MHB Prime Subfiellds - Lovett, Proposition 7.1.3 ....

[Math Amateur]

I am reading Abstract Algebra: Structures and Applications" by Stephen Lovett ...

I am currently focused on Chapter 7: Field Extensions ... ...

I need help with the proof of, or at least some remarks concerning, Proposition 7.1.3 ...Proposition 7.1.3 plus some introductory remarks (proof?) reads as follows:
https://www.physicsforums.com/attachments/6543

In the above text from Lovett we read the following:"... ... However, the multiplication on these elements as defined by distributivity gives this set of elements the structure of $$\mathbb{F}_p = \mathbb{Z} / p \mathbb{Z}$$. ... ... " ... ... BUT ... the subfield contains elements $$0, 1, 2, 3, 4, 5, \ ... \ ... \ (p -1)$$ ... and being a field, it contains divisions of these elements such as $$1/2, 3/5 \ ... \ ... \ ...$$... so how can this subfield be equal to $$\mathbb{Z} / p \mathbb{Z}$$ ... ... ?
Hope someone can help ...

Peter

 

[Evgeny.Makarov]

Are you aware that $\mathbb{Z} / p \mathbb{Z}$ is a field for a prime $p$ and therefore contains multiplicative inverses of all nonzero elements?

 

[Math Amateur]

Are you aware that $\mathbb{Z} / p \mathbb{Z}$ is a field for a prime $p$ and therefore contains multiplicative inverses of all nonzero elements?

Thanks Evgeny ... hmmm ... certainly seems that Lovett assumed I was aware of that ... ...

I was dimly aware ... but uncertain of how and why it all worked out ...

Intend to read material on finite fields in some of my texts ...

Peter

 

[Euge]

Hi Peter,

I would suggest looking back at your earlier posts on finite fields, in particular, one of your featured posts on finite fields.

 

[HallsofIvy]

For example, if p= 5 them F_p has the set {0, 1, 2, 3, 4} with "multiplication" being "multiplication modulo 5". In particular, 2*3= 6= 1 (mod 5) so "1/2" is "3" and "1/3" is "2". 4*4= 16= 1 (mod 5) so 4 is its own multiplicative inverse: "1/4" is "4".

 

[Math Amateur]

Hi Peter,

I would suggest looking back at your earlier posts on finite fields, in particular, one of your featured posts on finite fields.

Yes ... good suggestion Euge ...

On my excursion into finite fields, I obviously did not spend enough time and effort on the topic ...

Peter

- - - Updated - - -

For example, if p= 5 them F_p has the set {0, 1, 2, 3, 4} with "multiplication" being "multiplication modulo 5". In particular, 2*3= 6= 1 (mod 5) so "1/2" is "3" and "1/3" is "2". 4*4= 16= 1 (mod 5) so 4 is its own multiplicative inverse: "1/4" is "4".

Thanks HallsofIvy ... your post was most helpful ...

Peter