Why $\mathbb{Z}_p$ can't have proper subfields?

Question

From the notes I'm studying from I read that " $\mathbb{Z}_p=\mathbb{F}_p$ has no proper subfield."

The rationale is: "assuming $\mathbb{K}$ is a subfield of a finite field $\mathbb{Z}_p= \mathbb{F}_p$, $p$ prime, then $\mathbb{K}$ must contain 0, and 1, and so all other elements of $\mathbb{F}_p$ by the closure of $\mathbb{K}$ under addition. Therefore, it follows that $\mathbb{Z}_p=\mathbb{F}_p$ contains no proper subfield"

But thinking about $\mathbb{Z}_5 = \{0,1,2,3,4\}$ and $\mathbb{Z}_3 = \{0,1,2\}$

Isn't $\mathbb{Z}_3$ a proper subfield of $\mathbb{Z}_5$ since it contains 0, 1 and is closed under addition and multiplication?

Answer 1

It’s a proper subset, but it’s not a subfield at all: for example, $2^2=1$ in $\Bbb F_3$, and $2^2=4\ne 1$ in $\Bbb F_5$, so they don’t have the same operations. Similarly, $1+2=0$ in $\Bbb F_3$, but $1+2=3\ne 0$ in $\Bbb F_5$.

Answer 2

Any proper subfield of $\;\Bbb Z_p\;$ would have to be, in particular, a sugbroup of its additive group. But this is a group of order a prime $\;p\;$ , and as such it has no proper divisors, so by Lagrange's Theorem that's impossible.