Riemann surface
matlab

Let $V$ be the parabola in $\mathbb{R}^2$ given by the equation $y=x^2$. Let $P=\left(a, a^2\right)$ be a point of $V$. (I don’t mean that you should choose a specific value of $a$.)
a) Find a polynomial $f$ so that $V=\mathbb{V}(f)$. [Hint: this is as easy as it looks. Don’t look for anything tricky here.]

Solution: $f=y-x^2$.

b) Find a polynomial $\ell$ so that $\mathbb{V}(\ell)$ is the tangent line to $V$ at $P$.

Solution: We use methods from calculus. Since $\frac{d}{d x} x^2=2 x$, the slope of the tangent line at $P$ is 2a. So the tangent line is
$$y-a^2=2 a(x-a), \quad \text { i.e. } \quad y=2 a x-a^2 .$$
So $\ell=2 a x-y-a^2$.

c) Prove directly that $\langle\ell, f\rangle$ is not a radical ideal. That is, find a polynomial $g$ such that some power of $g$ is in $\langle\ell, f\rangle$ but $g$ itself is not. Be sure to show all your work: prove that some power of $g$ is in this ideal (what power?), and prove that $g$ itself is not in the ideal. [Hint: look at vertical lines for one possible answer.]

Solution:
\begin{aligned} \left\langle 2 a x-y-a^2, y-x^2\right\rangle & =\left\langle y-x^2,\left(2 a x-y-a^2\right)+\left(y-x^2\right)\right\rangle \ & =\left\langle y-x^2, 2 a x-x^2-a^2\right\rangle \ & =\left\langle y-x^2, x^2-2 a x+a^2\right\rangle \ & =\left\langle y-x^2,(x-a)^2\right\rangle \end{aligned}
Take $g=x-a$. Then we have just shown that $g^2 \in\langle\ell, f\rangle$. We have to show that $g$ itself is not in $\langle\ell, f\rangle$. But $\langle\ell, f\rangle=\left\langle y-x^2,(x-a)^2\right\rangle$, and the equation
$$h_1\left(y-x^2\right)+h_2(x-a)^2=x-a$$
can be rewritten as
$$\left(h_2-h_1\right) x^2-2 a h_2 x+h_1 y+h_2 a^2=x-a,$$

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