Constructing prime ideal of tensor product from two prime ideals

Solution 1:

The scheme-theoretic fibre product $\operatorname{Spec} M \times_{\operatorname{Spec} R} \operatorname{Spec} N$ is not the same space as the topological [set-theoretic] fibre product $|\operatorname{Spec} M| \times_{|\operatorname{Spec} R|} |\operatorname{Spec} N|$. What you probably meant is: describe the map $$\operatorname{Spec} M \times_{\operatorname{Spec} R} \operatorname{Spec} N = \operatorname{Spec} (M \otimes_R N) \to |\operatorname{Spec} M| \times_{|\operatorname{Spec} R|} |\operatorname{Spec} N|.$$ Let $M \stackrel i\to M \otimes_R N \stackrel j\leftarrow N$ be the natural inclusions.

Lemma. The fibre above the pair $(\mathfrak p,\mathfrak q)$ is the spectrum of $\kappa(\mathfrak p) \otimes_{R} \kappa(\mathfrak q)$.

Proof 1. Let $\mathfrak x \subseteq M \otimes_R N$ be a prime ideal. We will show that $(i^{-1} \mathfrak x, j^{-1} \mathfrak x) = (\mathfrak p, \mathfrak q)$ if and only if the natural map $M \otimes_R N \to \kappa(\mathfrak x)$ factors as $M \otimes_R N\to\kappa(\mathfrak p) \otimes_R \kappa(\mathfrak q) \to \kappa(\mathfrak x)$.

Clearly if the map factors through $\kappa(\mathfrak p) \otimes_R \kappa(\mathfrak q)$, then $i^{-1}\mathfrak x = \mathfrak p$, by commutativity of the diagram $$\begin{array}{ccc}M & \stackrel i \to & M \otimes_R N \\ \downarrow & & \downarrow \\ \kappa(\mathfrak p) & \to &\ \kappa(\mathfrak p) \otimes_R \kappa(\mathfrak q).\end{array}$$ Similarly, $j^{-1}\mathfrak x = \mathfrak q$, which proves one implication.

Conversely, assume $(\mathfrak p, \mathfrak q) = (i^{-1}\mathfrak x, j^{-1} \mathfrak x)$. Then $\mathfrak p \otimes 1$ and $1 \otimes \mathfrak q$ map to $0$ in $\kappa(\mathfrak x)$, so we already get a factorisation $$M \otimes_R N \to M/\mathfrak p \otimes_R N/\mathfrak q \stackrel f\to \kappa(\mathfrak x).$$ We can extend this to $\kappa(\mathfrak p) \otimes_R \kappa(\mathfrak q)$ by setting \begin{align*} \kappa(\mathfrak p) \otimes_R \kappa(\mathfrak q) &\to \kappa(\mathfrak x)\\ \frac{a}{b} \otimes \frac{c}{d} &\mapsto \frac{f(a \otimes c)}{f(b \otimes d)} = \frac{f(a \otimes 1)}{f(b \otimes 1)} \frac{f(1 \otimes c)}{f(1 \otimes d)}, \end{align*} noting that $f(b \otimes d) = f(b \otimes 1) f(1 \otimes d) \neq 0$ for $b \not \in \mathfrak p$, $d \not \in \mathfrak q$ (check that the map is bilinear). $\square$

Proof 2. For every field $K$, there is a correspondence $$(\operatorname{Spec} M \times_{\operatorname{Spec} R} \operatorname{Spec} N)(K) = \operatorname{Spec} M(K) \times_{\operatorname{Spec} R(K)} \operatorname{Spec} N(K).$$ Thus, if you have primes $\mathfrak p \subseteq M$, $\mathfrak q \subseteq N$, you get homomorphisms \begin{align*} M \to \kappa(\mathfrak p), & & N \to \kappa(\mathfrak q). \end{align*} For any field $K$, any homomorphism $\kappa(\mathfrak p) \to K \leftarrow \kappa(\mathfrak q)$ gives a $K$-point of the fibre product, which necessarily lies in the fibre over $(\mathfrak p, \mathfrak q)$. Thus, the $K$-points of $\operatorname{Spec}(\kappa(\mathfrak p) \otimes_R \kappa(\mathfrak q))$ are given by pairs of homomorphisms $\kappa(\mathfrak p) \to K \leftarrow \kappa(\mathfrak q)$, i.e. by homomorphisms $$\kappa(\mathfrak p) \otimes_R \kappa(\mathfrak q) \to K.$$ This also proves the claim (modulo some bookkeeping). $\square$

Example. Let $R = k$ be a field, and $M$ and $N$ finite type $k$-algebras. Then for maximal ideals $\mathfrak p \subseteq M$, $\mathfrak q \subseteq N$, the fields $\kappa(\mathfrak p)$ and $\kappa(\mathfrak q)$ are finite extensions of $k$ (this is the weak Nullstellensatz). Then $\kappa(\mathfrak p) \otimes_k \kappa(\mathfrak q)$ is an Artinian ring, but it can be nonreduced, reducible (i.e. non-local), or both. For example:

• If both $\kappa(\mathfrak p)$ and $\kappa(\mathfrak q)$ are separable over $k$, then $\kappa(\mathfrak p) \otimes_k \kappa(\mathfrak q)$ is a product of finite separable extensions. (Key word: étale algebras.)
• If $\kappa(\mathfrak p)$ and $\kappa(\mathfrak q)$ are the same purely inseparable extension, then $\kappa(\mathfrak p) \otimes_k \kappa(\mathfrak q)$ has nilpotents (exercise: do this for the easiest example $\mathbb F_p(T) \subseteq \mathbb F_p(T^{\tfrac{1}{p}})$.)
• The ring $\kappa(\mathfrak p) \otimes_k \kappa(\mathfrak q)$ is a field if and only if $\kappa(\mathfrak p)$ and $\kappa(\mathfrak q)$ are linearly disjoint over $k$. In this case, it is just the compositum.

One could try to take $K$ to be their compositum inside some fixed algebraic closure. For this to make sense, you have to choose embeddings $\kappa(\mathfrak p) \to \bar k \leftarrow \kappa(\mathfrak q)$; even the isomorphism type of $K$ might depend on these (if $\kappa(\mathfrak p)$ and $\kappa(\mathfrak q)$ are not Galois), and even if it doesn't, the actual $K$-point you get in the fibre product still depends on this choice.

Remark. If $\mathfrak p$ and $\mathfrak q$ are not maximal, the story gets even more complicated. In this case, the transcendence degrees of the residue fields of the points in the fibre can be different!

Example. Consider for example the ring $$k(x) \otimes_k k(y).$$ It has prime ideals with residue field of different transcendence degree. For example, there are morphisms \begin{align*} k(x) \otimes_k k(y) &\to k(x,y)\\ x &\mapsto x,\\ y &\mapsto y, \end{align*} and \begin{align*} k(x) \otimes_k k(y) &\to k(t)\\ x &\mapsto t,\\ y &\mapsto t. \end{align*} These give two different points of $\mathbb A^1 \times \mathbb A^1$ whose image in both $\mathbb A^1$s is the generic point. The first one corresponds to the generic point in $\mathbb A^2$, and the second one to the prime ideal $(x-y) \subseteq k[x,y]$, i.e. to the generic point of the diagonal in $\mathbb A^1 \times \mathbb A^1$.

Solution 2:

The image of $\operatorname{Spec} (M \otimes_R N)$ in the absolute fiber product is the set-theoretic fiber product. It is enough to prove the following. Suppose I have primes $(p,q) \in \operatorname{Spec} M \times \operatorname{Spec} N$ that map down to $r$ in $\operatorname{Spec} R$. Then we have maps $k(r) \to k(p)$ and $k(r) \to k(q)$. By the universal property of the tensor product of algebras, we have a map

$$\operatorname{Spec} k(p) \otimes_{k(r)} k(q) \to \operatorname{Spec} (M \otimes_R N).$$

The left hand side is not the empty scheme by faithful flatness (every map of fields is faithfully flat). So there is some prime on the left mapping to the right. This prime automatically maps to $p$ and $q$ and we are done.

Solution 3:

Remy's answer is very nice. And I found it not so immediate without writing down the details, so I write here how the factorization deduces the lemma in Remy's answer:

1. Given a prime $\xi\subset M\underset{R}{\otimes}N$, and we have the factorization $\psi :M\underset{R}{\otimes}N\overset{\psi_1}{\rightarrow} k(p)\underset{R}{\otimes}k(q) \overset{\psi_2}{\rightarrow} k(\xi)$, we associate $\xi$ with $\psi_2^{-1}(0)$ which is a prime in $k(p)\underset{R}{\otimes}k(q)$.

2. Consider the map $\psi_1: M \underset{R}{\otimes}N \rightarrow k(p)\underset{R}{\otimes}k(q)$, and we associate a prime $\zeta$ in $k(p)\underset{R}{\otimes}k(q)$ to $\psi_1^{-1}(\zeta).$

We prove these two maps are inverse to each other.

3. Start from $\xi \subset M\underset{R}{\otimes}N$. By the property of localization and quotient, $\xi = \psi^{-1}(0) = \psi_{1}^{-1}\psi_2^{-2}(\xi). $ This proves one direction.

4. Start from $\zeta \subset k(p)\underset{R}{\otimes}k(q)$. Since $\psi_1$ s subjective, $\psi_1^{-1}(\zeta) = \psi_1^{-1}\psi_2^{-1}(0) \implies \psi_1\psi_1^{-1}(\zeta) =\psi_1 \psi_1^{-1}\psi_2^{-1}(0)\implies\zeta = \psi_2^{-1}(0).$

This fills the gap.