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Let $f_1,\dots,f_n\in\mathbb Z[X]$ be non-constant polynomials (not necessarily distinct). Is it true that $f_1(X_1),\dots,f_n(X_n)$ is a regular sequence in $\mathbb Z[X_1,\dots,X_n]$?

The trivial case (that suggested me this question) is $f_1(X)=\cdots=f_n(X)=X$ when we get the regular sequence $X_1,\dots,X_n$.

  • Can you explain your notation? What do you mean by $f_i(X_i)$? Did you mean that $f_i \in \mathbb{Z}[X_i] \subseteq \mathbb{Z}[X_1,\dots, X_n]?$ Also, please explain the notation in your trivial case. – Youngsu Aug 04 '13 at 16:48
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    For question 2, the ideal generated by the $f_i(X_i)$ will not be prime in general. Consider $f_1(x)=x^2+1$, $f_2(x)=(x+1)^2+1$. Then $(x_1+x_2+1)(x_1-x_2-1)\in (f_1(x_1),f_2(x_2))$, but $x_1+x_2+1,x_1-x_2-1\not\in (f_1(x_1),f_2(x_2))$. – Julian Rosen Aug 04 '13 at 17:22
  • After a second thought I've realized that the answer to the present question is positive when the polynomials are monic. –  Aug 05 '13 at 09:24

1 Answers1

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The answer is yes if $f_i(X_i)\in\mathbb Z[X_i]$ is non-zero and primitive (gcd of the coefficients is $1$) for all $i$.

Lemma. Let $R$ be a commutative ring. Let $f(X)\in R[X]$ be a regular element such that $R[X]/(f(X))$ is flat over $R$ (e.g. if $f(X)$ is monic, but this is not a necessary condition). Then for any flat $R$-algebra $A$, $f(X)$ (viewed in $A[X]$) is a regular element of $A[X]$ and $A[X]/(f(X))$ is flat over $R$.

Proof: The first part comes from the flatness of $A[X]$ over $R[X]$. The second part comes from the fact $A[X]/(f(X))=A\otimes_R R[X]/(f(X))$ is flat over $R[X]/(f(X))$.

Corollary. Let $f_i(X)\in R[X]$, $i=1, \dots, n$, satisfy the conditions of the above lemma. Then $f_1(X_1), \dots, f_n(X_n)$ is a regular sequence in $R[X_1, \dots, X_n]$.

Proof: Let $A_n:=R[X_1,\dots, X_n]/(f_1(X_1), \dots, f_n(X_n))$ is flat over $R$. If $n\ge 2$, we have $A_n=A_{n-1}[X_n]/(f_n(X_n))$. The above lemma implies (by induction on $n$) that $A_{n-1}$ is flat over $R$ and $f_n(X_n)$ is a regular element in $A_n$.

Now in the case $R=\mathbb Z$, $f_i(X)\ne 0$ is equivalent to $f_i(X)$ regular, and $f_i(X)$ primitive is equivalent to $\mathbb Z[X]/(f(X))$ torsion-free (equivalently, flat) over $\mathbb Z$.

Remark. Clearly one can not remove the condition being primitive (consider $2X$ and $2$). But if we suppose all $f_i(X)$ are irreducible, and no prime number occurs twice in the sequence, then I think we still have a regular sequence (I didn't check the details).

Remark'. As for the flatness of $R[X]/(f(X))$ over an arbitrary $R$, if $f(X)$ is primitive (in the sense that its coefficients generate the unit ideal of $R$), then $R[X]/(f(X))$ is flat over $R$. Conversely, if $\mathrm{Spec}(R)$ is connected, then the flatness implies $f=0$ or $f$ is primitive.