Maximal degree of irreducible polynomials This is a question I have thought about for a while. We know that every polynomial $p \in \mathbb C[z]$ can be written as a product of monomials
$$p(z) = a \displaystyle\prod_{i=1}^n(z-z_i).$$
Now for real polynomials, this isn't the case since $x^2+1$ does not have a real root.
So this got me interested in the number which can be defined as
$$\alpha_n(k) := \max_{p \in k[x_1, \cdots, x_n] \text{ and } p \text{ is irreducible} } \deg(p).$$
Are there any known methods to compute the number or bounds on $\alpha_n(k)$? which is known.
Obviously if $k$ is algebraically closed, then $\alpha_1(k)$ is just 1.
I have looked at fields like $\mathbb F_{p^i}$, but did not really get a picture of what makes $\alpha_n(k)$ a particular number.
 A: The case $n\ge2$ and a part of the case $n=1$ were already addressed by @RaviFernando. This answer finishes the case $n=1$. Let $k$ be a field, $k\subseteq\overline{k}$ an algebraic closure of $k$ and $k\subseteq k^s\subseteq\overline{k}$ the separable closure of $k$ inside $\overline{k}$. The extension $\overline{k}/k^s$ is purely inseparable and the extension $k^s/k$ is separable. If $k$ is perfect, then $k^s=\overline{k}$, which is the case already taken care of. If $K/k$ is an algebraic extension and $x\in K$, the minimal polynomial of $x$ over $k$ will be denoted $P_{x/k}\in k[T]$. First, some lemmata.
Lemma: Let $K/k$ be a purely inseparable field extension with $K$ algebraically closed. Then, either $[K\colon k]=1$ or $[K\colon k]=\infty$. In the latter case, there are $x\in L$ with $\deg(P_{x/k})$ arbitrarily large.
Proof: If $\mathrm{char}(k)=0$, necessarily $[K\colon k]=1$. Let $\mathrm{char}(k)=p$, $p$ prime. If $[K\colon k]>1$, there exists an $x\in K\setminus k$ of positive exponent $n\ge1$ (meaning $n$ is the smallest integer such that $x^{p^n}\in k$). Then, for each $m\ge0$, there exists a $p^m$-th root $x_m\in K$ of $x$, since $K$ is algebraically closed. Now, $x_m^{p^{n+m}}=(x_m^{p^m})^{p^n}=x^{p^n}\in k$ and $x_m^{p^{n+m-1}}=(x_m^{p^m})^{p^{n-1}}=x^{p^{n-1}}\not\in k$ (here, we need $n\ge1$). By the general theory of pure inseparability, this implies $P_{x_m/k}=T^{p^{n+m}}-x^{p^n}\in k[T]$, in particular $\deg(P_{x_m/k})=p^{n+m}$. Since $m\ge0$ was arbitrary, this grows arbitrarily large. Furthermore, $[K\colon k]\ge[k(x_m)\colon k]=\deg(P_{x_m/k})=p^{n+m}\rightarrow\infty$.
Lemma: Let $k$ be a field. Then, $[\overline{k}\colon k]\ge a_1(k)\ge[k^s\colon k]$.
Proof: Let $P\in k[x]$ be irreducible. There exists a root $\alpha\in\overline{k}$ of $P$, since $\overline{k}$ is algebraically closed. Then, since $P=P_{\alpha/k}$, we have that $[\overline{k}\colon k]\ge[k(\alpha)\colon k]=\deg(P_{\alpha/k})=\deg(P)$. Taking the supremum over all irreducible polynomials, we obtain $[\overline{k}\colon k]\ge a_1(k)$. For the other inequality, there is nothing to prove if $a_1(k)=\infty$, so assume $a_1(k)=n<\infty$. Now, since $k^s/k$ is separable, any finite subextension is simple by the primitive element theorem. Thus, if $k\subseteq K\subseteq k^s$ with $K/k$ finite, $K=k(\alpha)$ for some $\alpha\in k^s$ and $[K\colon k]=[k(\alpha)\colon k]=\deg(P_{\alpha/k})\le n$. This shows that $[k^s\colon k]<\infty$, since an infinite algebraic extension contains subextensions of arbitrarily large finite degree. Then, applying the result to $k^s/k$ itself, we obtain $[k^s\colon k]\le n=a_1(k)$, as desired.
Theorem: Let $k$ be a field. Then, $a_1(k)=1$ iff $k$ is algebraically closed, $a_1(k)=2$ iff $k$ is real closed and $a_1(k)=\infty$ otherwise.
Proof: Obviously, $a_1(k)=1$ if $k$ is algebraically closed and $a_1(k)=2$ if $k$ is real closed. If $[k^s\colon k]=\infty$, $a_1(k)=\infty$ by the second Lemma, so $k$ is neither algebraically closed nor real closed and we're good. Let's assume that $[k^s\colon k]<\infty$. By the first Lemma, either $[\overline{k}\colon k^s]=1$ or $[\overline{k}\colon k^s]=\infty$. In the first case, $\overline{k}=k^s$, so $\overline{k}/k$ is finite. The Artin-Schreier theorem now tells us that $k$ is either algebraically closed or real closed and we're good (furthermore, the inequalities in the second Lemma become equalities). Let's assume that $[\overline{k}\colon k^s]=\infty$. For each sufficiently large $n$, we can find $x\in\overline{k}$, such that $\deg(P_{x/k^s})=p^n$. Then, since $P_{x/k^s}\mid P_{x/k}$ in $k^s[T]$, we obtain $a_1(k)\ge\deg(P_{x/k})\ge\deg(P_{x/k^s})=p^n\rightarrow\infty$.
A: If $n \geq 2$, then $\alpha_n(k) = \infty$ regardless of the field $k$.  Proof:  for every $f \in k[x]$, the polynomial $x_1 - f(x_2) \in k[x_1, \dots, x_n]$ has degree $1$ in $x_1$, so any factorization must have the form $$
\big(a(x_2, \dots, x_n) x_1 + b(x_2, \dots, x_n)\big) \cdot c(x_2, \dots, x_n),
$$
where $a(x_2, \dots, x_n) \cdot c(x_2, \dots, x_n) = 1$.  This forces the second factor to be a unit.  So every polynomial of the form $x_1 - f(x_2)$ is irreducible, and their (total) degrees can be arbitrarily large.  Moral of the story:  it's way too easy for multivariable polynomials to be irreducible.
So the only real question is when $n = 1$.  Here there are three cases.  If $k$ is algebraically closed, then of course $\alpha_1(k) = 1$.  If $k$ is real closed, then its algebraic closure is a quadratic extension, so $\alpha_1(k) = 2$.  Otherwise, the Artin-Schreier theorem says that $\overline{k}$ has infinite degree over $k$, so there exist arbitrarily large finite extensions of $k$.  If $k$ happens to be perfect (e.g. if $\mathrm{char} \: k = 0$ or $k$ is finite), then all finite extensions of $k$ are separable.  In this case, the primitive element theorem provides us with elements of $\overline{k}$ with arbitrarily high degree over $k$, and taking their minimal polynomials shows that $\alpha_1(k) = \infty$.
To avoid the assumption that $k$ is perfect, see @Thorgott's answer.
