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Pinter's textbook "A book of abstract algebra" asks to prove the following:

There are infinitely many irreducible polynomials in $\mathbb{Z}_{5} [x]$.

Here's my attempt to prove it any field $\mathbb{F}$:

Let $F$ be a field. Suppose that there are finitely many irreducible polynomials in $F[x]$, say, $r_1 , r_2 , \ldots , r_n $. Then $r_1 r_2 \ldots r_n +1 \in F[x]$. Hence, by the factorisation into irreducible polynomial theorem tells us that $$r_1 r_2 \ldots r_n +1 = k( r_1 r_2 \ldots r_n)$$

for some $k \in \mathbb{F}$. But then $r_1 r_2 \ldots r_n$ would be an associate of $1$. Thus each of $r_i$ must be a constant polynomial which contradicts the definition of a irreducible polynomial.

But I suppose this is not what the author wanted. I was wondering if there's any way to generate such irreducible polynomials over $\mathbb{Z}_{5}$. It looks to me as if $x^n+1$ where $n$ is even is a irreducible polynomial but I couldn't prove it. Any hints?

ashK
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    $(2)^2 + 1 = 4 + 1 = 5 = 0$; also if $n = 5k$ then $x^n +1 = x^{5k} + 1^5 = (x^k+1)^5$ – dcolazin Mar 29 '19 at 07:35
  • Also, why $r_1 \ldots r_n + 1 = k(r_1 \ldots r_n)$? Doesn't the theorem say $r_1 \ldots r_n + 1 = k(r_1^{e_1} \ldots r_n^{e_n})$? – dcolazin Mar 29 '19 at 10:20
  • Also worth noting is that, since $\mathbb{Z}5$ is a field, and hence a UFD, then $\mathbb{Z}_5[x]$ is a UFD and thus 'irreducible' and 'prime' are equivalent. So, we could also just use Euclid's classical proof (briefly): Assume $p_1,p_2,..., p_m$ completes a list of primes, and consider $p=\prod{i=1}^m p_i$. Either $p+1$ is prime, thus giving one more not on the list, or $p+1$ has a prime factor $q$. Either $q$ is on our list, or is again a new prime. If $q$ is on our list, then $q|p$ and $q|p+1 \Rightarrow q|p-(p+1)=1$ which is a contradiction. Thus $k[x]$ has infinitely many irreducibles. – Christopher.L Mar 29 '19 at 10:56
  • @dcolazin that's right. i guess the proof needs to be fixed. – ashK Mar 29 '19 at 12:35
  • @Christopher.L i had kept in mind while proving this – ashK Mar 29 '19 at 12:35

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Suppose that there are finitely many irreducible polynomials $p_1, \ldots, p_n$ over a finite field $\mathbb{F}$. Because every polynomial has finite solutions in the algebraic closure, then all the solutions of $p_1, \ldots, p_n$ are $a_1, \ldots, a_k \in \bar{\mathbb{F}}$. Then in $\mathbb{F}[a_1, \ldots, a_k]$ all the polynomials split (if $p \in \mathbb{F}[a_1, \ldots, a_k][x]$ is irreducible of degree $\geq 2$, call $b \in \bar{\mathbb{F}}$ a solution of $p$, then the minimal polynomial in $\mathbb{F}$ of $b$ is irreducible, so $b \in \mathbb{F}[a_1, \ldots, a_k]$, which is absurd). Then $\mathbb{F}[a_1, \ldots, a_k]$ is a finite field and is algebraically closed, which is absurd.

dcolazin
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