# Flat algebra over polynomial ring

Let $$k$$ be a field, and let $$f : k[t_1,\dots, t_r]\to k[x_1,\dots, x_n]$$ be the $$k$$-algebra map defined by $$f(t_i) = f_i\in k[x_1,\dots, x_n].$$ Suppose that the $$f_i$$ are algebraically independent over $$k,$$ so that the map $$f$$ is an injection. To show that $$f$$ is flat, it suffices to show that $$f_\mathfrak{m} : k[t_1,\dots, t_r]\to k[x_1,\dots, x_n]_\mathfrak{m}$$ is flat for $$\mathfrak{m}\subseteq k[x_1,\dots, x_n]$$ an arbitrary prime ideal. If $$(f_1,\dots, f_r)\subseteq\mathfrak{m},$$ then $$f_\mathfrak{m}$$ is flat if and only if the $$f_i$$ form a regular sequence in $$k[x_1,\dots, x_n]_\mathfrak{m}$$ (c.f. Eisenbud's Commutative Algebra, exercise 18.18).

In general, we cannot expect that the $$f_i$$ form a regular sequence in every $$k[x_1,\dots, x_n]_\mathfrak{m}.$$ However, suppose that if they do fail to form a regular sequence, they fail in a "silly" way: by some of the $$f_i$$ becoming units in $$k[x_1,\dots, x_n]_\mathfrak{m}.$$ My question is, does flatness still hold if some of these $$f_i$$ become units? In particular, can we generalize exercise 18.18 of Eisenbud's Commutative Algebra to say:

Let $$(R,\mathfrak{m})$$ be a Noetherian local ring containing a field $$k,$$ and let $$x_1,\dots, x_r\in\mathfrak{m}$$ and $$u_1,\dots, u_s\in R^\times$$ be two sequences of elements. Then $$i : k[x_1,\dots, x_r, u_1, \dots, u_s]\to R$$ is flat if $$x_1,\dots, x_r$$ is a regular sequence.

Edit: Above I am happy to interpret $$k[x_1,\dots, x_r, u_1, \dots, u_s]$$ as either the $$k$$-subalgebra of $$R$$ generated by the $$x_i$$ and $$u_i$$ or as a polynomial ring over $$k$$ in the $$x_i$$ and $$u_i$$ (mapping to $$R$$ by sending each variable to the corresponding element). In the situation I am concerned with, these coincide, so I would be interested in results about either situation.

Edit 2: As noted in the comments to Eric's answer below, this problem could also be stated as follows:

Suppose that $$k$$ is a field, and that $$(R,\mathfrak{m})$$ is a Noetherian local ring containing $$A := k[t_1^{\pm1},\dots, t_s^{\pm1}].$$ Let $$x_1,\dots, x_r$$ be a sequence of elements in $$\mathfrak{m}.$$ Then $$i : A[x_1,\dots, x_r]\to R$$ is flat if $$x_1,\dots, x_r$$ is a regular sequence.

The map $$i$$ above will be flat if and only if $$A[x_1,\dots, x_n]_\mathfrak{n}\to R$$ is flat, where $$\mathfrak{n} = i^{-1}(\mathfrak{m}).$$ Moreover, assuming that $$R$$ is Cohen-Macauley (which I am happy to do), this map is flat if and only if $$\dim R = \dim A[x_1,\dots, x_r]_\mathfrak{n} + \dim R/\mathfrak{n}R,$$ but I'm not sure how to get a handle on $$A[x_1,\dots, x_r]_{\mathfrak{n}}$$ (or $$\dim R/\mathfrak{n}R$$) in general. (The proof of the exercise mentioned above does not generalize immediately to this situation: $$\mathfrak{n}$$ need not be simply $$(x_1,\dots, x_n)A[x_1,\dots, x_r];$$ there could be contributions coming from the $$t_i.$$)

As an example, we might consider the injection $$k[u]\to k[x]_{(x)},$$ where $$u = 1/(1+x).$$ In this case the preimage of $$(x)$$ is $$(1 - u),$$ so $$k[u]\to k[x]_{(x)}$$ is flat because $$k[u]_{(1 - u)}\to k[x]_{(x)}$$ is an isomorphism. The result is true in this very simple example, but I'm not sure that it suggests a more general approach.

• What do you mean by $k[x,u]$ with $u\in R^{\times }$? – abx Oct 22 '19 at 12:47
• @abx: I guess it's the k-subalgebra of R generated by x and u. For example, if $R=k[[x, y]]$, $i$ could be $i: k[x,1/(1-y)] \to k[[x,y]]$. – tj_ Oct 22 '19 at 14:53
• I mean on the left hand side that that $x$ and $u$ are formal variables mapping to the specified elements of $R,$ although I would also be happy interpreting the left hand side as the $k$-subalgebra of $R$ generated by those elements. In the cases I'm interested in, the subalgebra generated by the elements will be a polynomial algebra on those elements. – Stahl Oct 22 '19 at 21:31

Let's call $$A = k[x_1, \dots, x_r; u_1, \dots, u_s]$$. If $$u_1, \dots, u_s \in R^\times$$, then $$i: A \to R$$ factors through the localization $$A_u$$ with respect to $$u = u_1 \cdots u_s$$. Localization is flat, so I believe the condition reduces to the statement of the exercise, as stated in Eisenbud.
A bit more precisely, suppose $$0 \to N' \to N \to N'' \to 0$$ is an exact $$A$$-sequence, and we want to check that applying $$-\otimes_A R$$ leaves this sequence exact. As stated above, the $$u_i$$ are units in $$R$$ so $$-\otimes_A R = (-\otimes_A A_u) \otimes_{A_u}R$$. Therefore, we can first localize $$0 \to N'_u \to N_u \to N''_u \to 0$$, and now $$-\otimes_{A_u} R$$ will leave the sequence exact iff the $$x_j$$ form a regular sequence.
• This might work! But, I have to think about whether the proof of the exercise will apply to $A_u\to R,$ as $A_u$ is not necessarily going to be of the form $k'[x_1,\dots, x_r]$ for some $k'$ contained in $R.$ Is this obvious? – Stahl Oct 23 '19 at 6:03
• @Stahl: If I'm not mistaken, $A_u \cong k[u_i, u_i^{-1}][x_j]$. – tj_ Oct 23 '19 at 8:24
• @tj_ that's true, but $k[x,x^{-1}]$ is not the same as $k(x)$ (and similarly for more variables): the latter contains things like $1/(x-1)$, but the former does not. – Stahl Oct 23 '19 at 15:03
• @Stahl: That's why I have written $k[u_i^\pm]$ and not $k(u_i)$. But maybe Eisenbud can be adjusted for $k[u_i^\pm]$ in place of a field (I don't know if it can). – tj_ Oct 23 '19 at 20:59
• @tj_ Ah, I see. Yes, I've been wondering this as well. It doesn't seem unreasonable that if $R$ is a local noetherian ring containing $A_u$ then the map $A_u[\{x_j\}_j]\to R$ will be flat iff the $x_j$ form a regular sequence in $R,$ but I haven't been able to write down a proof of that statement. – Stahl Oct 23 '19 at 21:15