# Base change to algebraic closure commutes with quotient of polynomial ring by maximal ideal

Let $$k$$ be a field, $$R:=k[x_1, \cdots , x_n]$$ and $$\mathfrak m$$ be a maximal ideal such that $$R/\mathfrak m$$ is a finite separable field extension of $$k$$. Consider the algebraic closure $$\overline k$$ of $$k$$ and let $$\mathfrak m_1, \dots , \mathfrak m_r$$ be the maximal ideals of the ring $$\overline R :=\overline k \otimes_k R = \overline k[x_1, \cdots , x_n]$$ lying over $$\mathfrak m$$.

In an article I have been reading, the following isomorphism is used: $$\overline k \otimes_k R/\mathfrak m \cong \overline R \big/ \bigcap_{1 \leq j \leq r} \mathfrak m_j$$

Since I have not been able to find any reference for this, I have trying to justify the same myself:

In the forward direction, I could make the usual map work, namely, the one sending each element $$(\alpha, f + \mathfrak m) \in \overline k \times R/\mathfrak m \text{ (for }f \in R\text{) }\hspace{2mm}\text{ to }\hspace{2mm} \alpha f+ \bigcap_{i=1}^r \mathfrak m_i \in R \big/ \bigcap_{1 \leq j \leq r} \mathfrak m_j.$$ Indeed it turns out to be well-defined since each $$\mathfrak m_i$$ lies over $$\mathfrak m$$ and $$k$$-bilinearity is checked easily, whereupon the universal property of the tensor product applies.

However it is while showing the inverse isomorphism that I have gotten stuck. My idea was to try and get a map $$\phi: \overline R \longrightarrow \overline k \otimes_k R/\mathfrak m$$ whose kernel contains the intersection $$\bigcap_{j=1}^r \mathfrak m_j$$, and to do the same I tried sending each $$\alpha \otimes f \in \overline k \otimes_k R = \overline R$$ to $$\alpha \otimes \overline f = \alpha \otimes (f + \mathfrak m) \in \overline k \otimes_k R/\mathfrak m$$. However, in order to be able to use the Mapping Property and get my desired map $$\phi$$, I seem to need to have $$\bigcap_{i=1}^r \mathfrak m_i \subset \mathfrak m$$, which in this case becomes equivalent to $$\bigcap_{i=1}^r \mathfrak m_i = \mathfrak m$$).

And it is not clear to me why this last equality should hold; perhaps I am missing something obvious or some property of the ring extension $$R \subset \overline R$$ is at work here. I tried to use the weak Nullstellensatz to note that each $$\mathfrak m_i$$ must be of the form $$(x_1-a_{i1}, \cdots , x_n-a_{in})$$ for some $$(a_{i1}, \cdots , a_{in}) \in \overline k^n$$, but this hasn't proven to be useful yet. I am not sure whether the condition of $$R/\mathfrak m$$ being separable over $$k$$ will be helpful here.

I would really appreciate a proof (perhaps one using the above ideas?) or reference. Thank you.

• Prime ideals lying over $\mathfrak m$ are just prime ideals of $\overline k\otimes_k(R/\mathfrak m)$, i.e., an extension of scalar of a field extension.
– Z. M
May 3, 2021 at 9:45
• @Z. M Thank you for your response. I have not seen this exact fact although it intuitively seems clear to me why it should be true. Can you please give a reference to that end? Thanks again. May 3, 2021 at 10:25
• Let $f\colon A\to B$ be a map of rings and $M\subseteq A$ be a maximal ideal. Then for any prime ideal $P\subseteq B$, $f^{-1}(P)\subseteq A$ is a prime ideal. We have $M=f^{-1}(P)$ if and only if $M\subseteq f^{-1}(P)$ (by maximality of $M$) if and only if $f(M)\subseteq P$. Such $P$ corresponds bijectively to prime ideals of $B/f(M)\cong (A/M)\otimes_{A,f}B$.
– Z. M
May 4, 2021 at 9:32
• Thank you for the clarification. I was wondering if you could kindly help me resolve this mathoverflow.net/questions/391769/… as well. May 4, 2021 at 11:06
• @Z.M Thanks again. I just had one last question: the article that I have been reading mentions that this isomorphism needs the separability of the extension $R/\mathfrak m$ over $k$, a consequence of which is that $\overline k \otimes_k R/\mathfrak m$ is a reduced ring. This isomorphism in my question is part of the proof of a bigger proposition which assumes the aforementioned separability and in addition, it seems that it is the only place where the said separability should be used. So where is the separability of the extension $R/\mathfrak m$ over $k$ getting used in this argument? Thanks. May 5, 2021 at 9:52