# 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

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.

• There are six bullet points in your question. Which of them have you tried to establish? Which are you having trouble with? Show us what you've done so far and perhaps we can help. – Ethan Bolker Mar 3 '17 at 0:14
• @ethan-bolker I have just started out with proofs and am trying to grasp the concept as of yet. – YFP Mar 3 '17 at 0:21
• Have you shown yet that it is a ring? – Dean Young Mar 3 '17 at 0:33
• This is much too big a question to ask all at once. I suggest you delete this question. Then try to prove that addition modulo $n$ is commutative. Post your proof here in a new question and ask us to check it for you.. – Ethan Bolker Mar 3 '17 at 0:34
• @EthanBolker I can only make sense of the fact that certain collections of real number are also fields by manipulating using the above properties. Other than that I have no clue where to begin whatsoever. This is one of the challenging questions marked in my text book and is more than likely to show up on the midterm tomorrow.. – YFP Mar 3 '17 at 0:51

(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).

(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: $$xy\equiv xy'\equiv 1\implies$$ $$\implies y\equiv y(1)\equiv y(xy')\equiv (yx)y'\equiv (1)y'\equiv y'.$$