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I know this question has been asked before and I understand that it can be proved using the same sort of proof as the one used to show that there's infinite primes, but are there other ways of showing this? Perhaps a counter example?

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  • $\begingroup$ Here: math.stackexchange.com/q/80389/2838 $\endgroup$
    – J126
    Dec 10 '18 at 22:03
  • $\begingroup$ For linking purposes here is a prior thread with Euclid's proof and a variant using the strong divisibility sequence $\,(x^n-1)/(x-1)\ $ $\endgroup$ Dec 10 '18 at 22:07
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One nice approach is to actually count the irreducible polynomials of given degree. For example, see the answers to this question: How many irreducible polynomials of degree $n$ exist over $\mathbb{F}_p$? (this question uses prime fields, but exactly the same arguments and formulas work if $p$ is a power of a prime)

In particular, there is at least one irreducible polynomial of any given degree. Since there are infinitely many degrees, there are infinitely many irreducible polynomials.

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Suppose there are only finitely many irreducible polynomials. Consider the splitting field $K$ of their product, which is finite dimensional over $k$, hence finite.

Suppose $K'$ is an algebraic extension field of $K$; if $b\in K'$, then $b$ is algebraic over $K$, hence also over $k$, so its minimal polynomial over $k$ is irreducible. But in $K$ there is a root of every irreducible polynomial in $k[x]$. Hence $b\in K$.

Therefore $K$ is algebraically closed.

Let $K=\{a_1=0,a_2=1,a_3,\dots,a_n\}$. The polynomial $$ (x-a_1)(x-a_2)(x-a_3)\dots(x-a_n)+1 $$ has no root in $K$.

Contradiction.

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This is a bit of an overkill, and I'm not sure if you can make it non-circular, but here's one way to convince yourself that this is true.

For every finite field extension $K\supseteq F$, there is an irreducible $p\in F[x]$ such that $K$ is isomorphic (over $F$) to $F[x]/p$.

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No, there exist irreducible polynomial in every degree, due to the following formula:

Let $p$ be a prime number, $q=p^n$. In $ \mathbf F_q[X]$, we have the factorisation: $$ X^{q^n}-X=\prod_{d\mid n}\prod_{\:\deg P=d,\\\text{ irreducible}}\mkern-12mu P$$

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So after more reading, I believe I can show it is false by a counter-example.

Consider f(x) = xn + 3 ∈ ℤ5[x] with n ∈ ℕ. Then Eisenstein's criterion (with p = 3) says that f(x) is irreducible in ℚ[x] and therefore, f(x) is irreducible in ℤ5[x] since ℤ5[x] ⊂ ℚ[x]. This gives infinitely many irreducible polynomials in ℤ5[x] because there's infinite possibilities for the value of the exponent n.

Therefore, the initial statement is false.

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