Finite Integral Domains .... Adkins & Weintraub, Proposition 1.5 ....

[Math Amateur]

I am reading "Algebra: An Approach via Module Theory" by William A. Adkins and Steven H. Weintraub ...

I am currently focused on Chapter 2: Rings ...

I need help with an aspect of the proof of Proposition 1.5 ... ...

Proposition 1.5 and its proof read as follows:https://www.physicsforums.com/attachments/7924
At the end of the above proof from Adkins and Weintraub we read the following:

" ... ... and hence \(\displaystyle \phi_a (R) = R\). In particular, the equation \(\displaystyle ax = 1\) is solvable for every \(\displaystyle a \neq 0\) and \(\displaystyle R\) is a field. ... ... "
Can someone please explain to me how the conclusion that "the equation \(\displaystyle ax = 1\) is solvable for every \(\displaystyle a \neq 0\) and \(\displaystyle R\) is a field" follows from the arguments preceding it ...

Basically I do not understand how the arguments before this statement lead to the conclusion ...Help will be much appreciated ...

Peter

 

[castor28]

Hi Peter,

Assume that $R$ contains $n$ elements. If $R$ is an integral domain and $a\ne0$, the $n$ elements $ax_1,\ldots,ax_n$ are all distinct.

As $|R|=n$, these elements are just the elements of $R$ in a different order; in particular, one of these elements is equal to $1$.

Now, $ax_i=1$ means that $x_i$ is a multiplicative inverse of $a$; as such an inverse exists for each $a\ne0$, $R$ is a field.

 

[Math Amateur]

Hi Peter,

Assume that $R$ contains $n$ elements. If $R$ is an integral domain and $a\ne0$, the $n$ elements $ax_1,\ldots,ax_n$ are all distinct.

As $|R|=n$, these elements are just the elements of $R$ in a different order; in particular, one of these elements is equal to $1$.

Now, $ax_i=1$ means that $x_i$ is a multiplicative inverse of $a$; as such an inverse exists for each $a\ne0$, $R$ is a field.

Thanks castor28 ...

Most helpful ...

Peter