Working with finite fields of form Z_p

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Homework Help Overview

The discussion revolves around properties of finite fields, specifically Z_p, where p is an odd prime. The original poster attempts to prove that not every element of Z_p is a square of some element in Z_p, leading to various explorations of congruences and properties of squares in modular arithmetic.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of treating congruences as equalities and question the validity of dividing by p in Z_p. They explore specific cases in Z_3 and Z_5 to identify which elements are squares and which are not.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on examining specific finite fields and suggesting the use of counting logic to analyze the number of square roots available in Z_p. There is a recognition of the need to clarify understanding of the properties of Z_p.

Contextual Notes

There is an ongoing discussion about the constraints of working within Z_p, particularly regarding the treatment of zero and the implications of dividing by p. Participants are also considering the limitations of square roots in the context of finite fields.

Syrus
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Homework Statement



Let p be an odd prime. Then Char(Z_p) is nonzero.

Prove: Not every element of Z_p is the square of some element in Z_p.

Homework Equations


The Attempt at a Solution



I first did this, but i was informed by a peer that it was incorrect because I was treating the congruency as an equality:

Suppose not. Then every element of Z_p is the square of some element in Z_p. Take 1. Since in mod p: 1 = (p-1)2 = 12, it follows that 0 = p2 - 2p, and hence p = 2, a contradiction.

What can i do to find a correct way of proving this?
 
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In a finite field, the group of units is cyclic. This gives you a quick way of proving the result.
 
It's probably even easier if you simply observe that if i^2=k then (-i)^2=k. But since your first proof is so bogus, you probably really don't understand what Z_p is. Try doing a warmup and figuring out what elements of Z_3, Z_5 and Z_7 are squares and which aren't.
 
Syrus said:
Suppose not. Then every element of Z_p is the square of some element in Z_p. Take 1. Since in mod p: 1 = (p-1)2 = 12, it follows that 0 = p2 - 2p, and hence p = 2, a contradiction.

Are you allowed to divide by p in Zp?

For example suppose we're in Z5. Now

5 * 2 = 10 = 0 (mod 5)

and 5 * 3 = 15 = 0 (mod 5).

So since 5*2 = 5*3 we conclude that 2 = 3.

What's wrong with that proof?
 
Yes, I suppose that's silly... dividing by p (which is congruent to zero). I have done what dick suggested and discovered that for Z_3, 2 is not a square. For Z_5, neither 2 nor three are squares. I am trying to work off this for now but can't seem to generalize these cases.
 
Last edited:
Syrus said:
Yes, I suppose that's silly... dividing by p (which is congruent to zero). I have done what dick suggested and discovered that for Z_3, 2 is not a square. For Z_5, neither 2 nor three are squares. I am trying to work off this for now but can't seem to generalize these cases.

Ok, I think you got what Z_p means. Now use counting logic. If every element of Z_p were a square then for every element z of Z_p there is an element k such that k^2=z. If 1^2=1 and (-1)^2=(p-1)^2=1 do you have enough elements of Z_p to give each element a square root?
 

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