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.

  • $\begingroup$ What do you mean by $k[x,u]$ with $u\in R^{\times }$? $\endgroup$
    – abx
    Oct 22, 2019 at 12:47
  • $\begingroup$ @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 $i: k[x,1/(1-y)] \to k[[x,y]]$. $\endgroup$
    – tj_
    Oct 22, 2019 at 14:53
  • $\begingroup$ 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. $\endgroup$
    – Stahl
    Oct 22, 2019 at 21:31

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – Stahl
    Oct 23, 2019 at 6:03
  • $\begingroup$ @Stahl: If I'm not mistaken, $A_u \cong k[u_i, u_i^{-1}][x_j]$. $\endgroup$
    – tj_
    Oct 23, 2019 at 8:24
  • $\begingroup$ @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. $\endgroup$
    – Stahl
    Oct 23, 2019 at 15:03
  • $\begingroup$ @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). $\endgroup$
    – tj_
    Oct 23, 2019 at 20:59
  • $\begingroup$ @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. $\endgroup$
    – Stahl
    Oct 23, 2019 at 21:15

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