[proofplan] We use the definitions of a [Weierstrass change of variables](/page/Weierstrass%20Change%20of%20Variables) and a [long Weierstrass equation](/page/Long%20Weierstrass%20Equation): substitute the change of variables into the original equation and expand in the new variables. The leading quadratic and cubic terms acquire the same invertible sixth-power factor from the scaling parameter, so division by that factor restores the normalized long Weierstrass form. All remaining terms are collected by monomial type, and the resulting coefficients lie in $k$ because the scaling parameter is invertible in $k$. [/proofplan]

[step:Introduce the coefficients and transformed variables and substitute into the equation] Let $a_1,a_2,a_3,a_4,a_6 \in k$ denote the coefficients of the original long Weierstrass equation \begin{align*} y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6. \end{align*} Let $u \in k^\times$ and $r,s,t \in k$ be the parameters of the Weierstrass change of variables. Write $X$ for the transformed variable $x'$ and $Y$ for the transformed variable $y'$. In the [polynomial ring](/page/Polynomial%20Ring) $k[X,Y]$, define the pullback expressions $x_{\mathrm{old}} := u^2X+r$ and $y_{\mathrm{old}} := u^3Y+u^2sX+t$ for the original affine coordinates $x$ and $y$. Substituting $x=x_{\mathrm{old}}$ and $y=y_{\mathrm{old}}$ into the original equation produces an equation between two polynomials in $k[X,Y]$ after expansion and collection of terms. [/step]

[step:Compute the transformed left-hand side by monomial type]Expanding the left-hand side in $k[X,Y]$ gives \begin{align*} y^2 = u^6Y^2+2u^5sXY+2u^3tY+u^4s^2X^2+2u^2stX+t^2. \end{align*} \begin{align*} a_1xy = a_1u^5XY+a_1u^3rY+a_1u^4sX^2+a_1u^2(t+rs)X+a_1rt. \end{align*} \begin{align*} a_3y = a_3u^3Y+a_3u^2sX+a_3t. \end{align*} Therefore \begin{align*} y^2+a_1xy+a_3y = u^6Y^2+u^5(2s+a_1)XY+u^3(2t+a_1r+a_3)Y+u^4(s^2+a_1s)X^2+u^2(2st+a_1t+a_1rs+a_3s)X+t^2+a_1rt+a_3t. \end{align*}[/step]

[guided]Let $a_1,a_2,a_3,a_4,a_6 \in k$ be the coefficients of the original long Weierstrass equation \begin{align*} y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6. \end{align*} Throughout this guided computation, $X$ denotes the new variable $x'$ and $Y$ denotes the new variable $y'$. Thus the Weierstrass change of variables is being written as $x=u^2X+r$ and $y=u^3Y+u^2sX+t$ in the polynomial ring $k[X,Y]$. The point of this expansion is to check that no unwanted powers of $Y$ appear. Since the change of variables is affine-linear in $Y$, the only possible quadratic term in $Y$ comes from $y^2$, and it is \begin{align*} (u^3Y)^2=u^6Y^2. \end{align*} The mixed $XY$ terms come from the product of $u^3Y$ with $u^2sX$ in $y^2$, and from the product of $u^2X$ with $u^3Y$ in $xy$: \begin{align*} 2(u^3Y)(u^2sX)+a_1(u^2X)(u^3Y) =u^5(2s+a_1)XY. \end{align*} The terms linear in $Y$ come from the product of $u^3Y$ with $t$ in $y^2$, from the product of $r$ with $u^3Y$ in $xy$, and from $a_3y$: \begin{align*} 2(u^3Y)t+a_1r(u^3Y)+a_3(u^3Y) =u^3(2t+a_1r+a_3)Y. \end{align*} The remaining terms contain only $X$ or are constant: \begin{align*} u^4(s^2+a_1s)X^2 +u^2(2st+a_1t+a_1rs+a_3s)X +t^2+a_1rt+a_3t. \end{align*} Thus the transformed left-hand side has exactly one $Y^2$ term, one possible $XY$ term, one possible $Y$ term, and pure $X$ terms that can be moved to the right-hand side.[/guided]

[step:Compute the transformed right-hand side by monomial type] Expanding the right-hand side in $k[X,Y]$ gives \begin{align*} x^3 = u^6X^3+3u^4rX^2+3u^2r^2X+r^3. \end{align*} \begin{align*} a_2x^2 = a_2u^4X^2+2a_2u^2rX+a_2r^2. \end{align*} \begin{align*} a_4x = a_4u^2X+a_4r. \end{align*} Hence \begin{align*} x^3+a_2x^2+a_4x+a_6 = u^6X^3+u^4(a_2+3r)X^2+u^2(a_4+2a_2r+3r^2)X+a_6+a_4r+a_2r^2+r^3. \end{align*} [/step]

[step:Move pure $X$ terms to the right and divide by the common leading factor]The substituted equation is equivalent to \begin{align*} u^6Y^2+u^5(2s+a_1)XY+u^3(2t+a_1r+a_3)Y = u^6X^3+u^4(a_2+3r-s^2-a_1s)X^2+u^2(a_4+2a_2r+3r^2-2st-a_1t-a_1rs-a_3s)X+a_6+a_4r+a_2r^2+r^3-t^2-a_1rt-a_3t. \end{align*} Since $u \in k^\times$, the element $u^6$ is invertible in $k$. Dividing by $u^6$ gives \begin{align*} Y^2+a_1'XY+a_3'Y=X^3+a_2'X^2+a_4'X+a_6', \end{align*} where \begin{align*} a_1'=\frac{a_1+2s}{u}. \end{align*} \begin{align*} a_2'=\frac{a_2+3r-s^2-a_1s}{u^2}. \end{align*} \begin{align*} a_3'=\frac{a_3+a_1r+2t}{u^3}. \end{align*} \begin{align*} a_4'=\frac{a_4+2a_2r+3r^2-2st-a_1t-a_1rs-a_3s}{u^4}. \end{align*} \begin{align*} a_6'=\frac{a_6+a_4r+a_2r^2+r^3-t^2-a_1rt-a_3t}{u^6}. \end{align*} Each numerator belongs to $k$, and each denominator is invertible in $k$, so \begin{align*} a_1',a_2',a_3',a_4',a_6' \in k. \end{align*}[/step]

[guided]Let $a_1,a_2,a_3,a_4,a_6 \in k$ be the coefficients of the original long Weierstrass equation \begin{align*} y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6. \end{align*} Throughout this guided computation, $X$ denotes the new variable $x'$ and $Y$ denotes the new variable $y'$. The right-hand side contributes only pure powers of $X$, because $x=u^2X+r$ contains no $Y$ term. Expanding in the polynomial ring $k[X,Y]$ gives \begin{align*} x^3 = u^6X^3+3u^4rX^2+3u^2r^2X+r^3, \end{align*} \begin{align*} a_2x^2 = a_2u^4X^2+2a_2u^2rX+a_2r^2, \end{align*} and \begin{align*} a_4x = a_4u^2X+a_4r. \end{align*} Adding the constant term $a_6$ and collecting equal powers of $X$ yields \begin{align*} x^3+a_2x^2+a_4x+a_6 = u^6X^3+u^4(a_2+3r)X^2+u^2(a_4+2a_2r+3r^2)X+a_6+a_4r+a_2r^2+r^3. \end{align*} Now we compare this with the left-hand-side expansion. The terms involving $Y$ must stay on the left, because the target long Weierstrass form has $Y^2$, $XY$, and $Y$ on the left. The pure $X$ terms from the left-hand side are moved to the right by subtraction, giving \begin{align*} u^6Y^2+u^5(2s+a_1)XY+u^3(2t+a_1r+a_3)Y = u^6X^3+u^4(a_2+3r-s^2-a_1s)X^2+u^2(a_4+2a_2r+3r^2-2st-a_1t-a_1rs-a_3s)X+a_6+a_4r+a_2r^2+r^3-t^2-a_1rt-a_3t. \end{align*} The definition of a Weierstrass change of variables requires $u \in k^\times$, so $u$ is invertible in $k$. Hence $u^6$ is also invertible in $k$. Dividing every term by $u^6$ normalizes the coefficients of $Y^2$ and $X^3$ to $1$, which is exactly what is needed for long Weierstrass form. We obtain \begin{align*} Y^2+a_1'XY+a_3'Y=X^3+a_2'X^2+a_4'X+a_6', \end{align*} where the transformed coefficients are defined by \begin{align*} a_1'=\frac{a_1+2s}{u}, \end{align*} \begin{align*} a_2'=\frac{a_2+3r-s^2-a_1s}{u^2}, \end{align*} \begin{align*} a_3'=\frac{a_3+a_1r+2t}{u^3}, \end{align*} \begin{align*} a_4'=\frac{a_4+2a_2r+3r^2-2st-a_1t-a_1rs-a_3s}{u^4}, \end{align*} and \begin{align*} a_6'=\frac{a_6+a_4r+a_2r^2+r^3-t^2-a_1rt-a_3t}{u^6}. \end{align*} Each numerator is an expression formed from elements of $k$ using addition and multiplication, so each numerator lies in $k$. Since $u \in k^\times$, each denominator $u,u^2,u^3,u^4,u^6$ is invertible in $k$. Therefore \begin{align*} a_1',a_2',a_3',a_4',a_6' \in k. \end{align*}[/guided]

[step:Return to the original primed notation] Replacing $X$ by $x'$ and $Y$ by $y'$, the transformed equation is \begin{align*} (y')^2+a_1'x'y'+a_3'y'=(x')^3+a_2'(x')^2+a_4'x'+a_6', \end{align*} with $a_1',a_2',a_3',a_4',a_6' \in k$. This is a long Weierstrass equation over $k$, so the Weierstrass change of variables preserves long Weierstrass form. [/step]