I can intuitively understand that there are infinitely many irreducible polynomials in $\Bbb{F}_p[X]$, where $p$ is a prime number, but I'm having a hard time actually proving it. What proof strategy should I take for this?
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3Remember Euclid. – Karanko Apr 06 '15 at 05:25
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I was thinking about his proof of infinitely many primes... – Patrick Shambayati Apr 06 '15 at 05:26
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... and what happened? – Karanko Apr 06 '15 at 05:27
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Suppose all primes are on this list: $p_1,...,p_r$. Then $p_1p_2\cdots p_r+1$ is also prime thus there are infinitely many – Patrick Shambayati Apr 06 '15 at 05:27
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I couldn't apply it to this! I got lost in abstraction – Patrick Shambayati Apr 06 '15 at 05:27
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By the way, $p_1p_2...p_r+1$ is not necessarily prime. What the proof shows is that it is not divisible by any $p_1,p_2,...,p_r$. But it must be divisible by some prime. – Karanko Apr 06 '15 at 05:28
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I tried multiplying together two irreducible polynomials and adding 1 to the result, then trying to find a contradiction in the attempt to factor that result. Using the polynomial division algorithm, in specific cases I can get the desired result... but I don't know how to extend it to the general case – Patrick Shambayati Apr 06 '15 at 05:29
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Oh yes. I remember. So it may or may not be divisible by a prime strictly less that $p_1\cdots p_r+1$? – Patrick Shambayati Apr 06 '15 at 05:29
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The only abstraction needed is to replace $p_i$ by $P_i(x)$. But you need to remember well the proof for the integer primes. – Karanko Apr 06 '15 at 05:30
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irreducible polynomials $\equiv$ prime numbers ! – Fardad Pouran Apr 06 '15 at 05:50
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How does this proof sound? – Patrick Shambayati Apr 06 '15 at 06:54
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Suppose there are finitely many irreducible polynomials in $\Bbb{F}_p[X]$, and let them be the monic polynomials $\pi_1,\pi_2,\cdots,\pi_r$. Clearly, $\pi_1\pi_2\cdots\pi_r+1$ is not equal to the elements of the list $\pi_1,\pi_2,\cdots,\pi_r$ because $\deg(\pi_1\pi_2\cdots\pi_r+1)>\deg(\pi_i)$ for all $i\in[1,r]$. Then $\pi_1\pi_2\cdots\pi_r+1$ cannot be an irreducible element because it is not in that set. This is a contradiction! (continued....) – Patrick Shambayati Apr 06 '15 at 06:54
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1(continued: ) Every element $\pi_i\in {\pi_1,\cdots,\pi_r}$ has remainder 1 after division with $\pi_1\pi_2\cdots\pi_r+1$. Thus, $\pi_1\pi_2\cdots\pi_r+1$ is also irreducible. Therefore, there are infinitely many irreducible elements in $\Bbb{F}_p[X]$. – Patrick Shambayati Apr 06 '15 at 06:54
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Assume that there are finitely many (monic) irreducible polynomials in $\mathbf F_{p}[X]$,$p_{1}(x), p_{2}(x), \dots, p_{n}(x)$. Consider $f(x)=p_{1}(x)p_{2}(x) \cdots p_{n}(x)+1$. Now $f(x)$ is not divisible by any irreducible, and hence is irreducible and is not in the list of irreducibles. Contradiction.
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@AndréNicolas Why? It's known that any reducible polynomial is divisibile by an irreducible polynomial; since $f$ isn't, it must be irreducible. After all, this is a proof by contradiction, isn't it? I would make it into a direct proof, and in this case your remark would apply. – egreg Apr 06 '15 at 14:04
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Mimic Euclid: $\, f_{n+1}\! = 1+f_1\cdots f_n\,$ is an infinite sequence of coprimes $\,f_i.\,$ Choosing $\,g_i\,$ to be an irreducible factor of $\,f_i\,$ yields an infinite sequence of coprime (so nonassociate) irreducibles.
Alternatively, recall that the sequence of polynomials $\rm\:f_n = (x^n\!-\!1)/(x\!-\!1)\:$ is a strong divisibility sequence, i.e. $\rm\:(f_m,f_n) = f_{(m,n)}$ in $\rm\mathbb Z[x].\:$ Hence the subsequence with prime indices yields an infinite sequence of pairwise coprime polynomials. Further the linked proof shows the gcd has linear (Bezout) form $\rm\:(f_m,f_n) = f_{(m,n)}\! = g\, f_m + h\, f_n,\,$ $\rm\, g,h\in\mathbb Z[x],\:$ so said coprimality persists in every ring $\,R[x].\,$ Thus, for each prime $\rm\:q,\:$ choosing a factor of $\rm\:f_q\:$ irreducible in $R[x]$ yields infinitely many irreducible polynomials, pariwise nonassociate (being pairwise coprime).
Bill Dubuque
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I arrived here since somebody asked the same question.
Other than mimicking Euclid's proof on the existence of infinitely many primes we can proceed as follows to show the existence of infinitely many irreducible monic polynomials in $\mathbb{F}[X]$ where $\mathbb{F}$ is any field.
If $\mathbb{F}$ is infinite the infinitely many polynomials $$X-\alpha,\qquad\alpha\in\mathbb{F}$$ are irreducible.
If $\mathbb{F}=\mathbb{F}_q$ is finite $q=p^r$ and $p\neq2$ the cyclic group $\mathbb{F}_q^\times$ has even order and thus there exists
$\beta\in\mathbb{F}_q$ which is not a square. This implies that the polynomial $X^2-\beta$ is irreducible. Therefore $\mathbb{F}_q$ is not algebraically closed and its algebraic closure $\overline{\mathbb{F}}$ must be infinite. Since every element $x\in\overline{\mathbb{F}}$ has only finitely many conjugates there are infinitely many irreducible polynomials in $\mathbb{F}_q[X]$ obtained as minimal polynomials of elements in $\overline{\mathbb{F}}$.If $p=2$ proceed as above using the polynomial $X^2+X+1$ if $q=2$ and a polynomial of the form $X^d-\beta$ if $q>2$ and $d$ divides $q-1$ where $\beta$ is not a $d$-th power in $\mathbb{F}_q$.
Andrea Mori
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