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以下是一些我们可以帮助您解决的问题：
仿射簇和射影簇：仿射簇和射影簇的定义、性质和分类，如代数簇的维度、奇异点的性质等。
态射和概型：态射的定义、性质和应用，概型的概念、构造和性质，如环上的概型、射影概型等。
切空间和余切空间：切空间和余切空间的定义和性质，切矢量和余切矢量的作用和计算。
仿射和射影坐标环：仿射和射影坐标环的定义和性质，坐标环的代数性质和几何性质。
代数曲线和代数曲面：代数曲线和代数曲面的定义和性质，代数曲线和曲面上的函数环和理想。
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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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