Prove that ℤ/nℤ is a field- prove closed under addition

Question

I have been asked to prove the following:

Let $\mathbb{F}_n$ = {[0],[1],...,[n-1]} be an $n$ point set, where $n$ is a positive integer. We impose addition and multiplication as follows:

$$[a] + [b] = [a + b], [a] * [b] = [ab]$$

where $[x] = [y]$ iff $n$ divides $(x-y)$.

i) Prove that $\mathbb{F}_n$ is a field when $n$ is a prime number.

Right now, I am working on proving that addition is closed. I am kind of uncomfortable with my work so far and any guidance or verification would be greatly appreciated.

Here is what I have so far:

Closed under addition:

Let $a = nx, b = ny$ where $a,b \in n\mathbb{Z}, x, y \in \mathbb{Z}$ and $n$ is prime.

$$[a] + [b]$$

$$= [nx] + [ny]$$

$$= [nx + ny]$$

$$=[n(x+y)]$$

$$n(x+y) = z \ \bmod n,$$ by definition of $[n(x+y)].$

$$z = nj + n(x+y),$$ by division algorithm.

$$z = n(j+x+y)$$

By definition, $n\mid z$, so $[a + b] \in n\mathbb{Z}$

Answer 1

It looks like you're confused about what set you're talking about. The set you want to show is closed under addition is $\mathbb{F}_n$, not $n\mathbb{Z}$. So you want to start with arbitrary $[a],[b]\in\mathbb{F}_n$ (so, according to your definition of $\mathbb{F}_n$, $a$ and $b$ are integers between $0$ and $n-1$ inclusive). Then you want to show $[a]+[b]=[a+b]$ is also an element of $\mathbb{F}_n$. Again, according to your definition of $\mathbb{F}_n$, this means you need to show there is some integer $c$ between $0$ and $n-1$ such that $[c]=[a+b]$.

(Incidentally, you should probably also show that addition is well-defined: in other words, that there is only one $c$ between $0$ and $n-1$ such that $[c]=[a+b]$. Otherwise, the given definition of "$[a]+[b]$" may not describe a unique element of $\mathbb{F}_n$. That may be something you are allowed to assume as known, though. In fact, you may be allowed to assume as known that addition and multiplication are binary operations on the set $\mathbb{F}_n$, so you do not have to prove closure or well-definedness at all.)

Answer 2

For every two residue classes, $\overline{a},\overline{b}\in\mathbb{Z}/n\mathbb{Z}$, we can pick representatives in $\mathbb{Z}$, say $a$ and $b$. Since $\mathbb{Z}$ is closed under addition, $a+b$ is an element of $\mathbb{Z}$. Since $\mathbb{Z}/n\mathbb{Z}$ is literally what we get when partitioning $\mathbb{Z}$ using the equivalence relation $$a\sim b\iff a\equiv b\pmod n,$$ we know that $a+b$ is an element of exacly one residue class, so $\overline{a+b}$ is a well-defined element of $\mathbb{Z}/n\mathbb{Z}$.

---

Of course this assumes you've already proven that addition $$\overline{a}+\overline{b}=\overline{a+b}$$ is well-defined.

Answer 3

Say $(a+b)\cong z \pmod n$... Now say $a\cong x \pmod n$ and $b \cong y \pmod n$.

It is easy to see $x+y \cong z \pmod n$.

Namely, $\exists k,l,m \in \mathbb Z: (a+b)=z+k\cdot n, a=x+l\cdot n \text { and } b=y+m\cdot n$...

Now $a+b=(x+y)+(l+m)\cdot n \implies z+k\cdot n=(x+y)+ (l+m)\cdot n \implies z-(x+y)=(l+m-k)\cdot n$... Of course, $l+m-k\in \mathbb Z$... So $x+y\cong z \pmod n$...