Showing ${y=x^2}$ has dimension ${1}$ in ${\mathbb{A}^2(\mathbb{R})}$ from definition? I would like to show that the variety $X$ defined by $X=V(y-x^2)$ in $\mathbb{A}^2(\mathbb{R})$  has dimension $1$ using the definition that the dimension of a variety is the maximal length of chains of non-empty distinct proper sub-varieties. Of course it's easy to see the dimension is at least $1$ (as picking any single point on ${y=x^2}$ gives at least $1$ such sub-variety) but I'm not sure how to go about proving this is maximal. Any tip would be appreciated.
 A: Here is another possible solution. $V(y-x^2)$ is isomorphic to $\Bbb{A}^1_{\Bbb{R}}$ by the map $t\mapsto (t,t^2)$. Dimension is an isomorphism invariant, and so if you know that $\Bbb{A}^1_{\Bbb{R}}$ is dimension $1$, you are done.
With this being said, the solution using the Hauptidealsatz is more general.
A: So you agree that the affine plane $\mathbf{A}^2(\mathbf{R})$ has dimension $2$? Then Krulls Hauptidealsatz tells you that the considering the zero locus $X=V(x^2-y)\subset \mathbf{A}^2$ defined by the equation (= principal proper ideal) $(y-x^2) \subset \mathbf{R}[x,y]$ has exactly 1 dimension less, hence is of dimension 1.
More concretely, you are asking what is $$\text{dim}(\mathbf{R}[x,y]/(x^2-y))$$
So this ring has obviously dimension $\leq 2$. Now it is not zero dimensional, because you have a non trivial chain of primes $(x^2-y) \subseteq (x-a,y-a^2)$ for $a \in \mathbf{R}$ (these maximal ideals are the points on the parabola). It also doesn't have dimension 2 because you loose a dimension by Krulls Hauptidealsatz.
So by imposing this relation on $x,y$ you lose one dimension. You get the parabola X, a curve in the plane!
A: Edit: OK so I think my idea works now. In fact you can show that any proper algebraic subset of ${V(y - f(x))}$, where ${f(x)}$ is some polynomial in ${k[x]}$, is either empty or finite. From this it follows that the dimension of any hypersurface of the form ${y = f(x)}$ where ${f \in k[x]}$ is $1$.
Suppose ${V_* \subseteq V(y - f(x))}$ is a non-empty algebraic subset (since ${\emptyset}$ is of course always an algebraic subset). Write ${V_* = V(S)}$ for some set of polynomials ${S \subset k[x,y]}$. Suppose ${F\in S}$ is some polynomial in $S$. Then we know that every point in ${V_*}$ is of the form ${(a,f(a))}$ for some $a$, and ${F(a,f(a)) = 0}$. Note that ${F(x,f(x))}$ is actually a polynomial now in one variable. Thus if ${F(x,f(x))}$ has degree ${n > 0}$, then ${V_*}$ can thus contain at most $n$ points. Otherwise, if ${F(x,f(x))}$ is identically $0$ for every ${F \in S}$, then it must be the case that ${V(y-f(x)) \subseteq V_*}$. And so ${V_* = V(y-f(x))}$. And we are done.
