[proofplan] We work over the [algebraic closure](/page/Algebraic%20Closure) $K := \overline{k}$, because nonsingularity is being tested geometrically. The point at infinity is checked directly from the homogeneous partial derivatives. On the affine chart $Z \ne 0$, the singularity equations reduce to the condition that the cubic polynomial $g(t)=t^3+at+b$ and its formal derivative have a common root. We then eliminate that root explicitly and show that such a common root exists exactly when $4a^3+27b^2=0$, which is equivalent to $\Delta(E)=0$ because $\operatorname{char}(k)\ne 2$. [/proofplan]

[step:Compute the homogeneous partial derivatives and isolate the point at infinity] Let $K := \overline{k}$. Define the homogeneous polynomial map $F: K^3 \to K$ by \begin{align*} F(X,Y,Z)=Y^2Z - X^3 - aXZ^2 - bZ^3. \end{align*} The base-changed curve $E_K \subset \mathbb{P}^2_K$ is the zero locus of $F$. A point $P=[X:Y:Z]\in E_K$ is singular precisely when, for any nonzero representative $(X,Y,Z)\in K^3$, all three partial derivatives of $F$ vanish at $(X,Y,Z)$. The partial derivative with respect to $X$ is \begin{align*} F_X(X,Y,Z) = -3X^2 - aZ^2. \end{align*} The partial derivative with respect to $Y$ is \begin{align*} F_Y(X,Y,Z) = 2YZ. \end{align*} The partial derivative with respect to $Z$ is \begin{align*} F_Z(X,Y,Z) = Y^2 - 2aXZ - 3bZ^2. \end{align*} Now suppose $P=[X:Y:Z]\in E_K$ satisfies $Z=0$. Since $F(X,Y,0)=-X^3$, the condition $P\in E_K$ gives $X=0$. Hence $P=[0:Y:0]$ with $Y\ne 0$, so $P=[0:1:0]$. At this point, \begin{align*} F_Z(0,Y,0)=Y^2\ne 0. \end{align*} Thus the unique point of $E_K$ with $Z=0$ is nonsingular. [/step]

[step:Reduce affine singularities to common roots of $g$ and $g'$]It remains to study points with $Z\ne 0$. On the affine chart $Z\ne 0$, define affine coordinates $x := X/Z$ and $y := Y/Z$. Equivalently, every point in this chart has a representative $(x,y,1)\in K^3$. Define the polynomial map $g: K \to K$ by \begin{align*} g(t)=t^3 + at + b. \end{align*} Its formal derivative is the polynomial map $g': K \to K$ given by \begin{align*} g'(t)=3t^2 + a. \end{align*} For a point $(x,y,1)$, the equation $F(x,y,1)=0$ is \begin{align*} y^2 = x^3 + ax + b = g(x). \end{align*} The first partial derivative equation is \begin{align*} F_X(x,y,1)=0 \iff -3x^2-a=0 \iff g'(x)=0. \end{align*} The second partial derivative equation is \begin{align*} F_Y(x,y,1)=0 \iff 2y=0 \iff y=0, \end{align*} where the final equivalence uses $\operatorname{char}(K)\ne 2$. Therefore any affine singular point must have $y=0$, $g(x)=0$, and $g'(x)=0$. Conversely, if $x\in K$ satisfies $g(x)=0$ and $g'(x)=0$, then $(x,0,1)\in E_K$, and \begin{align*} F_X(x,0,1)=0,\qquad F_Y(x,0,1)=0. \end{align*} Moreover $g'(x)=0$ gives $a=-3x^2$, while $g(x)=0$ gives \begin{align*} 0=x^3+ax+b=x^3-3x^3+b=-2x^3+b, \end{align*} so $b=2x^3$. Hence \begin{align*} F_Z(x,0,1)=-2ax-3b=6x^3-6x^3=0. \end{align*} Thus $E_K$ has an affine singular point if and only if there exists $x\in K$ such that \begin{align*} g(x)=0,\qquad g'(x)=0. \end{align*}[/step]

[guided]We now translate the geometric singularity condition into a one-variable algebra condition. The affine chart $Z\ne 0$ allows us to use the representative $(x,y,1)$, where \begin{align*} x := X/Z,\qquad y := Y/Z. \end{align*} On this chart the equation of the curve is \begin{align*} F(x,y,1)=y^2-x^3-ax-b=0, \end{align*} or equivalently \begin{align*} y^2=g(x), \end{align*} where $g: K \to K$ is the affine cubic polynomial defined by \begin{align*} g(t)=t^3+at+b. \end{align*} A singular point must make all partial derivatives vanish. The derivative with respect to $Y$ gives \begin{align*} F_Y(x,y,1)=2y. \end{align*} Since $\operatorname{char}(K)\ne 2$, the scalar $2$ is invertible in $K$, so $F_Y(x,y,1)=0$ is exactly the condition $y=0$. The derivative with respect to $X$ gives \begin{align*} F_X(x,y,1)=-3x^2-a. \end{align*} Because $g': K \to K$ is the formal derivative of $g$ and is given by \begin{align*} g'(t)=3t^2+a, \end{align*} the equation $F_X(x,y,1)=0$ is exactly $g'(x)=0$. Combining these equations with the curve equation gives \begin{align*} 0=y^2=g(x), \end{align*} so every affine singular point produces an element $x\in K$ satisfying \begin{align*} g(x)=0,\qquad g'(x)=0. \end{align*} Conversely, suppose $x\in K$ satisfies $g(x)=0$ and $g'(x)=0$. Then $(x,0,1)$ lies on the curve because $F(x,0,1)=-g(x)=0$. Also \begin{align*} F_X(x,0,1)=-(3x^2+a)=-g'(x)=0,\qquad F_Y(x,0,1)=0. \end{align*} It remains only to check $F_Z$. From $g'(x)=0$ we get $a=-3x^2$. Substituting this into $g(x)=0$ gives \begin{align*} 0=x^3+ax+b=x^3-3x^3+b=-2x^3+b, \end{align*} so $b=2x^3$. Therefore \begin{align*} F_Z(x,0,1)=0-2ax-3b=-2(-3x^2)x-3(2x^3)=6x^3-6x^3=0. \end{align*} Thus $(x,0,1)$ is singular. This proves that affine singularities are exactly common roots of $g$ and $g'$.[/guided]

[step:Eliminate the common root condition for the depressed cubic] We prove that there exists $x\in K$ with $g(x)=g'(x)=0$ if and only if \begin{align*} 4a^3+27b^2=0. \end{align*} First suppose $x\in K$ satisfies $g(x)=0$ and $g'(x)=0$. From $g'(x)=0$, \begin{align*} a=-3x^2. \end{align*} Substituting this into $g(x)=0$ gives \begin{align*} 0=x^3+ax+b=x^3-3x^3+b=-2x^3+b, \end{align*} so $b=2x^3$. Hence \begin{align*} 4a^3+27b^2 =4(-3x^2)^3+27(2x^3)^2 =-108x^6+108x^6 =0. \end{align*} Conversely, suppose \begin{align*} 4a^3+27b^2=0. \end{align*} If $a=0$, then $27b^2=0$. Since $\operatorname{char}(K)\ne 3$, the scalar $27$ is nonzero, so $b=0$. Then $x=0$ satisfies \begin{align*} g(0)=0,\qquad g'(0)=0. \end{align*} Now assume $a\ne 0$. Since $\operatorname{char}(K)\ne 2,3$, the scalars $2$ and $3$ are invertible, so define \begin{align*} x := -\frac{3b}{2a}\in K. \end{align*} Using $27b^2=-4a^3$, we compute \begin{align*} x^2 =\frac{9b^2}{4a^2} =\frac{9}{4a^2}\left(-\frac{4a^3}{27}\right) =-\frac{a}{3}. \end{align*} Therefore \begin{align*} g'(x)=3x^2+a=3\left(-\frac{a}{3}\right)+a=0. \end{align*} Also, \begin{align*} g(x) =x^3+ax+b =x\left(-\frac{a}{3}\right)+ax+b =\frac{2a}{3}x+b =\frac{2a}{3}\left(-\frac{3b}{2a}\right)+b =0. \end{align*} Thus $g$ and $g'$ have a common root in $K$ exactly when $4a^3+27b^2=0$. [/step]

[step:Translate the common root criterion into the discriminant criterion] From the previous steps, $E_K$ is singular if and only if \begin{align*} 4a^3+27b^2=0. \end{align*} Since $\operatorname{char}(K)\ne 2$, the scalar $-16$ is nonzero in $K$. Therefore \begin{align*} \Delta(E)=-16(4a^3+27b^2) \end{align*} vanishes in $K$ if and only if $4a^3+27b^2$ vanishes in $K$. Because $k\subset K$ is a [field extension](/page/Field%20Extension), an element of $k$ is zero in $K$ if and only if it is zero in $k$. Hence $E_{\overline{k}}$ is nonsingular if and only if \begin{align*} \Delta(E)\ne 0. \end{align*} This proves the theorem. [/step]