Let $A\subset B$ be an integral ring extension, and $p\in\operatorname{Spec}A$. Then there exists $q\in\operatorname{Spec}B$ such that $q\cap A=p$.

Question

I need an example of this theorem. I could think of trivial example if $B=A$. But any suggestions for a non trivial example? How is this theorem true in case $A = \mathbb C[x,y]/\left<y^2-x^3\right>$ ?

Answer 1

One class of nontrivial examples is given by extensions of rings of integers in number fields. For example, let $A = \Bbb{Z},$ and let $K/\Bbb{Q}$ be a finite extension. The integral closure of $\Bbb{Z}$ in $K$ is denoted $\mathcal{O}_K,$ and in fact, any prime $p\in\Bbb{Z}$ factors uniquely in $\mathcal{O}_K$ as a product of prime ideals $p\mathcal{O}_K = \mathfrak{p}_1^{e_1}\dots\mathfrak{p}_r^{e_r}.$ For any $\mathfrak{p}_i,$ we have $\mathfrak{p}_i\cap\Bbb{Z} = (p).$

Let's look at $A = \Bbb{C}[x,y]/(y^2 - x^3).$ The integral closure of $A$ in $\mathrm{Frac}(A)\cong\Bbb{C}(t)$ is $\Bbb{C}[t],$ the inclusion being given by \begin{align*} \Bbb{C}[x,y]/(y^2 - x^3)&\to\Bbb{C}[t]\\ x + (y^2 - x^3) &\mapsto t^2,\\ y + (y^2 - x^3) &\mapsto t^3. \end{align*} (To see this, note that $A\cong\Bbb{C}[t^2,t^3]$ via $x\mapsto t^2,$ $y\mapsto t^3,$ that $\Bbb{C}[t]$ is integrally closed, and that $t$ must be in the integral closure of $A$.) Take $B = \Bbb{C}[t];$ as $B$ is the integral closure of $A,$ the inclusion $A\to B$ is an integral ring extension. So for any prime ideal $\mathfrak{p}\subseteq A,$ there exists a prime ideal $\mathfrak{q} = (f)\subseteq\Bbb{C}[t]$ such that $\mathfrak{q}\cap A = \mathfrak{p}.$ For example, suppose $\mathfrak{p} = (x - a^2, y - a^3).$ Then you can check that $(t - a)\cap A = (x - a^2, y - a^3).$ (This may be easier to see using the isomorphism $A\cong\Bbb{C}[t^2,t^3]$: then $\mathfrak{p} = (t^2 - a^2, t^3 - a^3)$ and $(t-a)\cap A = (t^2 - a^2, t^3 - a^3).$)

Geometrically, this property means that the induced map of schemes $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$ is surjective.