Three axes are not three coplanar lines

The rings $A_V=k[x,y,z]\mathrm{/}(xy,yz,zx)$ and $A_W=k[u,v]\mathrm{/}I$ are commutative $k$-algebras presented as quotients of polynomial rings by the ideals of [F4], and $V=\mathrm{Spec}{A}_V$ and $W=\mathrm{Spec}{A}_W$ are affine $k$-schemes with structure maps to $\mathrm{Spec}k$ by [F10]; the two ideals are the actual ideals of the presentation, not a replacement of them.

The $k$-rational points of $V$: for $(a,b,c)\in k^3$ the evaluation homomorphism at $(a,b,c)$ is the unique $k$-algebra map with $x\mapsto a$, $y\mapsto b$, $z\mapsto c$ by [F2], its kernel $(x-a,y-b,z-c)$ is maximal by [F7], and it kills $(xy,yz,zx)$ exactly when $ab=bc=ca=0$ because a ring map is multiplicative [F1]; then [F8] factors it through a $k$-algebra map $A_V\to k$, which by [F9] is a $k$-rational point of $V$, and conversely every $k$-rational point arises from such a triple by [F9] and [F2]. If $ab=bc=ca=0$ and $a\mathrm{\ne }0$, then $b=a^{-1}(ab)=0$ and $c=a^{-1}(ca)=0$ by [F17], and symmetrically; hence at most one coordinate is nonzero, and the $k$-rational points of $V$ are exactly the origin together with the points of the three coordinate axes.

The $k$-rational points of $W$: by [F9] and [F2] a $k$-rational point $q$ of $W$ corresponds to a pair $(a,b)\in k^2$ whose evaluation map $k[u,v]\to k$ has $g_i(a,b)=0$ for every $i$, because [F8] factors this evaluation through $A_W$ exactly when it kills $I=(g_1,\dots ,g_r)$ as a multiplicative map [F1]; for the particular list $(uv(u-v))$ the condition reads $ab(a-b)=0$, which holds exactly when $a=0$ or $b=0$ or $a=b$ by [F17], so the $k$-rational points of that $W$ are exactly the points of the three lines $u=0$, $v=0$ and $u=v$.

The tangent bound in the plane: for every $k$-rational point $q$ of $W$ the ideal $I$ has the finite generating list $(g_1,\dots ,g_r)$, so [F6] gives a canonical $k$-linear isomorphism $T_{q}W\cong \ker J(q)$ with $J(q)$ the $r\times 2$ matrix of [F5]; the kernel of the $k$-linear map $J(q)$ is a linear subspace of $k^2$ by [F15], whence ${\dim }_k\ker J(q)\le {\dim }_k{k}^2=2$ by [F16] and [F14]; therefore ${\dim }_k{T}_{q}W\le 2$ at every $k$-rational point of $W$, independently of the list and of the field.

The tangent space of $V$ at the origin: the Jacobian matrix of the list $(xy,yz,zx)$ at $o$ is the $3\times 3$ matrix with rows $(y,x,0)$, $(0,z,y)$, $(z,0,x)$ by [F5], and every entry is $0$ at the origin, so $J(o)$ is the zero matrix; by [F6] the tangent space is $T_{o}V\cong \ker (0:k^3\to k^3)=k^3$, and ${\dim }_k{T}_{o}V=3$ by [F13] and [F14], since the standard basis of $k^3$ has three elements and ${\dim }_k{k}^3=3$.

No isomorphism: suppose $\phi :V\to W$ is an isomorphism of $k$-schemes over $k$; its inverse ${\phi }^{-1}$ is again a $k$-morphism by [F10], and $q:=\phi (o)$ is a $k$-rational point of $W$ because $o$ is a $k$-rational point of $V$ and $\phi$ is a $k$-morphism, composition of morphisms being [F11]; applying the identity and chain rules of [F12] to ${\phi }^{-1}\circ \phi ={\mathrm{id}}_V$ and $\phi \circ {\phi }^{-1}={\mathrm{id}}_W$ at the $k$-rational points $o$ and $q$ gives $d_q{\phi }^{-1}\circ d_o\phi ={\mathrm{id}}_{T_{o}V}$ and $d_o\phi \circ d_q{\phi }^{-1}={\mathrm{id}}_{T_{q}W}$, so $d_o\phi :T_{o}V\to T_{q}W$ is a $k$-linear isomorphism; hence ${\dim }_k{T}_{q}W={\dim }_k{T}_{o}V=3$ by [F13] and step 3.1, contradicting ${\dim }_k{T}_{q}W\le 2$ from step 2.3. Thus there is no isomorphism $V\to W$ over $k$.

Milne's three-line configuration: take the single equation $g_1=uv(u-v)$, so $r=1$ and $W=\mathrm{Spec}(k[u,v]\mathrm{/}(uv(u-v)))$; the three factors $u$, $v$ and $u-v$ are linear forms, hence homogeneous of degree $1$ with no constant term by [F3], so each of the three lines $u=0$, $v=0$, $u=v$ passes through the origin; the lines are pairwise distinct, witnessed by the $k$-rational points of step 2.2 that lie on one line and on no other: $(0,1)$ has $u=0$ while $v=1$ and $u-v=-1$, $(1,0)$ has $v=0$ while $u=1$ and $u-v=1$, and $(1,1)$ has $u=v$ while $u=1$ and $v=1$, all these values being nonzero by [F17]; so $W$ is the union of the three distinct lines $u=0$, $v=0$ and $u=v$ through the origin, and step 4.1 shows that $V$ is not isomorphic to it over $k$.

Boundary and scope dispositions: $V$ and $W$ are nonempty, the origin being a $k$-rational point of both by steps 2.1 and 2.2, and the particular three-line $W$ of step 5.1 contains the three lines; the zero cases are the zero Jacobian matrix of step 3.1 with kernel $k^3$, the zero vector of each tangent space, the empty list $r=0$ for which $W={\mathbb{A}}_k^2$ and the bound of step 2.3 still holds, and the vanishing coordinate cases $a=0$, $b=0$, $a=b$ of step 2.2; there is one equation $uv(u-v)$ and one Jacobian row in the configuration of step 5.1, the general bound of step 2.3 using the displayed finite list; the degenerate configurations, such as a repeated line or the whole plane, are not excluded but only strengthen the bound, which uses no distinctness of the lines, distinctness being verified separately in step 5.1 for the three-line instance; the endpoints of the comparison are the two ambient dimensions three and two, realized as ${\dim }_k{T}_{o}V=3$ in step 3.1 and as the bound ${\dim }_k{T}_{q}W\le 2$ in step 2.3; no Axiom of Choice or dependent choice is used, since the finite lists, the evaluation points, the matrices and the isomorphism are single given data and every cited supplier, including the functoriality lemma [F12] and the Jacobian kernel theorem [F6], is choice-free; and no biconditional is asserted as a claim, the only equivalences used being the correspondence between $k$-rational points and $k$-algebra maps of [F9] and the one-directional contradiction argument of step 4.1.