I am motivated by the case of $x^2+y^2$. It is irreducible in $\mathbb{R}[x,y]$ but reducible in $\mathbb{C}[x,y]$, where it factors as $x^2+y^2=(x+iy)(x-iy)$. In any algebraically closed field $k$, $x^2+1 \in k[x]$ has roots in $k$ which we may denote as $\pm \sqrt{-1}= \pm i$. Hence $x \pm iy \in k[x,y]$ and $x^2+y^2$ has the same factorization as over $\mathbb{C}$.
This makes me wonder if the factorization of a polynomial is the same over any algebraically closed field. I can't immediately think of any counterexamples. What if one restricts attention to algebraically closed fields of characteristic 0?
Sorry if I am missing something obvious, and thanks for any input.
Best Answer
If you can consider two algebraically closed fields $\mathbb{F}_{1}$ and $\mathbb{F}_{2}$ to have a nontrivial intersection containing the coefficients of your polynomial, then you can consider the smallest algebraically closed subfield containing all of these coefficients $k \subseteq \mathbb{F}_{1} \cap \mathbb{F}_{2}$, your polynomial will factor completely over $k$. So yes in this case case your polynomial will factor uniquely (however you mean this).
But in general, it doesn't make sense to think of this question in a more broad sense. For example, you can't in general think of the roots of $x^{2}+1$ in an arbitrary field as being complex numbers. In $\overline{\mathbb{F}_{2}}$, this polynomial will factor as $(x+1)^{2}$ while in $\overline{\mathbb{F}_{5}}$ it will factor as $(x-2)(x-3)$.
Over $\overline{\mathbb{F}_{3}}$, the roots are not in the prime field, so here we could call the roots of this polynomial $\pm \mathrm{i}$, and say $x^{2} + 1 = (x+\mathrm{i})(x-\mathrm{i})$. But this isn't the complex number $\mathrm{i}$; we would not think of $\overline{\mathbb{F}_{3}}$ as having any elements in common with $\mathbb{C}$ so it is pretty meaningless to say that this factorization is "the same as" the factorization over $\mathbb{C}$. In fact, it would not even be wise to consider the coefficients of $x^{2}+1 \in \overline{\mathbb{F}_{3}}[x]$ to be the same as the coefficients of $x^{2}+1 \in \mathbb{C}$, though they can be loosely connected through the canonical homomorphisms of $\mathbb{Z}$ into these fields.