The tangent space of the parabola at a general point and the local parameter

Setup: $k$ is a field and $a\in k$; the iterated ring $k[x,y]=k[x][y]$ is a commutative ring by [F1]; the polynomial $f=y-x^2$ is nonzero because in its unique expansion [F2] the coefficient of $y$ is $1\mathrm{\ne }0$ in $k$ by [F18]; $I=(f)$ is the principal ideal generated by $f$ in the sense of [F3], $A=k[x,y]\mathrm{/}I$ is the quotient ring, $C=\mathrm{Spec}A\subseteq {\mathbb{A}}_k^2$ is the corresponding closed subscheme, and the point $P=(a,a^2)$ satisfies $f(P)=a^2-a^2=0$; the evaluation map $\epsilon :k[x,y]\to k$, $g\mapsto g(a,a^2)$ has kernel $(x-a,y-a^2)$ by [F7].

The substitution homomorphism and the structure of $A$: by [F11] with the coefficient map $k\to k[x]$ and assigned images $x\mapsto x$, $y\mapsto x^2$ there is a unique $k$-algebra homomorphism $\sigma :k[x,y]\to k[x]$ with $\sigma (x)=x$ and $\sigma (y)=x^2$, and $\sigma$ is surjective because it is the identity on the subring $k[x]$; for an arbitrary $g\in k[x,y]=k[x][y]$, division by the monic polynomial $f=y-x^2$ of degree one in $y$ over the commutative ring $k[x]$ gives, by [F10], unique $q,r\in k[x,y]$ with $g=qf+r$ and $r=0$ or ${\mathrm{deg}}_{y}r<1$, so that $r\in k[x]$, and applying $\sigma$ gives $\sigma (g)=r$ because $\sigma$ fixes $k[x]$ pointwise and $\sigma (f)=x^2-x^2=0$; hence $g\in \ker \sigma$ if and only if $r=0$, if and only if $g\in (f)=I$, so $\ker \sigma =I$ and by [F13] the quotient $A=k[x,y]\mathrm{/}I$ is isomorphic to $k[x]$ as a $k$-algebra, with $\bar{x}\mapsto x$ and $\bar{y}\mapsto x^2$.

Consequences for $A$: $k$ is an integral domain by [F18], so $k[x]$ is an integral domain by [F12] and hence $A\cong k[x]$ is an integral domain; the element $t=\bar{x}-a\in A$ is nonzero, because its image $x-a$ in $k[x]$ has $x$-coefficient $1\mathrm{\ne }0$ in the unique expansion [F2]; therefore $t$ is a nonzerodivisor in $A$, so $tc=0$ with $c\in A$ forces $c=0$.

The point $P$ and its maximal ideal: since $f(P)=0$ by step 1.1, the polynomial $f$ lies in the kernel $(x-a,y-a^2)$ of $\epsilon$, so [F8] factors $\epsilon$ uniquely through the quotient as a $k$-algebra homomorphism $\delta :A\to k$; $\delta$ is surjective because $\epsilon$ is, so by [F13] one has $A\mathrm{/}\mathfrak{m}\cong k$ for $\mathfrak{m}=\ker \delta$, and $\mathfrak{m}$ is the image of $(x-a,y-a^2)$ in $A$, namely $(\bar{x}-a,\bar{y}-a^2)$; in $A$ one has $\bar{y}={\bar{x}}^2$, so distributing over the difference [F1] gives $\bar{y}-a^2={\bar{x}}^2-a^2=(\bar{x}-a)(\bar{x}+a)$, and hence $\mathfrak{m}=(\bar{x}-a)$ is the principal ideal generated by $t=\bar{x}-a$ in the sense of [F3]; by [F9] the $k$-algebra homomorphism $\delta$ exhibits $P$ as the $k$-rational point of $C$ with maximal ideal $\mathfrak{m}$.

The cotangent space $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$: with $t=\bar{x}-a$ and $\mathfrak{m}=(t)$, the product ideal ${\mathfrak{m}}^2=\mathfrak{m}\cdot \mathfrak{m}$ equals $(t^2)$ by [F4]; the multiplication map $\mu :A\mathrm{/}(t)\to \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$, $[c]\mapsto [ct]$, is well defined because $c\in (t)$ implies $ct\in (t^2)={\mathfrak{m}}^2$, is surjective because $\mathfrak{m}=(t)$ consists exactly of the elements $ct$, and is injective because $ct=t^{2}d$ for some $d\in A$ gives $t(c-td)=0$, which the domain $A$ with $t\mathrm{\ne }0$ of step 3.1 turns into $c=td\in (t)$; hence $\mu$ is an isomorphism of $k$-vector spaces, and since the surjective map $\delta :A\to k$ of step 4.1 has kernel $(t)$, [F13] gives $A\mathrm{/}(t)\cong k$, so that $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\cong k$ is one-dimensional with $k$-basis the class $[t]$ of the local parameter $t=\bar{x}-a$.

The Jacobian matrix and the tangent space: $I=(y-x^2)$ has the finite generating list $(f)$, and every element of $I$ is a multiple of $f$ with $f(P)=0$, so the Jacobian matrix of this list at $P=(a,a^2)$ is the $1\times 2$ matrix of formal partial derivatives $(\mathrm{\partial }f\mathrm{/}\mathrm{\partial }x,\mathrm{\partial }f\mathrm{/}\mathrm{\partial }y)$ evaluated at $P$ by [F5]; since $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }x=-2x$ and $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }y=1$, the matrix is $(-2a,1)$ with the integer $2$ read in $k$, and a vector $(u,v)\in k^2$ lies in its kernel exactly when $-2au+v=0$, that is $v=2au$, so $\ker J=\{(u,2au):u\in k\}=k\cdot (1,2a)$ is one-dimensional because the first coordinate $1$ of $(1,2a)$ is nonzero in $k$ by [F18]; by [F6] the coordinate-velocity map gives a canonical $k$-linear isomorphism $T_{P}C\cong \ker J$, hence $T_{P}C\cong \{(u,2au):u\in k\}$; in the translated coordinates $(X-a,Y-a^2)=(u,v)$ the tangent line of the parabola at $P$ is therefore $v=2au$, that is $Y-a^2=2a(X-a)$, Milne's equation (20) for $F=Y-X^2$; if the characteristic is $2$, so that $2=0$ in $k$ by [F17], the same formula gives the row $(0,1)$, the kernel $k\cdot (1,0)=\{(u,0)\}$, and the horizontal tangent line $Y=a^2$.

Intrinsic cotangent space and duality: the local ring of $C$ at $P$ is ${\mathcal{O}}_{C,P}=A_{\mathfrak{m}}$ with maximal ideal $\mathfrak{n}=\mathfrak{m}{A}_{\mathfrak{m}}$, and the intrinsic cotangent space is $C_{P}C=\mathfrak{n}\mathrm{/}{\mathfrak{n}}^2$ by [F14]; the maximal ideal $\mathfrak{m}$ satisfies $A\mathrm{/}\mathfrak{m}=k$ via the structure map by step 4.1, so [F15] with $r=1$ applies and its canonical map ${\theta }_1:\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\to \mathfrak{n}\mathrm{/}{\mathfrak{n}}^2$ is an isomorphism, giving $C_{P}C\cong \mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\cong k$ with the class of the local parameter $t$ as a $k$-basis; by [F16] the tangent space is the dual $T_{P}C={\mathrm{Hom}}_k(C_{P}C,k)$, so $T_{P}C\cong {\mathrm{Hom}}_k(k,k)\cong k$ is one-dimensional, in agreement with the kernel $k\cdot (1,2a)$ of step 5.2.

Conclusion: for the parabola $C=\mathrm{Spec}(k[x,y]\mathrm{/}(y-x^2))$ over any field $k$ and every $a\in k$, at $P=(a,a^2)$ the tangent space is $T_{P}C\cong \{(u,2au):u\in k\}=k\cdot (1,2a)$, equivalently the tangent line $Y-a^2=2a(X-a)$ through $P$, and the maximal ideal $\mathfrak{m}=(\bar{x}-a)$ is generated by the local parameter $\bar{x}-a$, whose class is a $k$-basis of $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\cong C_{P}C\cong k$; both spaces are one-dimensional, in every characteristic, with no perfectness, algebraic-closedness or reducedness hypothesis.

Boundary and scope dispositions: $C$ is nonempty because $P$ is a $k$-rational point of it by step 4.1, $k$ is nonempty since it is a field with $0\mathrm{\ne }1$ by [F18], and the cotangent space $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2\cong k$ is nonzero by step 5.1, so no object here is empty and the case of an empty instance does not arise; the zero cases are the vertex $a=0$, where the tangent space is $k\cdot (1,0)=\{(u,0)\}$ and the tangent line is $Y=0$, the vanishing $f(P)=0$ of step 1.1, the vanishing of the entries $-2a$ and $2a$ when $a=0$, and the nonzero class $[t]$ of step 5.1; there is one equation, one generator of $I$, one Jacobian row, one local parameter $t$ generating $\mathfrak{m}$, one basis class $[t]$ and one point $P$, and the tangent space has dimension one rather than zero; the degenerate case is characteristic $2$, where $(-2a,1)=(0,1)$, $T_{P}C=k\cdot (1,0)$ and the formula $v=2au$ collapses to $v=0$, and since no characteristic hypothesis is used or needed elsewhere, all fields and all $a$ are covered; the endpoints of the parameter are $a=0$, the vertex, where the tangent is horizontal, and the values with $2a\mathrm{\ne }0$, where it is not, while the degree endpoints of $f=y-x^2$ are its least degree $1$ and top degree $2$, differentiated in step 5.2 to the constant $1$ and the linear polynomial $-2x$; the two directions of the membership equivalence $\ker \sigma =I$ are proved in step 2.1, the forward one because $\sigma (g)=0$ forces $r=0$ and hence $g=qf\in I$, and the reverse one because $g=hf$ gives $\sigma (g)=\sigma (h)\sigma (f)=0$, while both inclusions of the ideal equality $\mathfrak{m}=(\bar{x}-a)$ are proved in step 4.1, since $t$ is the first generator and conversely every element of $(x-a,y-a^2)$ is $t(a_1+(\bar{x}+a)a_2)$, so the two iff axes are checked rather than vacuous; no Axiom of Choice or dependent choice is used, since $a$, $f$, the substitution $\sigma$, the division, the class $[t]$ and the Jacobian row are all exhibited explicitly and every cited supplier is choice-free, with no def-axiom-of-choice dependency declared.