by definition, is a commutative division ring. And thus, $\displaystyle \alpha\ne 0\implies \alpha^{-1}$ exists. So $\displaystyle \alpha\cdot \beta=0\implies \beta=0\cdot\alpha^{-1}$. But, it is relatively easy to see that $\displaystyle \alpha^{-1}\cdot 0=\alpha^{-1}\left(0+0\right)=\alpha^{-1}\cdot 0+\alpha^{-1}\cdot 0\implies \alpha^{-1}\cdot 0=0$ and thus $\displaystyle \beta=0$.
The hint wasn't directed at you...

9. thanks for all the help guys