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The following argument classifies affine conics over an algebraically closed field $k$ with char$(k)\neq 2.$

But I don't see where it uses the hypothesis that char$(k)\neq 2.$ Is it an unnecessary hypothesis?

Let $k$ be an algebraically closed field with char$(k)\neq2$ and let $Q(x,y)$ be an irreducible polynomial of degree 2 in $k[x,y].$

We write $Q(x,y)=ax^2+bxy+cy^2+dx+ey+f$ for some $a,\ldots,f \in k.$

Since $k$ is algebraically closed, we have $$Q(x,y)=L_1L_2+L_3+f$$ where $L_3=dx+ef$ and $L_1,L_2$ are linear forms in $k[x,y]$ such that $L_1L_2=ax^2+bxy+cy^2.$

We split into two cases: (1) $L_1\sim L_2$ and (2) $L_1\not\sim L_2.$

(1) If $L_1\sim L_2,$ then $L_2=\lambda L_1$ for some $\lambda \in k^\times$ and we have $Q(x,y)=\lambda L_1^2 + L_3 + f.$

Since $k$ is algebraically closed and $Q(x,y)$ is irreducible in $k[x,y],$ it follows that $L_1\not\sim L_3.$

Therefore $(u,v)=(\sqrt{-\lambda}L_1,\,L_3+f)$ is an affine change of coordinates such that $$Q(x,y)=v-u^2.$$

(2) If $L_1\not\sim L_2,$ then there exist $\lambda,\mu \in k$ such that $L_3=\lambda L_1+\mu L_2.$

As such, we have $Q(x,y)=(L_1+\mu)(L_2+\lambda)-(\lambda\mu-f).$

Since $Q(x,y)$ is irreducible in $k[x,y],$ it follows that $\lambda\mu-f\neq 0.$

Therefore $(u,v)=\left(\dfrac{L_1+\mu}{\sqrt{\lambda\mu-f}}\,,\dfrac{L_2+\lambda}{\sqrt{\lambda\mu-f}}\right)$ is an affine change of coordinates such that $$Q(x,y)\sim uv-1.$$

If $L_1=\alpha x+ \beta y$ and $L_2=\gamma x + \delta y,$ then $L_1\sim L_2$ if and only if $\alpha\delta-\beta\gamma=0.$

Moreover, since $L_1L_2=ax^2+bxy+cy^2,$ we have $$(\alpha\delta-\beta\gamma)^2=(\alpha\delta+\beta\gamma)^2-4(\alpha\gamma)(\beta\delta)=b^2-4ac.$$ It thus follows that $$k[x,y]/(Q(x,y))\cong \left\{\begin{array}{ll} k[t] & \text{if } b^2=4ac,\\ k[t,t^{-1}] & \text{if } b^2\neq 4ac. \end{array}\right.$$

user350031
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1 Answers1

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Expanding my comment in to an answer:

Here are some lemmas used in the proof. All of these are true no matter the characteristic of the field.

Lemma 1: Let $k$ be an algebraically closed field. Then any polynomial in one variable over $k$ factors completely in to a product of (not necessarily distinct) linear factors.

Proof: This is the definition of $k$ being algebraically closed.

Lemma 2: Let $k$ be an algebraically closed field. Then any homogeneous polynomial in two variables over $k$ factors completely in to (not necessarily distinct) homogeneous linear factors.

Proof: Let $f(x,y)$ be our polynomial. Write $f(x,y)=x^ng(x,y)$ with $n$ maximal. Now $g(x,y)$ is not divisible by $x$. If $g=1$, we are done. If not, then dehomogenize $g$ with respect to $x$, apply lemma 1, and rehomogenize to get the result.

Lemma 3: A linear endomorphism of $k[x_1,\cdots,x_n]$ given by $x_i\mapsto \sum a_{ij}x_j$ is an automorphism iff the matrix $[a_{ij}]$ is invertible.

Proof: The inverse transformation is given by the inverse of the above matrix, and composing the transformations corresponds to multiplying matrices.

Lemma 4: An affine endomorphism of $k[x_1,\cdots,x_n]$ given by $x_i\mapsto \sum a_{ij}x_j + c_i$ is an automorphism iff the matrix $[a_{ij}]$ is invertible.

Proof: It's enough to show that every translation $x_i\mapsto x_i+c_i$ is an endomorphism, as the composition of endomorphisms are endomorphisms, and the linear portion is taken care of by lemma 3. But this is clear, as the inverse of a translation $x_i\mapsto x_i+c_i$ is $x_i\mapsto x_i-c_i$.

We factor $f=L_1L_2+L_3+c$ by lemma 2 applied to the degree two portion.

In case 1, we see that $L_3\not \sim L_1$ otherwise $f=p(L_1)$ for some quadratic $p\in k[t]$, which would imply it must be reducible by lemma 1. The stated affine coordinate change is then an isomorphism because the determinant of the relevant matrix mentioned in lemma 4 is $\sqrt{-\lambda}\neq 0$.

In case 2, we note that the first claim about writing $L_3$ as a linear combination of $L_1=Ax+By$ and $L_2=Cx+Dy$ follows from the fact that the vectors $(A,B)$ and $(C,D)$ are linearly independent - $L_1\not\sim L_2$ iff $\det\begin{pmatrix} A & B \\ C & D \end{pmatrix} \neq0$, so this means we can write $L_3=Ex+Fy$ as the desired linear combination. We then apply lemma 4.

In the final statement, it is clear that characteristic two is never used, and the classification exactly matches that of the fine answer of Takumi Murayama written expressly for the characteristic two case.

Thus the proof detailed in the question is valid in any characteristic.

KReiser
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