Distinct subrings of cardinality $p$ of the field $\mathbb F_{p^2} $ This particular question was asked in masters entrance of a university for which I am preparing.

Question Let $p$ be a prime number. How many distinct subrings ( with unity) of cardinality $p$ does the field $\mathbb F_{p^2}$ have?


*

*$0$


*$1$


*$p$


*$p^{2} $
Unfortunately, I am unable to even start this question as such kind of questions were not done during Abstract Algebra course. So, I am unable to show any attempt.
Also, this question was asked before here:
Number of subrings of a Field
But I had some questions in proof of Answer, so I asked user who answered to supply the proof. But as he didn't supplied the proof, so I have no option other than asking it here.

Question in the answer of User who answered the question in link are :
1.why subring will always be a vector space over prime field?

2.Can you please prove it. Also, how does dimension of $F_{p^2}$ over prime field must be 2.

Also if someone wants to answer with a different approach, that also would be really appreciated.

But kindly  give a proof of a result which you are assuming.
 A: A subring contains $1$, so it contains $2$, $3, \ldots, p-1$ so all of the subfield $\Bbb F_p$. As this subfield is a ring and has $p$ elements, that's all the subring can be. So the second answer $1$ is correct: the unique subring of $p$ elements is $\Bbb F_p$, the prime subfield.
A: Alright so let's take a unital subring $R$ of $\mathbb F_{p^2}$. It's unital so it contains $1$. It's $1$ so it contains $1 + 1$, $1 + 1 + 1$, ..., $p - 1$. These are $p$ elements, so since any subring must contain these, a subring of order $p$ must as well. Thus order $p$ subring must equal exactly $\{1, \dots, p\} = \mathbb F_{p} \subseteq \mathbb F_{p}$ so there is only one such subring.
A: Let $\mathbb{F}_p$ be the prime field of characteristic $p$. Then $\mathbb{F}_{p^2}$ contains $\mathbb{F}_p$. Define the operation $$\cdot:\mathbb{F}_p\times \mathbb{F}_{p^2}\to \mathbb{F}_{p^2}$$ given by
$$(r,s)\mapsto r\cdot s=rs.$$
Then $\cdot$ makes $\mathbb{F}_{p^2}$ a vector space over $\mathbb{F}_p$, which can be proven by restricting the ring axioms of $\mathbb{F}_{p^2}$ when one of the elements lies in $\mathbb{F}_p$.
Also, a vector space of dimension $d$ over $\mathbb{F}_p$ has exactly $p^d$ elements (Why?). Since $\mathbb{F}_{p^2}$ is the field with $p^2$ elements, then $\dim_{\mathbb{F}_p}\mathbb{F}_{p^2}=2$.
Now, if $S$ is a subring of $\mathbb{F}_{p^2}$, then $1\in S$, therefore $\mathbb{F}_p\subseteq S$ (Since $\mathbb{F}_p$ is just a cyclic group additively). The same argument which makes $\mathbb{F}_{p^2}$ a vector space over $\mathbb{F}_p$ now makes $S$ a vector subspace. Now, $S$ is a vector subspace of $\mathbb{F}_{p^2}$ which contains $\mathbb{F}_p$, so if $S$ has $p$ elements then $S$ must be $\mathbb{F}_p$.
