A nonradical ideal need not enlarge every tangent space

The ring $R=k[x,y]=k[x][y]$ is a commutative ring and an integral domain in which $1\mathrm{\ne }0$, every element of $R$ has a unique expression as a finite sum $f={\sum }_j{a}_j(x)y^j$ with coefficients $a_j\in k[x]$, and for every commutative ring $S$, every unital ring homomorphism $\phi :k\to S$ and every pair $s_1,s_2\in S$ there is a unique unital ring homomorphism $R\to S$ extending $\phi$ with $x\mapsto s_1$ and $y\mapsto s_2$.

The ideals $I=(x^2,xy)$ and $J=(x)$ of $R$ satisfy $I\subseteq J$ with $x^2,xy\in I$ and $x\in J$; every element of $I$ is a finite sum $a{x}^2+bxy$ and every element of $J$ is a multiple $xh$ with $a,b,h\in R$, and every ideal of $R$ is an additive subgroup closed under multiplication by elements of $R$.

The universal property gives unique unital ring homomorphisms ${\mathrm{e}\mathrm{v}}_{(0,1)},{\mathrm{e}\mathrm{v}}_{(0,0)},{\mathrm{e}\mathrm{v}}_{(1,0)}:R\to k$ with values $x\mapsto 0$, $y\mapsto 1$; $x\mapsto 0$, $y\mapsto 0$; $x\mapsto 1$, $y\mapsto 0$, and a unique homomorphism $\mathrm{e}\mathrm{v}:R\to k[y]$ with $x\mapsto 0$ and $y\mapsto y$; the restriction of $\mathrm{e}\mathrm{v}$ to $k[x]$ is the unique homomorphism $k[x]\to k$ with $x\mapsto 0$, so $\mathrm{e}\mathrm{v}(f)={\sum }_j{a}_j(0)y^j$ for $f={\sum }_j{a}_j(x)y^j$, where $a_j(0)$ denotes the image of $a_j$ under that map; since ${\mathrm{e}\mathrm{v}}_{(1,0)}(x)=1\mathrm{\ne }0$ and the unique homomorphism $k[y]\to k$ with $y\mapsto 1$ sends $y$ to $1\mathrm{\ne }0$, both $x\mathrm{\ne }0$ in $R$ and $y\mathrm{\ne }0$ in $k[y]$.

Each of ${\mathrm{e}\mathrm{v}}_{(0,1)}$, ${\mathrm{e}\mathrm{v}}_{(0,0)}$ and $\mathrm{e}\mathrm{v}$ sends $x$ to $0$ and hence kills $x^2$ and $xy$; their kernels are additive subgroups closed under multiplication by ring elements, hence ideals, so these kernels contain the generators $x^2,xy$ and therefore contain $I=(x^2,xy)$, and they contain $x$ and therefore contain $J=(x)$; in particular $I\subseteq \ker \mathrm{e}\mathrm{v}$.

The inclusion $I⊊J$ is strict: $x\in J$ by step 1.2 and $x\mathrm{\notin }I$, since $x\in I$ would give $x=a{x}^2+bxy=x(ax+by)$ for some $a,b\in R$, whence $x(1-ax-by)=0$; $R$ is an integral domain with $x\mathrm{\ne }0$ by steps 1.1 and 1.3, so $ax+by=1$, and applying ${\mathrm{e}\mathrm{v}}_{(0,0)}$ gives $0=1$ in $k$, contradicting $1\mathrm{\ne }0$.

The kernel of $\mathrm{e}\mathrm{v}$ is $J=(x)$: if $f={\sum }_j{a}_j(x)y^j$ lies in the kernel, then ${\sum }_j{a}_j(0)y^j=0$ in $k[y]$ by step 1.3, so every coefficient $a_j(0)$ is zero; dividing $a_j$ by the monic polynomial $x$ gives $a_j=q_{j}x+r_j$ with $r_j$ a constant, and applying the homomorphism $k[x]\to k$ with $x\mapsto 0$ gives $r_j=a_j(0)=0$, so $a_j=q_{j}x\in (x)$ and $f=x{\sum }_j{q}_j{y}^j\in (x)$; the reverse inclusion is step 2.1.

The points $(0,1)$ and $(0,0)$: by step 2.1 the homomorphisms ${\mathrm{e}\mathrm{v}}_{(0,1)}$ and ${\mathrm{e}\mathrm{v}}_{(0,0)}$ kill $I$ and $J$, so by the quotient universal property they induce $k$-algebra homomorphisms $R\mathrm{/}I\to k$ and $R\mathrm{/}J\to k$; under the affine anti-equivalence these are $k$-rational points of $X=\mathrm{Spec}(R\mathrm{/}I)$ and of $Y=\mathrm{Spec}(R\mathrm{/}J)$ with coordinate tuples $(0,1)$ and $(0,0)$, so the Jacobian-kernel theorem applies to both schemes at both points; also $I$ and $J$ are proper, because $I\subseteq \ker {\mathrm{e}\mathrm{v}}_{(0,1)}=(x,y-1)$ and $J\subseteq (x,y-1)$ with $(x,y-1)$ maximal by the evaluation-ideal lemma.

The radical of $I$ is $\sqrt{I}=J=(x)$: if $f\in \sqrt{I}$ then $f^n\in I$ for some $n\ge 1$ by the definition of the radical, and $I\subseteq \ker \mathrm{e}\mathrm{v}$ by step 2.1, so $\mathrm{e}\mathrm{v}(f{)}^n=\mathrm{e}\mathrm{v}(f^n)=0$ in the integral domain $k[y]$ and hence $\mathrm{e}\mathrm{v}(f)=0$, which by step 3.1 gives $f\in (x)=J$; conversely $x^2\in I$ by step 1.2 gives $x\in \sqrt{I}$, and each $f=xh\in J$ has $f^2=x^2{h}^2\in I$ because $I$ is an ideal containing $x^2$, so $f\in \sqrt{I}$ and $J\subseteq \sqrt{I}$.

The Jacobian matrix of the two equations $x^2,xy$ has rows $({\mathrm{\partial }}_x{x}^2,{\mathrm{\partial }}_y{x}^2)=(2x,0)$ and $({\mathrm{\partial }}_{x}xy,{\mathrm{\partial }}_{y}xy)=(y,x)$; at the point $(0,1)$ these rows are $(0,0)$ and $(1,0)$, whose common kernel is $\{v:v_1=0\}=k\cdot (0,1)$, of dimension one, while at the origin both rows are zero and the kernel is all of $k^2$, of dimension two; since $(0,1)$ and $(0,0)$ are $k$-rational points of $X$ by step 3.2, the coordinate-velocity isomorphism of the Jacobian-kernel theorem gives $T_{(0,1)}X\cong k\cdot (0,1)$ and $T_{(0,0)}X\cong k^2$.

The Jacobian matrix of the single equation $x$ is the row $({\mathrm{\partial }}_{x}x,{\mathrm{\partial }}_{y}x)=(1,0)$, the same at $(0,1)$ and at $(0,0)$, with kernel $\{v:v_1=0\}=k\cdot (0,1)$, of dimension one; the Jacobian-kernel theorem, applicable by step 3.2, gives $T_{(0,1)}Y\cong k\cdot (0,1)\cong T_{(0,0)}Y$ for $Y=\mathrm{Spec}(R\mathrm{/}J)$.

Comparison of the two computations: at $(0,1)$ the two Jacobian kernels are the same subspace $k\cdot (0,1)$ of $k^2$, so under the coordinate-velocity identifications the tangent spaces $T_{(0,1)}X$ and $T_{(0,1)}Y$ coincide as one-dimensional spaces, and the nonradical ideal $I$ does not enlarge the tangent space there; at the origin the kernel $k^2$ of the Jacobian of $I$ strictly contains the kernel $k\cdot (0,1)$ of the Jacobian of $J$, so $T_{(0,0)}X=k^2$ is two-dimensional and strictly larger than the one-dimensional $T_{(0,0)}Y=k\cdot (0,1)$.

Local reason for the coincidence: the point $(0,1)$ is the closed point $\mathfrak{m}\mathrm{/}I$ of $X$ with $\mathfrak{m}=(x,y-1)=\ker {\mathrm{e}\mathrm{v}}_{(0,1)}$, and $y\mathrm{\notin }\mathfrak{m}$, since otherwise $1=y-(y-1)\in \mathfrak{m}$ against properness; hence $y$ is inverted in $R_{\mathfrak{m}}$ and is a unit of $R_{\mathfrak{m}}$, so $I{R}_{\mathfrak{m}}=J{R}_{\mathfrak{m}}$: the inclusion $I{R}_{\mathfrak{m}}\subseteq J{R}_{\mathfrak{m}}$ follows from $I\subseteq J$ in step 1.2, while $x=y^{-1}(xy)\in I{R}_{\mathfrak{m}}$ with $xy\in I$ gives $(x)R_{\mathfrak{m}}\subseteq I{R}_{\mathfrak{m}}$; the local rings of the two closed subschemes at the corresponding closed points are ${\mathcal{O}}_{X,x}\cong (R\mathrm{/}I{)}_{\mathfrak{m}\mathrm{/}I}\cong R_{\mathfrak{m}}\mathrm{/}I{R}_{\mathfrak{m}}=R_{\mathfrak{m}}\mathrm{/}J{R}_{\mathfrak{m}}\cong (R\mathrm{/}J{)}_{\mathfrak{m}\mathrm{/}J}\cong {\mathcal{O}}_{Y,x}$ by the affine stalk identification and the localisation-quotient isomorphism, and since the cotangent space and the tangent space at a point are computed from its local ring alone, this is why the tangent spaces at $(0,1)$ coincide, while at the origin the distinct tangent dimensions of steps 4.2 and 4.3 show that the local rings there are not isomorphic.

Conclusion: by steps 2.2 and 4.1 the ideal $I=(x^2,xy)$ is strictly smaller than its radical $\sqrt{I}=(x)=J$, so $Y=\mathrm{Spec}(R\mathrm{/}J)$ is the reduced subscheme $\mathrm{Spec}(R\mathrm{/}\sqrt{I})$ of $X=\mathrm{Spec}(R\mathrm{/}I)$; over an arbitrary field $k$, at the $k$-rational point $(0,1)$ the tangent spaces of $X$ and $Y$ coincide with the common line $k\cdot (0,1)$ of dimension one, while at the origin $T_{(0,0)}X=k^2$ strictly contains $T_{(0,0)}Y=k\cdot (0,1)$; hence a nonradical ideal does not always enlarge the tangent space, and the false claim stated above is refuted by this single ideal; the computation is valid in every characteristic, the integer coefficients of the formal derivatives being read in $k$ and $1\mathrm{\ne }0$ in $k$, and no reduction of $I$ is performed.

Boundary and scope dispositions: $X$ and $Y$ are nonempty, since $(0,1)$ and $(0,0)$ are $k$-rational points of both by step 3.2; the zero case appears at the origin, where the Jacobian rows of $I$ vanish identically and the kernel is all of $k^2$ by step 4.2, the zero velocity lying in every kernel as a subspace; $J$ is generated by the single element $x$ with the one-dimensional tangent spaces of step 4.3, while $I$ needs the two generators $x^2,xy$; the example is itself the degenerate case of a nonradical ideal, $I⊊\sqrt{I}$ by steps 2.2 and 4.1, with all nilpotents of $R\mathrm{/}I$ retained and no reduction performed, the equality at $(0,1)$ in step 5.1 showing that the nilpotent direction is invisible there; the points $(0,1)$ and $(0,0)$ are the two extreme cases $x=0,y=1$ and $x=y=0$ of the reduced line $x=0$, and characteristic $2$ is included because the entry $2x$ of the Jacobian is evaluated at $x=0$, where it vanishes by the formal monomial rule in every characteristic; no Axiom of Choice or dependent choice is used, all objects being exhibited explicitly from the single field $k$.