Exercise II-3 from Eisenbud & Harris' The Geometry of Schemes

Question

I am having difficulty with the following exercise and would appreciate any help and explanations. Exercise II-3 on page 52 of The Geometry of Schemes by Eisenbud and Harris, but I give the setup first.

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The setup: We have $K$-algebra inclusions $K[x,y] \hookrightarrow K[x,y]_{(x,y)} \hookrightarrow K[[x,y]]$, where $K[x,y]$ is the commutative polynomial ring over $K$ in two variables, $K[x,y]_{(x,y)}$ is the polynomial ring localised at the maximal ideal $(x,y)$, and $K[[x,y]]$ the commutative formal power series ring in two variables. These maps induce maps on the spectra as follows: $$\operatorname{Spec}K[[x,y]]\to\operatorname{Spec}K[x,y]_{(x,y)}\to\operatorname{Spec}K[x,y].$$

Consider the prime ideal $(y^{2}-x^{3}-x^{2})$ of $K[x,y]$. This is still prime in $K[x,y]_{(x,y)}$. However, in $K[[x,y]]$ we now have $$y^{2}-x^{3}-x^{2}=(y-u)(y+u)$$ where $u=x+\frac{1}{2}x^2 - \frac{1}{8}x^3 + \frac{1}{16}x^4 - \cdots$.

Exercise II-3. (a) With $u=\sqrt{x^2 +x^3}$ as above, what is the image of $[(y-u)]$ in $\operatorname{Spec}K[x,y]$? (Hint: it's a prime ideal containing $y^2 -x^3 -x^2$.)

(b) Show that the image of the point $(y-\Sigma_{n\geq 1}\frac{x^n}{n!})$ of $\operatorname{Spec}K[[x,y]]$ is the generic point of $\mathbb{A}^2_K = \operatorname{Spec}K[x,y].$

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My attempt for (a): $K[x,y]_{(x,y)}$ is local with unique maximal ideal $(x,y)K[x,y]_{(x,y)}$ and has Krull dimension $2 = \operatorname{ht}(x,y)$. $(y-u)\cap K[x,y]_{(x,y)}$ is a prime ideal containing the prime ideal $(y^2 -x^3 -x^2)$, so we have the following chain of primes: $$(0)\subsetneq(y^2 -x^3 -x^2)\subseteq (y-u)\cap K[x,y]_{(x,y)}\subseteq (x,y)K[x,y]_{(x,y)}.$$ As this is happening in a ring of Krull dimension 2, this means we have only two options: either $(y^2 -x^3 -x^2)= (y-u)\cap K[x,y]_{(x,y)}$ or $(y-u)\cap K[x,y]_{(x,y)}= (x,y)K[x,y]_{(x,y)}$.

Now I claim that $$(y-u)\cap K[x,y]_{(x,y)} \neq(x,y)K[x,y]_{(x,y)}. $$ If we did have $(y-u)\cap K[x,y]_{(x,y)} =(x,y)K[x,y]_{(x,y)}$, then we would have $x\in(y-u)\cap K[x,y]_{(x,y)},$ so that $x\in ((y-u)\cap K[x,y]_{(x,y)})\cdot K[[x,y]]\subseteq (y-u)$. This would also yield $y= y-u+u= y-u +x+\frac{1}{2}x^2 - \frac{1}{8}x^3 + \frac{1}{16}x^4 - \cdots=y-u +x(1+\frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \cdots)\in (y-u).$ Therefore, we would have $(x,y)\subseteq (y-u)$ and hence $(x,y)=(y-u)$ as $(x,y)$ is maximal in $K[[x,y]].$ But $$ K[[x]] \cong K[[x,y]]/(y-u) = K[[x,y]]/(x,y) \cong K $$ yields a contradiction. Thus, $(y-u)\cap K[x,y]_{(x,y)} \neq(x,y)K[x,y]_{(x,y)}$ and so we must have $(y^2 -x^3 -x^2)= (y-u)\cap K[x,y]_{(x,y)}$. Finally, contracting back to $\operatorname{Spec}K[x,y]$ we have that $(y-u)\cap K[x,y]=(y^2 -x^3 -x^2)\cap K[x,y]=(y^2 -x^3 -x^2).$

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My question: is my argument correct? If so, great. If not, could you explain why and what the correct answer is?

For part (b) I don't know how to solve it and would appreciate any help with that.

Answer 1

Given that I am not an expert, I'll try to answer anyway. I have just attempted this exercise myself and I stumbled here when looking for a solution, so I'm not sure mine is correct.

I don't see flaws in your argument, but I propose a different one.

Key idea: the element $u=\sqrt{x^3+x^2}$ is in some sense algebraic over $K[x,y]$, with "minimum polynomial" $y^2-x^3-x^2$, while $w:=\sum_{n\geq 1}x^n/n!$ is not.

To make this precise, consider the fields of fractions $F:=Frac\,K[x]=K(x)$ and $L:=Frac\,K[[x]]$. Then the have an extension $F\subset L$ and it makes sense to ask whether an element $\alpha\in L$ is algebraic over $F$.

Consider a (prime) ideal of the form $(y-\alpha)\subset K[[x,y]]$. We have that a polynomial $f(x,y)\in K[x,y]$ is in $(y-\alpha)\cap K[x,y]$ if and only if $y-\alpha\,|\,f(x,y)$ in $K[[x,y]]$, i.e. if and only if $f(x,\alpha)=0$ (I am not sure about this part). So we are left to note that:

(a) since $p(y)=y^2-x^3-x^2$ is (this time in a precise sense) the minimum polynomial of $u\in L$ over $F$, we have that $p(y)\,|\,f(x,y)$, so $f(x,y)\in(y^2-x^3-x^2)$. On the other hand the other inclusion is obvious, so we have $(y-u)\cap K[x,y]=(y^2-x^3-x^2)$, as you concluded;

(b) since $w=\sum_{n\geq 1}x^n/n!\in L$ is not algebraic over $F$ (see for example here), we must have $f(x,y)=0$ and thus $(y-w)\cap K[x,y]=(0)$.

After this, I went on reading and the following lines seem to suggest working in this direction:

One problem with the scheme $Y$ above is that the points described in exercise II-3(b) are extraneous from an algebraic point of view. To avoid this, we may work with the spectrum $Z$ of the ring $H\subset K[[x,y]]$ of power series that satisfy algebraic equations over $K(x,y)$, the field of rational functions. Called the Henselization of $X$, the scheme $Z$ sits between $Y$ and $X$ [...]

I hope my solution correct, or at least useful to find a correct one.

Answer 2

The argument for (a) can be simpler. You want to show that $(y-u)\cap k[x,y]_{(x,y)}\neq (x,y)k[x,y]_{(x,y)}$. It suffices to show that $x\notin (y-u)\cap k[x,y]_{(x,y)}$.

All elements $r\in(y-u)$have the form $$r=(y-u)h$$ for some power series $h=h(x,y)$. Now write $u=x+f(x)$ and $h=a+g(x,y)$ where all terms of $f$ have degree $\geq 2$ and $g$ has no constant term. This yields $$\begin{align} r&=(y-x-f(x))(a+g(x,y)) \\ &=-ax+ay + (\text{terms of degree }\geq 2) \end{align}$$ Hence it is impossible that $r=x$.