Finding singular points of $x^2=x^4 +y^4$ Locate the singular point of $x^2= x^4 + y^4$, assuming that $\operatorname{char} k \neq 2$. 
I am using the following definition: 

Let $Y \subset A_k^n$ be an affine variety, and let $f_1, \dots, f_t
> \in A =k[x_1, \dots, x_n]$ be a set of generators for the ideal $Y$.
  $Y$ is nonsingular at a point $p \in Y$ if the rank of the Jacobian is
  $n-r$ where $r$ is the dimension of $Y$.

In the example given, $Y=Z(x^2 = x^4 + y^4)$, and $I(Y) = (x^2 - x^4 - y^4)$. 
I think $(x^2, y^4)$ should generates $(x^2 - x^4 - y^4)$. Evaluating the Jacobian at $(0,0)$ will give a matrix of rank $0$. 
However, I need to show that $\operatorname{dim} Y \neq 2$. The dimension of $Y$ is equal to the dimension of $A(Y)= k[x_1, x_2] /(x^2-x^4-y^4)$. 
How do I compute the dimension of this coordinate ring? I am having a hard time using the definition of dimension: the supremum of the heights of all prime ideals. 
Furthermore, is there a way of making sure that the origin is not the only singular point? 
 A: Note that $\dim k[x,y]=2$. A prime ideal $I\subset k[x,y]$ is of height 2 if $I$ is a maximal ideal. Now the ideal $I=(y^4+x^4-x^2)$ is clearly not maximal since for example $I\subset (x,y)$. This shows that $\mathrm{ht}\: I\neq 2$. On the other hand $\mathrm{ht}\: I\neq 0$ because $k[x_1, x_2]$ is an integral domain. This shows that $\mathrm{ht}\: I = 1$, or $\dim k[x,y]/I=1$, becuase:
If $R$ is a ring, $\mathfrak{a}$ an ideal, then there is a one-to-one order preserving correspondence between prime ideals of $R/\mathfrak{a}$ and prime ideals of $R$ that contain $\mathfrak{a}$.
In general we have Krull principal ideal theorem which says: If $R$ is a Neotherian ring, $a\in R$ neither a unit, nor a zero-divisor, then the principal ideal $I=(a)$ has height one.
We also have: Let $k$ be a field. $R$ an integral domain which is a finitely generated $k$-algebra. Then for any prime ideal $\mathfrak{p}\subset R$ we have
$
\mathrm{ht}\: \mathfrak{p}+\dim R/\mathfrak{p}=\dim R
$.
However in this simple example of yours, as I've shown you don't need to know either of these theorems. Basically in your case if we denote by $f=y^4+x^4-x^2$. Then a point $(a,b)\in Y$ is singular if and only if 
$$
\frac{\partial f}{\partial x}(a,b)=\frac{\partial f}{\partial y}(a,b)=0
$$
which means the singular point is $(0,0)$.
