We all know that all the irreducible polynomials in $\mathbb{C}[x]$ are linear and in $\mathbb{R}[x]$ they aren't more than 2 degree. However,in $\mathbb{Q}[x]$ we can have an irreducible polynomial in any degree.
So the question is: for any given $n\in\mathbb{Z}^+$ does there exist a field $K$ $(\mathbb{Q}\subseteq K\subseteq\mathbb{C})$ such that all irreducible polynomials in $K[x]$ have degree $\le n$, and there exists an irreducible polynomial of degree $n$?
Any number field $K$ (i.e., a finite extension of $\mathbb Q$) has irreducible polynomials of all degrees $n\ge1$. To see this, fix a prime ideal $\mathfrak p$ of the ring of integers $\mathcal O_K$. Since $\mathcal O_K/\mathfrak p$ is a finite field, there exists a monic irreducible polynomial $\overline f\in(\mathcal O_K/\mathfrak p)[x]$ of degree $n$. It lifts to a monic polynomial $f\in\mathcal O_K[x]$, which is also irreducible, as any splitting of $f$ reduces modulo $\mathfrak p$ to a splitting of $\overline f$. By Gauss’s lemma (which works for monic polynomials over any integrally closed domain), $f$ remains irreducible in $K[x]$ as well.
If $K$ is an arbitrary field, then there exists a maximal degree $n$ of irreducible polynomials over $K$ if and only if $K$ is algebraically closed (in which case $n=1$) or real-closed ($n=2$). Indeed, assume the maximal degree is $n$. Then $K$ is perfect: if $K$ is an imperfect field of characteristic $p$, there is an element $a\in K$ without a $p$th root, and then the polynomials $x^{p^k}-a$ are irreducible for all $k$. By Artin’s primitive element theorem, it follows that every finite extension of $K$ is simple, hence it is of degree at most $n$. Thus, if $L$ is an extension of $K$ of degree $n$ (which exists as there is an irreducible polynomial of degree $n$), then $L$ has no proper finite extension; i.e., $L$ is the algebraic closure of $K$. By the Artin–Schreier theorem, the algebraic closure of $K$ is a finite extension of $K$ only if $K$ is algebraically closed or real-closed.
