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Apparently this question is asked twice or more on MSE, for example here, but I struggle to get the answers.

In The Geometry of Schemes by Eisenbud and Harris,exercise II-20, I am asked to classify all schemes of degree 3 over $\mathbb R$ supported at the origin in $\mathbb A_\mathbb R^2$. (That is, schemes of the form $\text{Spec } \mathbb R[x_1,\ldots, x_n ] /I$.) There is a hint that there are two nonisomorphic schemes in the list whose complexification is $\text{Spec } \mathbb C[x,y]/(x^2,y^2 ,xy).$

I am struggling to see why algebraic closure of $\mathbb C$ and algebraic non-closure of $\mathbb R$ is so important here. Apparently, for the same question over $\mathbb C$, I can just consider the vector space dimension of $\mathbb C[x_1,\ldots,x_n]/ I$ for an ideal $I$ and reach the conclusion that $I$ can take only one of a few forms. In this process, algebraic completenes doesn't seem to be used. To be more specific, if $\mathbb C[x_1,\ldots,x_n]/ I$ has dimension $3,$ then any polynomial $P$ in $\mathbb C[x_1,\ldots,x_n]$ can be written in the form $$ P = \sum_{1\leq i\leq 3}\lambda_i P_i + b, \lambda_i \in \mathbb R, b \in I, $$ where $P_i$ are a "basis". The expansion is easily seen to be unique. So by expanding the polynomials $1,x_i$ and so on, we will see that we can simplify the basis $P_i$ into simple polynomials $1,x_1,$ etc. Hence the particular simple schemes we are looking for. Where is algebraic closure used? Why doesn't this work for $\mathbb R$?

And what is the list of schemes we are looking for?

Ma Joad
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  • When you say "algebraic completeness", do you mean "algebraic closure"? – Captain Lama Jun 18 '21 at 10:21
  • @CaptainLama Yes. Now corrected. – Ma Joad Jun 18 '21 at 10:22
  • @MaJoad - You find a usage of the Chinese Remaider Theorem at the following link that is related to your problem. It relates cofinite ideals to maximal ideals. https://math.stackexchange.com/questions/4171591/applications-of-the-chinese-remainder-theorem-to-the-study-of-the-hilbert-scheme – hm2020 Jun 20 '21 at 17:16
  • @MaJoad: It follows from the CRT that any cofinite ideal $I \subseteq A:=k[x_1,..,x_n]$ may be written as a product of cofinite ideals $\mathfrak{m}_i^{l_i} \subseteq I_i \subseteq \mathfrak{m}_i \subseteq A$ where $\mathfrak{m}_i$ is a maximal ideal. It follows

    $$A/I \cong A/I_1\cdots I_d \cong A/I_1 \oplus \cdots \oplus A/I_d.$$

    Note that using the CRT you do not need the base field to be algebraically closed (see the above linked proof).

     – hm2020 Jun 21 '21 at 10:47
  • You must "classify" maximal ideals $\mathfrak{m} \subseteq \mathbb{R}[x,y]$. A maximal ideal $\mathfrak{m}$ may have residue field the real numbers $k:=\mathbb{R}$ or the complex numbers $K$. In the case of the real numbers it follows $\mathfrak{m}=(x-a,y-b)$ for $(a,b) \in k^2$. In the case of the complex numbers several things may happen.

    Example: If $z:=a+ib \in K$ with $b\neq 0$, it follows

    $$p_z(x):=(x-z)(x-\overline{z})=x^2-2ax+a^2+b^2$$

     – hm2020 Jun 21 '21 at 10:48
  • is irreducible in $k[x]$. For any $b\in k$ it follows $\mathfrak{m}:=(p_z(x),y-b) \subseteq k[x,y]$ is a maximal ideal. with $K$ as residue field. – hm2020 Jun 21 '21 at 10:48
  • @MaJoad - note moreover that for any surjective map $\rho: \mathbb{R}[x,y] \rightarrow \mathbb{C}$ of $\mathbb{R}$-algebras, you may make a variable change, and in the new variables it follows $ker(\rho) =(p_z(x), y-b)$ for $b\in \mathbb{R}$ and $z\in \mathbb{C}-\mathbb{R}$. – hm2020 Jun 22 '21 at 10:19

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