# I Fermat's Little Theorem ... Anderson and Feil, Theorem 8.7 .

1. Feb 26, 2017

### Math Amateur

I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 8: Integral Domains and Fields ...

I need some help with an aspect of the proof of Theorem 8.7 (Fermat's Little Theorem) ...

Theorem 8.7 and its proof read as follows:

My questions regarding the above are as follows:

Question 1

In the above text from Anderson and Feil we read the following:

" ... ... Because a field has no zero divisors, each element of $S$ is non-zero ... "

Can someone please demonstrate exactly why this follows ... ?

Question 2

In the above text from Anderson and Feil we read the following:

" ... ... Because a field satisfies multiplicative cancellation, no two of these elements are the same .. ... "

Can someone please demonstrate exactly why it follows that no two of the elements of $S$ are the same .. ... "

Help will be appreciated ...

Peter

*** EDIT ***

oh dear ... can see that the answer to Question 1 is obvious ... indeed it follows from the definition of zero divisor .... apologies ... brain not in gear ...

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2. Feb 26, 2017

### Staff: Mentor

It's also obvious. If we assumed $[x\cdot n]=[x\cdot m]$ then what would this mean? You can use all field operations and again the lack of zero divisors. In general, the question to be answered is: why do fields allow multiplicative cancellations? The reason doesn't require this special field, so you may as well prove: $a\cdot m = b \cdot m \Longrightarrow a = b$ for $a,b \neq 0$ of course.

3. Feb 28, 2017

### Math Amateur

Thanks for the help fresh_42 ... grateful for your help ...

If $[ x \cdot n ] = [ x \cdot m ]$ then by cancellation in a field, we have n = m which cannot be the case for the members of the set S ...

Cancellation works in fields since every element of a field has an inverse ...

... so if $a \cdot m = b \cdot m$ then we can post-multiply by $m^{-1}$ and get $a = b$ ...

Hope that's right ...

Peter

4. Mar 1, 2017

### Staff: Mentor

Well, partially. You cannot rule out the case $m=0$ at prior and operate with $m^{-1}$.

Therefore wait as long as you can with additional assumptions.

This is more of a general advice. Instead operate without (multiplicative) inverse elements.