I am struggling to understand what the following question is asking me to prove. In particular, I am not understanding whether the "p" in $\mathbb{Z}/p\mathbb{Z}$ is a prime or something else altogether.
Question: Let $F = \mathbb{Z}/p\mathbb{Z}$ be a finite field and define the set $F[x]$ to consist of polynomials in the variable $x$ with coefficients taken from $F$. Show that $F[x]$ contains infinitely many elements (and therefore that $F[x]$ is a vector space that is not a subspace of $F^{n}$ for any $n$).
Thank you in advance for any help you can provide.
Since it is saying $F$ is a finite field, $p$ has to be a prime. This is the only way to form a finite field via the quotient $\mathbb{Z}/n\mathbb{Z}$ for an integer $n$. Now note the ring $F[x]$ is defined as the set $$F[x]=\{\sum_{i=0}^na_ix^i : a_i\in F, n\geq 0\}$$ or in other words, they are simply polynomials of degree $n$ with coefficients in our field, where $n$ can be any arbitrary nonegative integer.
Now, the dimension of a vectorspace $F^n$, i.e. the sets of $n$-tuples with entries from $F$, has dimension $n$. It has a basis $(1,0,0,...,0), (0,1,0,...,0),..., (0,0,0...,1)$. Since there are finitely many elements in $F$, there are only finitely many linear combinations of this basis. If you show that $F[x]$ has infinitely many elements (as you can by showing $a_ix^n\in F[x]$ for every $n\in \mathbb{N}$ by definition), it can not be any subset of $F^n$. If you do not believe this, how could you have an infinite subset of a finite set?
Hopefully this clears some things up.