[Math] Prove that for any prime $p$, the set $\mathbb{Z}_p$ with the addition mod $p$ and multiplication mod $p$, and congruence mod $p$, is a field

abstract-algebrafield-theoryproof-explanationproof-writing

So this is all that we are given:

  • F is closed under addition and multiplication, i.e. $\forall a, b \in F,$ we have $a + b$ and $a × b$ are also both in F.
  • commutitivity: $a + b = b + a$ and $a × b = b × a$
  • associativity: $a + (b + c) = (a + b) + c$ and $a × (b × c) = (a × b) × c$
  • unique neutral element property: there are neutral elements 0 and 1 for addition and multiplication.
  • Unique inverse element for each given element: and element a of F has unique additive inverse and
    a unique multiplicative inverse.
  • distributivity of multiplication over addition: $a × (b + c) = a × b + a × c$.

How do we prove that for a given number $n>1$ and defining $\mathbb{Z}_p= \{0, 1, 2, . . . , n − 1\}$, for any prime $p$, the set $\mathbb{Z}_p$ with
the addition mod p and multiplication mod p, and congruence mod p, is a field.

I'd really appreciate it if someone could help me with this proof.

Best Answer

(1). The additive and multiplicative identities are (obviously) $0$ and $1$ and are (obviously) unique. And $0\not \equiv 1\pmod p$. (That is necessary to say, because in a field the additive and multiplicative identities are not allowed to be equal).

(2). Additive inverses:

(2A). Existence: If $0< x<p$ then $0<p-x<p$ with $x+(p-x)\equiv 0.$ And $0+0\equiv 0.$

(2B). Uniqueness: If $x+y\equiv x+y'\equiv 0$ then $y-y'\equiv 0.$ but if $0\leq y<p$ and $0\leq y'<p$ then $|y-y'|<p.$ And if $p$ divides $|y-y'|$ with $0\leq |y-y'|<p$ then $y-y'=0 .$

(3). Multiplicative inverses:

(3A). Existence: For $1\leq x<p$ let $M=\min \{z:0<z\land \exists y\;(xy\equiv z \pmod p\}.$

First, by considering the case $y=1$ we have $M\leq x<p.$

Second, we will show that if $xy\equiv k$ with $2\leq k<p$ then there exists $y'$ and $k'$ with $0< k'<k$ and $xy'\equiv k'.$ From this we will conclude that if $2\leq k<p$ then $k\ne M.$

Thirdly, from the first and second parts above, we will have $M=1$, so $xy\equiv 1$ for some $y$.

To prove the second part: Suppose $xy\equiv k$ with $2\leq k<p.$ Let $m=\max \{m'>0: m'k<p\}.$ Then $mk<p\leq (m+1)k.$ But $p$ is prime and $m+1>1<k$ so we cannot have $(m+1)k=p.$ So $mk<p<(m+1)k.$

Let $y'=(m+1)y$ and $k'=(m+1)k-p.$ We have $0<k'<k'-mk=k.$ Now $$xy'=xy(m+1)\equiv k(m+1)\equiv k(m+1)-p=k'.$$

(3B). Uniqueness of multiplicative inverse: $$xy\equiv xy'\equiv 1\implies$$ $$\implies y\equiv y(1)\equiv y(xy')\equiv (yx)y'\equiv (1)y'\equiv y'.$$

Related Question