Let $k$ be an algebraically closed field (for me I am using $k=\mathbb C$). I know that $\mathrm{Spec} \, k[x]/(x^2)$ consists of simply the prime ideal $(x)$. Indeed, any ideal $\mathfrak p$ of $k[x]/(x^2)$ is an ideal of $k[x]$ such that $(x^2) \subset \mathfrak p$.
If we now consider $\mathrm{Spec} \, k[x,y]/(y^2)$, now the prime ideals of $k[x,y]$ are $(0)$, $(x-a,y-b)$ for $a,b \in k$ and irreducible polynomials $f(x,y)$ generating $(f(x,y))$.
Clearly $(y^2)\not\subset (0)$. As for the irreducible polynomials, we have $(y^2) \subset k[x,y]f(x,y)$, so I think it is right to say that the ideals in bijection with these are of the form $(a+f(x)y+g(x))$ where $a,b \in k$ and $f,g$ irreducible. I guess $(x-a,y-b)$ would also be prime ideals of the quotient ring since quotienting by them gives an integral domain.
Now I am interested in understanding the generalisation $\mathrm{Spec} \, k[x,y_1,y_2,\dots,y_n]/(y_1^2,\dots,y_n^2)$. In particular:
- Can we classify all elements of the spectrum of this ring, for $n \geq 1$?
- Can we visualise this scheme, and has it been studied in some context in the literature?
Best Answer
Let $R=\frac {k[y_1,y_2,\dots ,y_n]}{(y_1^2,y_2^2,\dots , y_n^2)} $
Then $\operatorname {Spec} \frac {k[x,y_1,y_2,\dots ,y_n]}{(y_1^2,y_2^2,\dots , y_n^2)}=\operatorname{Spec} R[x]= \mathbb A^1_R$, the affine line over the ring $R$. Observe that since $m= (\bar y_1,\bar y_2, \dots , \bar y_n )$, is a nilpotent maximal ideal of $R$, $\operatorname {Spec} R= \{m \}$, namely it is a fat point in the sense of Mumford.
If $p\in \mathbb A^1_R$, consider it's image in $\operatorname {Spec} R$ under the structure morphism $\mathbb A^1_R\xrightarrow{\pi} \operatorname{Spec} R$. Thus $\pi(p)=m$.
Since we have $R/m \cong k $, we see a one-one correspondence $$\operatorname{Spec} R[x] \leftrightarrow \operatorname {Spec }k[x]$$
Thus as $\textbf{sets}$ you have $\mathbb A^1_R =\mathbb A^1_k $
But of course the structure sheaves are different. $\mathbb A^1_R$ has nilpotents in the structure sheaf.