What do ideals of $\mathbb{C}[x,y]$ containing $(y-x^2)$ look like? I encountered this statement in my ring theory course in the section on the correspondence theorem:

Every ideal of the polynomial ring $\mathbb{C}[x, y]$ that contains $y − x^2$ has the form
$I = (y − x^2,p(x))$, for some polynomial $p(x)$.

In our course we have defined the ideal $(a_1,\dots,a_n)$ as
$$(a_1,\dots,a_n) = \{a_1x_1+\dots+a_nx_n\mid x_{i}\in R\}$$
where $R$ is the ring in question.
I fail to see why the statement is true, however. Consider the ideal $(y-x^2, y^2-x)$, which contains $(y-x^2)$. Isn't this in direct contradiction to the above? Why do we have to be limited only to a single variable polynomial?
 A: We have an isomorphism
$$
\mathbb C[x,y]/(y-x^2)\to\mathbb C[x];\quad x\mapsto x,\quad y\mapsto x^2.
$$
An ideal $I\subseteq\mathbb C[x,y]$ which contains $y-x^2$ corresponds to an ideal of $\mathbb C[x,y]/(y-x^2)$, hence to an ideal $J\subseteq \mathbb C[x]$.
Since $\mathbb C[x]$ is a PID, we can write $J=(p)$ for some $p$.
Then it follows that $I=(y-x^2,p(x))$.
About the example you mentioned, we have $(y-x^2,y^2-x)=(y-x^2,x^4-x)$.
A: Question: "Every ideal of the polynomial ring $\mathbb C[x,y]$ that contains $y−x^2$ has the form $I=(y−x^2,p(x))$, for some polynomial $p(x)$."
Answer: There is for every $F(x,y)\in k[x,y]$ an equality
$$
F(x,y)=F(x,x^2+(y-x^2))=F_0(x)+F_1(x)(y-x^2)+\dotsm +F_d(x)(y-x^2)^d
$$
hence
$$
F(x,y)=f_0(x)+f_1(x,y)(y-x^2)
$$
for polynomials $f_0(x) \in k[x]$ and $f_1(x,y)\in k[x,y]$. Hence for any ideal
$$
(y-x^2) \subseteq J:=(y-x^2,F_1(x,y),\dotsc,F_k(x,y))
$$
you may write
$$
F_i(x,y)=f_i(x)+g_i(x,y)(y-x^2)\tag{*}
$$
and hence
$$
J=(y-x^2,f_1(x),... ,f_k(x))
$$
is an equality of ideals. The ideal $I:=(f_1(x),..,f_k(x))\subseteq k[x,y]$ is generated by one element $(f(x))$ (here we use that $k[x]$ is a PID) and it follows $J=(y-x^2,f(x))$.
Hence you may choose $p(x):=f(x)$ where $(f_1(x),..,f_k(x))=(f(x)) \subseteq k[x]$ in the decomposition (*) - $f(x)$ is a generator for the ideal $(f_1,..,f_k) \subseteq k[x]$.
Example: The inclusion $(y-x^2) \subseteq (x-a,y-a^2)$ for $a\in k$.
We get
$$y-a^2=y-x^2+x^2-a^2=y-x^2+(x-a)(x+a)$$
hence
$$(y-a^2,x-a)=(y-x^2+(x-a)(x+a),x-a)=(y-x^2,x-a)$$
and hence
$$(y-x^2) \subseteq (y-a^2,x-a).$$
Hence since  $k$ is a field it follows $\mathfrak{m}:=(y-a^2,x-a)\in Spec(k[x,y])(k)$ is a $k$-rational point for every $a\in k$ contained in $V(y-x^2)$. It follows
$$(y-a^2,x-a)\in Spec(k[x,y]/(y-x^2))(k).$$
Hence your ideal $(y-x^2) \subseteq k[x,y]$ is contained in a 1-dimensional family of maximal ideals $(x-a,y-a^2)$ with $a\in  k$.
Question: "I fail to see why the statement is true, however. Consider the ideal $(y−x^2,y^2−x)$, which contains $(y−x^2)$. Isn't this in direct contradiction to the above? Why do we have to be limited only to a single variable polynomial?"
Example:  Let $F_1(x,y)=y^2-x$. You get
$$F_1=-x+(x^2+y-x^2)^2= x^4-x+(x^2+y)(y-x^2)=f_1(x)+g_1(x,y)(y-x^2)$$
with $f_1(x):=x^4-x, g_1(x,y):=x^2+y$. Hence
$$(y-x^2,y^2-x)=(y-x^2,x^4-x+(x^2+y)(y-x^2))=(y-x^2,x^4-x).$$
Hence using the above "algorithm" you may choose $f(x)=x^4-x$.
