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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.

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    $\begingroup$ $p$ would be a prime, otherwise ${\Bbb Z}/p{\Bbb Z}$ is not a field. $\endgroup$ – David Feb 26 '15 at 5:26
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    $\begingroup$ p is always a prime in number theory. $\endgroup$ – user4894 Feb 26 '15 at 5:33
  • $\begingroup$ Since you can have polynomials of every degree F[x] must be infinite and F^n is finite so F[x] can't be a subset of F^n. $\endgroup$ – Vicfred Feb 27 '15 at 2:20
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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.

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  • $\begingroup$ Would this be sufficient as proof then? For suppose $|F[x]| = n$, for some $n \in \mathbb{Z}$ and consider an element $p(x) \in F[x]$ that has at least one term with coefficient $n$. Then, since $F[x]$ is a vector space, it must be closed under vector addition. So adding $p(x) + p(x)$ would produce an element of $F[x]$ with at least one coefficient $2n$ which means $p(x) + p(x) \notin F[x]$ which would contradict the closure of F[x]. $\endgroup$ – letsmakemuffinstogether Feb 26 '15 at 6:05

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