Fibers of Spec maps: is $\operatorname{Spec} \Bbb C[x,y]/(x^3+x^2y+xy^2)\to\operatorname{Spec} \Bbb C[x]$ finite?

Question

Let $R = \mathbb{C}[x,y]/(x^3+x^2y+xy^2)$. Look at $R$ as an $\mathbb{C}[x]$ algebra via the trivial map (I presume $\mathbb{C}[x] \to R$ with $x\mapsto x$). Is this a finite algebra? Translate the algebra into a geometric map $$ V(x^3+x^2y+xy^2) \to \mathbb{C}^2 \to \mathbb{C}^1 $$ are the fibers all finite?

I'm a bit stuck with this problem. I think the map is not integral because the generating relation $x^3+x^2y+xy^2 =0$ gives a non monic polynomial in $y$ so $y$ is probably not integral but how can I prove this exactly?

I have more trouble with the second part. What are these maps? I know that the algebra induces a Spec map: $F:\operatorname{Spec} R \to \operatorname{Spec} \mathbb{C}[x]$ and I know that if the algebra would be finite this map would have just finite fibers. But how does this map connects to the map given aboth. What is the map aboth even doing and how can I check the fibers?

(Especially I'm not sure what is meant with $V(\cdot)$, we learnt this in two different contexts, sometimes it means the primes above the ideal and sometimes all point in $C^2$ vanishing on the ideal, what is meant here? Especially I know what the induced Spec map is in the first case, but not what the induced map should be in the second case).

Answer 1

Whether the fibers of this map are finite and whether this map is a finite map are actually two separate questions. The former condition is called "quasi-finiteness" and is not in general equivalent to the map being finite, as one can see from the example of $\Bbb C[x]\to \Bbb C[x,x^{-1}]$. This is not a finite map (any finite set of generators for $\Bbb C[x,x^{-1}]$ as a $\Bbb C[x]$-module must have a least power of $x$ in it, and there are powers of $x$ in $\Bbb C[x,x^{-1}]$ of arbitrarily negative degree) but it does have finite fibers: all fibers are either zero or one point. We always have that a finite map is quasi-finite, though.

Let's check if your map has finite fibers - if the answer is yes, we'll move on to checking if it is finite. To calculate the fiber above a point $x=a$, we take the fiber product of our map $\operatorname{Spec} \Bbb C[x,y]/(x^3+x^2y+xy^2)\to\operatorname{Spec} \Bbb C[x]$ with $\operatorname{Spec} \Bbb C[x]/(x-a)\to\operatorname{Spec}\Bbb C[x]$, giving $$\operatorname{Spec} \Bbb C[x,y]/(x^3+x^2y+xy^2) \times_{\operatorname{Spec} \Bbb C[x]} \operatorname{Spec} \Bbb C[x]/(x-a) \cong \operatorname{Spec} \Bbb C[y]/(a^3+a^2y+ay^2).$$ If $a\neq 0$, this is finite: we get two points (coming from the two solutions of a quadratic in $y$); if $a=0$ we get $\operatorname{Spec} \Bbb C[y]$ which is not finite. So your map is not quasi-finite and therefore not finite. (For more details on why the fiber product calculates the fiber, see for instance here.)

To confirm your thought that the map is not integral due to the generating relation, we show that $y\in\Bbb C[x,y]/(x^3+x^2y+xy^2)$ is not the root of any monic polynomial $p(t)\in (\Bbb C[x])[t]$: the ring homomorphism $\Bbb C[x][t] \to \Bbb C[x,y]$ by $t\mapsto y$ would send $p(t)$ to a polynomial in $\Bbb C[x,y]$ which is not divisible by $x$ (as $p$ is monic) and hence not in $(x^3+x^2y+xy^2)$. But the composition with the quotient map $\Bbb C[x,y]\to\Bbb C[x,y]/(x^3+x^2y+xy^2)$ should send $p(y)$ to zero - that's what it means to satisfy the polynomial, after all. This shows that $y\in\Bbb C[x,y]/(x^3,x^2y,xy^2)$ is not integral over $\Bbb C[x]$.

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As far as what is meant by $V(\cdot)$, the details depends on the context, but it's always the same underlying idea: stuff where $\cdot$ vanishes. In "classical" algebraic geometry, this is just the set of points that satisfies all the stuff in $\cdot$; with schemes this is the primes containing $\cdot$, and you can prove these are "the same": if $(a_1,\cdots,a_n)$ satisfies some equation $f(x_1,\cdots,x_n)$, then $f(x_1,\cdots,x_n)\subset (x_1-a_1,\cdots,x_n-a_n)$.