[proofplan] We reduce the projective question to an affine chart containing the point. In that chart, after translating the point to the origin, the curve is defined locally by one polynomial $g(u,v)$ in the local ring $\overline{k}[u,v]_{(u,v)}$. The point is nonsingular exactly when the image of the maximal ideal has one-dimensional cotangent space, and this happens precisely when $g$ has a nonzero linear term. Finally, the linear coefficients of $g$ are the affine partial derivatives, and Euler's homogeneous identity identifies the affine condition with the nonvanishing of at least one of $F_X(P),F_Y(P),F_Z(P)$. [/proofplan]

[step:Move to an affine chart and translate the point to the origin]Let $K:=\overline{k}$, and let $C_K:=C\times_{\operatorname{Spec} k}\operatorname{Spec} K$ denote the base change of $C$ to $K$. Since $P=[a:b:c]\in \mathbb{P}^2(K)$, at least one of $a,b,c$ is nonzero. We treat the case $c\neq 0$; the cases $a\neq 0$ and $b\neq 0$ are obtained by the same argument in the affine charts $X\neq 0$ and $Y\neq 0$. Choose the representative of $P$ with $c=1$, so $P=[\alpha:\beta:1]$ for some $\alpha,\beta\in K$. Define the dehomogenized polynomial map $f:K^2\to K$ by requiring that, for every $(x,y)\in K^2$, \begin{align*} f(x,y)=F(x,y,1). \end{align*} Since $P\in C(K)$, we have $f(\alpha,\beta)=F(\alpha,\beta,1)=0$. The affine chart $Z\neq 0$ has coordinate ring $K[x,y]$, and the chart intersection $C_K\cap \{Z\neq 0\}$ has coordinate ring $K[x,y]/(f)$. Hence the local ring of $C_K$ at $P$ in this chart is \begin{align*} \left(K[x,y]/(f)\right)_{(x-\alpha,y-\beta)}. \end{align*} Define the translated affine polynomial map $g:K^2\to K$ by requiring that, for every $(u,v)\in K^2$, \begin{align*} g(u,v)=f(\alpha+u,\beta+v). \end{align*} The $K$-algebra isomorphism $K[x,y]\to K[u,v]$ determined by $x\mapsto \alpha+u$ and $y\mapsto \beta+v$ identifies the maximal ideal $(x-\alpha,y-\beta)$ with $(u,v)$ and identifies the ideal $(f)$ with $(g)$. Let $A:=K[u,v]_{(u,v)}$ and $\mathfrak n:=(u,v)A$. Let $R:=A/(g)$ and $\mathfrak m:=\mathfrak n/(g)$. Then $R$ is the local ring of $C_K$ at $P$ in the affine chart $Z\neq 0$, and $\mathfrak m$ is its maximal ideal.[/step]

[guided]We work over $K=\overline{k}$ because the point $P$ is given over $\overline{k}$ and nonsingularity at a geometric point is computed on the base-changed curve $C_K:=C\times_{\operatorname{Spec} k}\operatorname{Spec} K$. Suppose first that $c\neq 0$. Scaling the homogeneous representative by $c^{-1}$ gives $P=[\alpha:\beta:1]$, where \begin{align*} \alpha=\frac{a}{c},\qquad \beta=\frac{b}{c}. \end{align*} On the affine chart $Z\neq 0$, the coordinates are $x=X/Z$ and $y=Y/Z$. The homogeneous equation $F=0$ becomes the affine equation \begin{align*} f(x,y)=F(x,y,1)=0. \end{align*} Thus the local behaviour of the projective curve near $P$ is the local behaviour of the affine plane curve $f=0$ near $(\alpha,\beta)$. To move the point to the origin, define the polynomial map $g:K^2\to K$ by \begin{align*} g(u,v)=f(\alpha+u,\beta+v). \end{align*} Then $g(0,0)=f(\alpha,\beta)=0$. The affine chart $Z\neq 0$ has coordinate ring $K[x,y]$, and inside this chart the curve is defined by $f=0$, so the local ring at $P$ is \begin{align*} \left(K[x,y]/(f)\right)_{(x-\alpha,y-\beta)}. \end{align*} Now change variables by the $K$-algebra isomorphism $K[x,y]\to K[u,v]$ sending $x$ to $\alpha+u$ and $y$ to $\beta+v$. Under this isomorphism, the ideal $(x-\alpha,y-\beta)$ becomes $(u,v)$, and the equation $f$ becomes $g$. Therefore the same local ring is \begin{align*} R=K[u,v]_{(u,v)}/(g). \end{align*} The maximal ideal of $K[u,v]_{(u,v)}$ is $\mathfrak n=(u,v)K[u,v]_{(u,v)}$, and its image $\mathfrak m=\mathfrak n/(g)$ is the maximal ideal of $R$. This local ring records exactly the infinitesimal structure of the curve at $P$.[/guided]

[step:Compute the cotangent space from the linear part of the local equation]Write the Taylor expansion of $g$ at $(0,0)$ in the form \begin{align*} g(u,v)=\lambda u+\mu v+h(u,v), \end{align*} where $\lambda:=\frac{\partial f}{\partial x}(\alpha,\beta)$, $\mu:=\frac{\partial f}{\partial y}(\alpha,\beta)$, and $h\in (u,v)^2K[u,v]$. Since $g\in \mathfrak n$, the cotangent space of $R$ at $P$ is \begin{align*} \mathfrak m/\mathfrak m^2\cong \mathfrak n/(\mathfrak n^2+(g)). \end{align*} Modulo $\mathfrak n^2$, the element $g$ has image $\lambda u+\mu v$. Therefore $\dim_K \mathfrak m/\mathfrak m^2=1$ if $(\lambda,\mu)\neq (0,0)$, while $\dim_K \mathfrak m/\mathfrak m^2=2$ if $(\lambda,\mu)=(0,0)$. The ring $A=K[u,v]_{(u,v)}$ is a regular local ring of dimension $2$, and $R=A/(g)$ is the local ring of a plane curve at the point $P$. The polynomial $g$ is nonzero: if $g=0$, then $f=0$, hence the dehomogenization $F(X,Y,1)$ is zero, and by homogeneity this would force $F=0$, contrary to the hypothesis. Also $g\in\mathfrak n$, so $g$ is not a unit in $A$. Since $A$ is a two-dimensional domain and $(g)$ is a nonzero principal ideal, [Krull's Principal Ideal Theorem](/page/Krull%27s%20Principal%20Ideal%20Theorem) gives that every minimal prime over $(g)$ has height at most $1$; because $g\neq 0$ and $A$ is a domain, height $0$ is impossible. Thus every minimal prime $\mathfrak p$ over $(g)$ has height $1$. Since $A$ is a regular local ring of dimension $2$, the quotient component $A/\mathfrak p$ has dimension $2-1=1$. Hence the components through $P$ have codimension $1$ in $\operatorname{Spec} A$, and $\dim R=1$. By the definition of a nonsingular point of a curve, $P$ is nonsingular precisely when the Noetherian local ring $R=\mathcal O_{C_K,P}$ is a [regular local ring](/page/Regular%20Local%20Ring). The [regular local ring criterion](/page/Regular%20Local%20Ring) says that this is equivalent to \begin{align*} \dim_K\mathfrak m/\mathfrak m^2=\dim R. \end{align*} Since $\dim R=1$, the preceding cotangent-space computation gives that $P$ is nonsingular if and only if \begin{align*} \left(\frac{\partial f}{\partial x}(\alpha,\beta),\frac{\partial f}{\partial y}(\alpha,\beta)\right)\neq (0,0). \end{align*}[/step]

[guided]The maximal ideal $\mathfrak m$ consists of germs of functions vanishing at the point. The quotient $\mathfrak m/\mathfrak m^2$ keeps only first-order information: terms of degree at least $2$ vanish in this quotient. Thus nonsingularity for a plane curve is detected by whether the local equation imposes one independent linear condition on the two first-order coordinates $u$ and $v$. The translated polynomial has the expansion \begin{align*} g(u,v)=\lambda u+\mu v+h(u,v), \end{align*} where $\lambda=\frac{\partial f}{\partial x}(\alpha,\beta)$, $\mu=\frac{\partial f}{\partial y}(\alpha,\beta)$, and every monomial of $h$ has total degree at least $2$. In the local ring $A=K[u,v]_{(u,v)}$, this says $h\in \mathfrak n^2$. The cotangent space of the quotient ring $R=A/(g)$ is obtained by quotienting the cotangent space of $A$ by the first-order part of the relation $g=0$: \begin{align*} \mathfrak m/\mathfrak m^2\cong \mathfrak n/(\mathfrak n^2+(g)). \end{align*} The [vector space](/page/Vector%20Space) $\mathfrak n/\mathfrak n^2$ has basis given by the classes of $u$ and $v$, so it is two-dimensional over $K$. The image of $g$ in this two-dimensional space is the linear form $\lambda u+\mu v$. If $(\lambda,\mu)\neq(0,0)$, this linear form spans a one-dimensional subspace, so quotienting by it leaves a one-dimensional cotangent space: \begin{align*} \dim_K\mathfrak m/\mathfrak m^2=1. \end{align*} If $(\lambda,\mu)=(0,0)$, then $g\in \mathfrak n^2$, so $g$ imposes no first-order relation and \begin{align*} \dim_K\mathfrak m/\mathfrak m^2=2. \end{align*} We now connect this computation to nonsingularity. The ambient local ring $A=K[u,v]_{(u,v)}$ is a regular local ring of dimension $2$, with maximal ideal generated by $u$ and $v$. The quotient $R=A/(g)$ is the local ring of the affine plane curve at the translated point. We must check that $R$ really has dimension $1$. First, $g\neq 0$: if $g=0$, then $f=0$, so $F(X,Y,1)=0$; since $F$ is homogeneous, all coefficients of $F$ would be zero, contradicting that $F$ is nonzero. Second, $g\in\mathfrak n$, so $g$ is not a unit. Since $A$ is a two-dimensional domain and $(g)$ is a nonzero principal ideal, [Krull's Principal Ideal Theorem](/page/Krull%27s%20Principal%20Ideal%20Theorem) implies that every minimal prime over $(g)$ has height at most $1$; height $0$ cannot occur because $g\neq0$ in the domain $A$. Hence every minimal prime $\mathfrak p$ over $(g)$ has height $1$. Because $A$ is regular local of dimension $2$, the quotient component $A/\mathfrak p$ has dimension $2-1=1$. Thus every component through the point has codimension $1$ in $\operatorname{Spec} A$, and $\dim R=1$. By the definition of a nonsingular point of a curve, the point $P$ is nonsingular exactly when the Noetherian local ring $R=\mathcal O_{C_K,P}$ is a [regular local ring](/page/Regular%20Local%20Ring). The [regular local ring criterion](/page/Regular%20Local%20Ring) says that a Noetherian local ring $(R,\mathfrak m)$ is regular exactly when the embedding dimension $\dim_K\mathfrak m/\mathfrak m^2$ equals the Krull dimension $\dim R$. Therefore this curve point is nonsingular exactly when \begin{align*} \dim_K\mathfrak m/\mathfrak m^2=1. \end{align*} Combining this criterion with the first-order computation above, the point is nonsingular exactly when the affine equation has a nonzero linear term at the point.[/guided]

[step:Relate the affine partial derivatives to the homogeneous gradient]By the definition $f(x,y)=F(x,y,1)$, the chain rule gives \begin{align*} \frac{\partial f}{\partial x}(\alpha,\beta)=F_X(\alpha,\beta,1). \end{align*} It also gives \begin{align*} \frac{\partial f}{\partial y}(\alpha,\beta)=F_Y(\alpha,\beta,1). \end{align*} It remains to compare these two partials with $F_Z(\alpha,\beta,1)$. Let $d$ be the degree of $F$. Since $F$ is homogeneous of degree $d$, [Euler's Homogeneous Identity](/page/Euler%27s%20Homogeneous%20Identity) gives \begin{align*} X F_X+Y F_Y+Z F_Z=dF. \end{align*} Evaluating at $(\alpha,\beta,1)$ and using $F(\alpha,\beta,1)=0$, we obtain \begin{align*} \alpha F_X(\alpha,\beta,1)+\beta F_Y(\alpha,\beta,1)+F_Z(\alpha,\beta,1)=0. \end{align*} Therefore, if $F_X(\alpha,\beta,1)=F_Y(\alpha,\beta,1)=0$, then $F_Z(\alpha,\beta,1)=0$. Conversely, if either $F_X(\alpha,\beta,1)$ or $F_Y(\alpha,\beta,1)$ is nonzero, then the homogeneous gradient is nonzero. Hence \begin{align*} \left(\frac{\partial f}{\partial x}(\alpha,\beta),\frac{\partial f}{\partial y}(\alpha,\beta)\right)\neq (0,0) \end{align*} if and only if \begin{align*} (F_X(\alpha,\beta,1),F_Y(\alpha,\beta,1),F_Z(\alpha,\beta,1))\neq (0,0). \end{align*}[/step]

[guided]The affine computation detects only the two derivatives with respect to the chart coordinates $x=X/Z$ and $y=Y/Z$. We must now explain why the missing homogeneous derivative $F_Z$ carries no additional independent first-order condition at a point of the curve. By definition of the dehomogenized map $f:K^2\to K$, $f(x,y)=F(x,y,1)$. Differentiating this identity with respect to the affine variables gives \begin{align*} \frac{\partial f}{\partial x}(\alpha,\beta)=F_X(\alpha,\beta,1) \end{align*} and \begin{align*} \frac{\partial f}{\partial y}(\alpha,\beta)=F_Y(\alpha,\beta,1). \end{align*} This is the ordinary chain rule applied to the map $(x,y)\mapsto (x,y,1)$, whose derivative has coordinate directions only in the $X$ and $Y$ variables. It remains to control $F_Z(\alpha,\beta,1)$. Let $d$ denote the degree of the homogeneous polynomial $F$. Since $F$ is homogeneous of degree $d$, [Euler's Homogeneous Identity](/page/Euler%27s%20Homogeneous%20Identity) gives \begin{align*} X F_X+Y F_Y+Z F_Z=dF. \end{align*} We may apply it here because $F$ is homogeneous by hypothesis. Evaluating at $(\alpha,\beta,1)$ and using that $P\in C(K)$, equivalently $F(\alpha,\beta,1)=0$, gives \begin{align*} \alpha F_X(\alpha,\beta,1)+\beta F_Y(\alpha,\beta,1)+F_Z(\alpha,\beta,1)=0. \end{align*} If $F_X(\alpha,\beta,1)=F_Y(\alpha,\beta,1)=0$, this equality forces $F_Z(\alpha,\beta,1)=0$. Conversely, if either $F_X(\alpha,\beta,1)$ or $F_Y(\alpha,\beta,1)$ is nonzero, then the triple of homogeneous partial derivatives is nonzero. Therefore \begin{align*} \left(\frac{\partial f}{\partial x}(\alpha,\beta),\frac{\partial f}{\partial y}(\alpha,\beta)\right)\neq (0,0) \end{align*} if and only if \begin{align*} (F_X(\alpha,\beta,1),F_Y(\alpha,\beta,1),F_Z(\alpha,\beta,1))\neq (0,0). \end{align*}[/guided]

[step:Check independence of the homogeneous representative and conclude]Let $r\in K^\times$. Since $F_X,F_Y,F_Z$ are homogeneous of degree $d-1$, we have \begin{align*} F_X(ra,rb,rc)=r^{d-1}F_X(a,b,c). \end{align*} Similarly, \begin{align*} F_Y(ra,rb,rc)=r^{d-1}F_Y(a,b,c). \end{align*} and \begin{align*} F_Z(ra,rb,rc)=r^{d-1}F_Z(a,b,c). \end{align*} Thus the condition that the triple $(F_X(P),F_Y(P),F_Z(P))$ is nonzero is independent of the chosen representative of $P$. In the chart $Z\neq 0$, the preceding steps prove that $P$ is nonsingular if and only if the homogeneous gradient at $P$ is nonzero. If instead $a\neq0$, choose the representative $P=[1:\beta:\gamma]$ and use the affine chart $X\neq0$ with coordinates $y=Y/X$ and $z=Z/X$. The dehomogenized equation is the map $f_X:K^2\to K$ given by $(y,z)\mapsto F(1,y,z)$, and the translated local equation at $(\beta,\gamma)$ has nonzero linear part exactly when at least one of $F_Y(1,\beta,\gamma)$ and $F_Z(1,\beta,\gamma)$ is nonzero. Evaluating Euler's homogeneous identity at $(1,\beta,\gamma)$ and using $F(1,\beta,\gamma)=0$ shows that, when those two derivatives vanish, $F_X(1,\beta,\gamma)$ also vanishes. Hence the same equivalence holds in the chart $X\neq0$. If $b\neq0$, choose the representative $P=[\alpha:1:\gamma]$ and use the affine chart $Y\neq0$ with coordinates $x=X/Y$ and $z=Z/Y$. The dehomogenized equation is the map $f_Y:K^2\to K$ given by $(x,z)\mapsto F(x,1,z)$, and the translated local equation at $(\alpha,\gamma)$ has nonzero linear part exactly when at least one of $F_X(\alpha,1,\gamma)$ and $F_Z(\alpha,1,\gamma)$ is nonzero. Evaluating Euler's homogeneous identity at $(\alpha,1,\gamma)$ and using $F(\alpha,1,\gamma)=0$ shows that, when those two derivatives vanish, $F_Y(\alpha,1,\gamma)$ also vanishes. Hence the same equivalence holds in the chart $Y\neq0$. Therefore, for every $P\in C(\overline{k})$, \begin{align*} P \text{ is nonsingular on } C_{\overline{k}}\quad\Longleftrightarrow\quad(F_X(P),F_Y(P),F_Z(P))\neq(0,0). \end{align*} This proves the Jacobian criterion for plane curves.[/step]

[guided]The condition $(F_X(P),F_Y(P),F_Z(P))\neq(0,0)$ is meaningful on projective points because it is independent of the homogeneous representative. If $r\in K^\times$, then the partial derivatives $F_X,F_Y,F_Z$ are homogeneous of degree $d-1$, so \begin{align*} F_X(ra,rb,rc)=r^{d-1}F_X(a,b,c), \end{align*} \begin{align*} F_Y(ra,rb,rc)=r^{d-1}F_Y(a,b,c), \end{align*} and \begin{align*} F_Z(ra,rb,rc)=r^{d-1}F_Z(a,b,c). \end{align*} Since $r^{d-1}\neq0$, the triple is zero for one representative exactly when it is zero for every representative. We have already proved the criterion in the chart $Z\neq0$. We now spell out the other charts rather than hiding the verification. If $a\neq0$, scale the representative to $P=[1:\beta:\gamma]$ and work in the chart $X\neq0$. The affine coordinates are $y=Y/X$ and $z=Z/X$, and the dehomogenized map $f_X:K^2\to K$ is defined by requiring that, for every $(y,z)\in K^2$, \begin{align*} f_X(y,z)=F(1,y,z). \end{align*} Repeating the local-ring and cotangent-space computation with variables $y-\beta$ and $z-\gamma$ shows that $P$ is nonsingular exactly when \begin{align*} \left(\frac{\partial f_X}{\partial y}(\beta,\gamma),\frac{\partial f_X}{\partial z}(\beta,\gamma)\right)\neq(0,0). \end{align*} By the chain rule this is exactly the condition that at least one of $F_Y(1,\beta,\gamma)$ and $F_Z(1,\beta,\gamma)$ is nonzero. If both vanish, Euler's homogeneous identity at $(1,\beta,\gamma)$ gives \begin{align*} F_X(1,\beta,\gamma)+\beta F_Y(1,\beta,\gamma)+\gamma F_Z(1,\beta,\gamma)=dF(1,\beta,\gamma)=0, \end{align*} so $F_X(1,\beta,\gamma)=0$. Thus the affine criterion in the chart $X\neq0$ is equivalent to the nonvanishing of the full homogeneous gradient. If $b\neq0$, scale the representative to $P=[\alpha:1:\gamma]$ and work in the chart $Y\neq0$. The affine coordinates are $x=X/Y$ and $z=Z/Y$, and the dehomogenized map $f_Y:K^2\to K$ is defined by requiring that, for every $(x,z)\in K^2$, \begin{align*} f_Y(x,z)=F(x,1,z). \end{align*} The same cotangent-space computation gives nonsingularity exactly when at least one of $F_X(\alpha,1,\gamma)$ and $F_Z(\alpha,1,\gamma)$ is nonzero. If both vanish, Euler's identity at $(\alpha,1,\gamma)$ gives \begin{align*} \alpha F_X(\alpha,1,\gamma)+F_Y(\alpha,1,\gamma)+\gamma F_Z(\alpha,1,\gamma)=dF(\alpha,1,\gamma)=0, \end{align*} so $F_Y(\alpha,1,\gamma)=0$. Hence the affine criterion in the chart $Y\neq0$ is also equivalent to the nonvanishing of the full homogeneous gradient. Since every projective point has at least one nonzero coordinate, the three chart computations cover all $P\in C(\overline{k})$. Therefore \begin{align*} P \text{ is nonsingular on } C_{\overline{k}}\quad\Longleftrightarrow\quad(F_X(P),F_Y(P),F_Z(P))\neq(0,0), \end{align*} which is the asserted Jacobian criterion for plane curves.[/guided]