The parabola y=x^2 has coordinate ring k[t] and isomorphic intrinsic geometry to the affine line

Let $k$ be an algebraically closed field and let $X=V(y-x^2)\subseteq {\mathbf{A}}_k^2\mathrm{.}$ Division by the monic polynomial $y-x^2$ in the variable $y$ writes every $f\in k[x,y]$ uniquely as $f=q(y-x^2)+r(x)\mathrm{.}$ If $f$ vanishes on $X$, then $r(t)=f(t,t^2)=0$ for every $t\in k$. The field $k$ is infinite, so $r=0$ and $f\in (y-x^2)$. Hence $I(X)=(y-x^2)$ and $k[X]=k[x,y]\mathrm{/}(y-x^2)\mathrm{.}$ Define $\varphi :k[t]\to k[X]$ by $\varphi (t)=\bar{x}$. By the universal property of the polynomial ring, this is a $k$-algebra homomorphism, and every element of $k[X]$ can be written using only $\bar{x}$ because $\bar{y}={\bar{x}}^2$.

Conversely, define $\psi :k[X]\to k[t]$ on polynomial classes by substituting $x\mapsto t$ and $y\mapsto t^2$. The relation $y-x^2$ maps to $0$, so $\psi$ is well defined. The composites satisfy $\psi (\varphi (t))=t,\varphi (\psi (\bar{x}))=\bar{x},\varphi (\psi (\bar{y}))=\varphi (t^2)={\bar{x}}^2=\bar{y},$ so $\varphi$ and $\psi$ are inverse isomorphisms. Hence $k[X]\cong k[t]=k[{\mathbf{A}}_k^1]\mathrm{.}$

Set-theoretically, the maps $t⟼(t,t^2),(x,y)⟼x$ identify $X$ with the affine line, showing that the parabola is cut out by a nontrivial equation in the plane but has the same intrinsic affine geometry as ${\mathbf{A}}_k^1$.