整数mod $p$ 形成场

The group of integers modulo $p$ is denoted $\mathbb{Z}/p\mathbb{Z}$. This notation is a little bit clunky, but there is a mathematically sound reason for it! Here, we will prove that $\mathbb{Z}/p\mathbb{Z}$ is not just a group but also a field.

To accomplish this, we will first show that $\mathbb{Z}/p\mathbb{Z}$ is a ring, which is more specific than a group but still more general than a field. A ring is basically an algebraic structure that takes the rules of addition and multiplication and abstracts them to mathematical objects in general, not just numbers. Basically, we need to show $3$ things:

$\mathbb{Z}/p\mathbb{Z}$ is an abelian group under addition: This is trivial, since the operation of addition is already known to be commutative and associative.

$\mathbb{Z}/p\mathbb{Z}$ is associative and has an identity element under multiplication: This is also easy, because the operation of multiplication is known to be associative. Also, $a\cdot a^{-1}=a^{-1}\cdot a=1$, meaning that $1$ is the identity of $\mathbb{Z}/p\mathbb{Z}$ under multiplication.

Multiplication in $\mathbb{Z}/p\mathbb{Z}$ is distributive with respect to addition: Written symbolically, this means that: $$a\cdot (b+c)=a\cdot b + a\cdot c$$ Once again, this follows directly from the fact that the standard operations of multiplication and addition, as we normally think of them (multiplication and addition of numbers), already satisfy the distributive property. Since $$a\cdot (b+c)=a\cdot b + a\cdot c$$ it is obvious that $$a\cdot (b+c)\equiv a\cdot b + a\cdot c\pmod{p}$$ Thus, multiplication is distributive with respect to addition.

From all of this, it follows that $\mathbb{Z}/p\mathbb{Z}$ is a ring.

Now that we have proved that $\mathbb{Z}/p\mathbb{Z}$ is a ring, we will prove that it is a field. A field is a ring whose elements other than the identity form an abelian group under multiplication. In this case, the identity element of $\mathbb{Z}/p\mathbb{Z}$ is $0$. In fact, the group of nonzero integers modulo $p$ under multiplication has a special notation: $(\mathbb{Z}/p\mathbb{Z})^{\times}$.

Consider any element $a\in (\mathbb{Z}/p\mathbb{Z})^{\times}$.

First, we know that $1\cdot a=a\cdot 1=a$. Thus, $(\mathbb{Z}/p\mathbb{Z})^{\times}$ has an identity element, namely $1$.

Next, we know that every integer that is not a multiple of $p$ has a unique inverse modulo $p$ (a proof can be found here). Therefore, for any element $a\in (\mathbb{Z}/p\mathbb{Z})^{\times}$, we can find an element $a^{-1}$ such that $a\cdot a^{-1}=a^{-1}\cdot a=1$, meaning that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ satisfies the inverse property.

Finally, since multiplication is both commutative and associative, we know that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ satisfies the necessary associativity property, and we know that it must be abelian.

Therefore, $(\mathbb{Z}/p\mathbb{Z})^{\times}$ must be an abelian group. We conclude that $\mathbb{Z}/p\mathbb{Z}$ is a field.