Let $f(x,y)=x^{2}+y^{2}$ viewed as an element of $k[x,y]$ where $k$ is an algebraically closed field with char distinct from $2$. How do we find the zero locus of $f$? Is there a way to simplify the set $\{(x,x\sqrt{-1}): x \in K\} \cup \{(x,-x\sqrt{-1}): x \in K\}$?
Zeros of a polynomials in a field with char distinct from 2
2
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algebraic-geometry
1 Answers
5
Not very much. Let me rewrite your answer in a more standard algebraic-geometric way.
I think it is instructive to first look at $g(x,y) = x^2 - y^2 = 0$. Over any base field this factors as $(x+y)(x-y) = 0$, so the solution set is reducible: it consists of the union of two lines $y = x$ and $y = -x$ meeting at $(0,0)$.
Your example is the same except that the factorization requires $i = \sqrt{-1} \in k$: $f(x,y) = x^2 + y^2 = (x+iy)(x-iy) = 0$. Since the characteristic is not $2$ and thus $i \neq - i$, the solution set is again reducible, consisting of the union of two lines $y = ix$ and $y = -ix$ meeting at $(0,0)$.