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Let $F$ be a field and $n \geq 2$. Must there exist an irreducible polynomial of degree $n$ in $F[X]$?

When $F=\mathbb{Q}$ the answer is certainly "yes," as you can apply Eisenstein's criterion to $X^n-2$, for example. When $F = \mathbb{Z}/p\mathbb{Z}$ it is well known that the answer is "yes" (there have been numerous StackExchange posts about that), although it takes more work to get there (and especially to actually construct an example).

On the other hand, if $F=\mathbb{R}$ then there do not exist irreducible polynomials of odd degree, and if $F$ is an algebraically closed field like $\mathbb{C}$, the situation is even worse--only polynomials of degree 1 are irreducible. (Thanks to Mindlack and Servaes and lhf for helping me remember this, and to PJTraill for suggesting the revised and more interesting question title.)

What about other $F$? When do irreducibles exist of every degree?

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    $\begingroup$ What about $F=\mathbb{R}$ or $F=\mathbb{C}$? $\endgroup$ – Mindlack Jan 10 at 17:25
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    $\begingroup$ Perhaps you mean ‘Over which fields are there irreducible polynomials of every degree?’, a slightly different question. $\endgroup$ – PJTraill Jan 10 at 19:57
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If $F$ is algebraically closed then every polynomial with coefficients in $F$ splits into a product of linear factors over $F$. So there do not exist any irreducible polynomials of degree $n>1$.

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There are no irreducible cubic polynomials over the real numbers. Nor of any odd degree greater than 1. This follows from the intermediate value theorem.

In fact, the fundamental theorem of algebra says that the only irreducible polynomials over the real numbers are those of degree 1 and those of degree 2 with negative discriminant.

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