If a polynomial in $B[x]$ is integral over $A[x]$, then are its coefficients integral over $A$?

Question

Suppose $A \subset B$ are integral domains, and $f = b_0 + b_1 x + \ldots + b_m x^m$, where $b_k \in B$. Is it true that if $f$ is integral over $A[x]$ then all the coefficients $b_k$ are integral over $A$?

Note that the other way is obvious, i.e. if all the coefficients $b_k$ are integral over $A$, then $f$ is integral over $A$ (e.g. see the proof here).

Below are some observations that I made.

Since $f$ is integral we have $f^n + f_{n-1} f^{n-1} + \dots + f_0 = 0$, $f_k \in A[x]$.

1. Plugging in $0$ we immediately obtain that $b_0$ is integral over $A$. Hence $f - b_0$ is integral over $A[x]$.
2. Similarly plugging in $1$ we obtain that $b_0 + \dots + b_m$ is integral over $A$.
3. Expanding $f^n$, it's clear that coefficient in front of $x^{kn}$ has the form $b_k ^ n + $ (terms of lower degree in $b_k$). The coefficient in front of $x^{kn}$ coming from $f_l f^l$ will be homogenous polynomial in $\{b_k\}$ of degree $l$. Thus equating the coefficient to zero from our original expression will give us a monic polynomial in all the other coefficients which vanishes at $b_k$.

Edit: see here for the definition of integral element. In particular, a polynomial $f \in B[x]$ is integral over $A[x]$ if there exist polynomials $f_0, \ldots, f_{n-1} \in A[x]$ such that $f^n + f_{n-1} f^{n-1} + \dots + f_0 = 0$

Answer 1

To add to @Rodion Zaytsev's answer, the statement for an arbitrary ring $A$ can readily be reduced to the Noetherian case, so the Noetherian case is really the essential one. If $S$ is a ring and $R$ is a subring, we will write $\mathrm{int}_R(S)$ for the integral closure of $R$ in $S$.

Recall we have to show that if $S = \mathrm{int}_B(A)$ is the integral closure of $A$ in $B$ and $S_1 = \mathrm{int}_{B[x]}(A[x])$,then $S_1 = S[x]$. The containment $S[x]\subseteq S_1$ is easy, so the main point is to show $S_1\subseteq S[x]$. This statement is "local" in that it can be checked at each element of $S_1$, i.e., we just need to show that any $f\in B[x]$ which is integral over $A[x]$ must have its coefficients integral over $A$. For convenience, we will write $\mathrm{co}(f)\subset B$ for the set of coefficients of a polynomial $f \in B[x]$. Thus if $f = \sum_{k=0}^N b_kx^k$ where $b_N \neq 0$ so that $\deg(f)=N$, then $\mathrm{co}(f) = \{b_0,b_1,\ldots,b_N\}$.

Now fix such an $f$, then we may choose a $g\in A[x][t]$, monic as a polynomial in $t$, with $g(f)=0$. Suppose $g = t^d+ \sum_{k=0}^{d-1} c_k(x).t^k$. Then each $c_k \in A[x]$ and so $C=\bigcup_{k=0}^{d-1} \mathrm{co}(c_k)$ is a finite subset of $A$. Let $A_g = \mathbb Z[x:x \in C]$. This is a Noetherian ring because if $N=|C|$ then we may write $A_g$ as a quotient of $\mathbb Z[x_1,\ldots,x_N]$ (and Hilbert's basis theorem shows that a polynomial ring over $\mathbb Z$ is Noetherian).

Now $A_g \subseteq A \subseteq B$ and hence the integral closure of $A_g$ in $B$ is contained in $S$. But $g \in A_g[x][t]$ by defintion, hence $f\in B[x]$ is integral over $A_g[x]$. Using the Noetherian case Rodion established in his answer, it follows that $\mathrm{co}(f) \subseteq \mathrm{int}_B(A_g)\subseteq \mathrm{int}_B(A) = S$ as required.

Finally, it might be worth noting that the argument above is in the same spirit as the proof that if one has rings $R_1\subseteq R_2 \subseteq R_3$, then if $R_2$ is integral over $R_1$ and $R_3$ is integral over $R_2$ then $R_3$ is integral over $R_1$.

Answer 2

Thanks to the comments below for pointing out gaps in previous attempts to tackle this question.

Let $S$ be the integral closure of $A$ in $B$, and let $S_1$ be the integral closure of $R_1 = A[x]$ in $R_2 = B[x]$. Then $S[x]$ is certainly integral over $R_1$ so that $S[x]\subseteq S_1$. Thus replacing $A$ with $S$ it suffices to show the following:

Claim: If $A$ is integrally closed in $B$ then $R_1=A[x]$ is integrally closed in $R_2 =B[x]$.

Proof of claim:

Suppose that $f \in B[x]$ lies in $\mathrm{int}_{R_2}(R_1)$. Then there exists $g \in R_1[t]$ with

$$ g(t) = t^d -\sum_{k=0}^{d-1} c_kt^k, $$

where $c_k \in A[x]$ (and we set $c_d =1$ as this will be convenient later). Let $\mathrm{deg}_{x}(\phi)$ denote the degree of an element of $\phi \in B[x]$, and set $m = \text{max} \{ \mathrm{deg}_x(c_k): 0 \leq k \leq d-1 \}$.

Pick $N>\max\{ m,d \}$ and consider $f_1 = f-x^N$. Now clearly $f_1$ is integral over $R_1$ if and only if $f$ is, and $f_1 \in R_1$ if and only if $f\in R_1$, hence it suffices to show that $f_1 \in A[x]$. Now $g(f) = g(f_1+x^N)=0$, and hence $$ (f_1+x^N)^d -\sum_{k=0}^{d-1} c_k (f_1+x^N)^k = 0. $$ As $f_1$ and $x^N$ both have degree $N$, all the terms in the expansion of $(f_1+x^N)^k$ have degree $kN$. It follows that if we set $r_k = c_kf_1^{-1}\left((f_1+x^N)^k-f_1^k -x^{Nk}\right)$, then the above equation can be rewritten in the form $$ \begin{split} g(x^N) &= f_1\left(f_1^{-1}(c_0- g(f_1))+ \sum_{k=1}^{d-1}r_k(f_1)\right) \\ &= f_1.h \end{split} $$

Thus we see that $f_1$ divides $g(x^N)$, a monic polynomial in $A[x]$. But then if $K\supseteq B$ is a splitting field for the roots of $g(x^n)$, the roots of $g(x^N)$ are all integral over $A$, and since $-f_1$ is also monic, with roots a subset of those of $g(x^N)$ it follows that the coefficients of $f_1$ are integral over $A[x]$. But $f_1 \in B[x]$ and $A$ is integrally closed in $B$, hence we must have $f_1\in A[x]$ as required.

The introduction of the additional term $x^N$ may seem eccentric, but it functions in a similar fashion to Nagata's proof of Noether normalization.

Answer 3

Here is a proof in case $A$ is Noetherian.

Note that the equivalent definition of an integral element consists in the following: $b\in B$ is integral over $A$ iff the subring $A[b]$ is finitely generated as an $A$ module (this is the second definition in the section "equivalent definitions" in the wikipedia page I linked).

Now if $f = b_0 + b_1 x + \ldots + b_m x^m \in B[x]$ is integral, then $f^n + f_{n-1} f^{n-1} + \ldots + f_0 = 0$ and thus $1, f, \ldots, f^{n-1}$ generate $A[x, f]$ as $A[x]$ module. Consider now an $A$ module $M$ which consists of all the coefficients of elements of $A[x, f]$. Indeed, this is an $A$ module, because if $b, c \in H$, then there exist polynomials in $A[x,f]$, $$g = \ldots + b x^{m} + \ldots, h = \ldots + c x^k + \ldots$$ so $$x^k g +x^m h =\ldots + (b + c)x^{m+k} + \ldots \in A[x,f]$$ i.e. $b+c \in M$. For any $a \in A$ we have $ag \in A[x,f]$ hence $ab \in M$.

$M$ is generated by the coefficients of $1, f, \ldots, f^{n-1}$ since any $g\in A[x, f]$ has the form $$f_0 + f_1 f + \ldots f_{n-1} f^{n-1}, f_k \in A[x]$$

Since $A\subset M$ and $f^N \in A[x,f]\ \forall N$, we have $b_m^N \in M$, so $A[b_m]\subset M$.

Since $A$ is noetherian and $M$ is finitely generated over $A$, so is the submodule $A[b_m]$, hence $b_m$ is integral.

We have thus proved that if $f\in B[x]$ is integral over $A[x]$ then its leading coefficient is integral over $A$. But then $f - b_m x^m$ is integral over $A[x]$ and we can reason by induction, proving that all coefficients are integral.

Note 1 I'd like to thank Darij Grinberg for spotting a mistake in my original argument and suggesting the way to correct it. He also pointed out in the comments, that there are proofs of this statement with minimal assumptions in

1. Proposition 12 in §V.1.3 of Bourbaki's Commutative Algebra
2. Theorem 2.3.2 in Irena Swanson and Craig Huneke, Integral Closure of Ideals, Rings, and Modules.

Note 2 Thanks to krm2233, I simplified my previous argument by using a more appropriate equivalent definition of integral element.

Answer 4

Exercise 5.9 in Atiyah-Macdonald has the statement as follows:

$A\subset B$ ( comm, $1$), $C$ the integral closure of $A$ in $B$, then $C[x]$ is the integral closure of $A[x]$ in $B[x]$.

The given hint is very clear and uses in an essential way the following fact

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($\simeq$ Ex. 5.8) $A$ ring comm, $1$, $f$, $g$, $h$ are monic polynomials in $A[x]$, $f\cdot g = h$, then every coefficient of $f$ is integral over the ring generated by the coefficients of $h$.

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Now the proof of 5.9 proceeds as follows: take $f\in B[x]$ integral over $A[x]$.

$$f^m + g_1 f^{m-1} + \cdots + g_m =0$$

From here we get

$$f( f^{m-1} + g_1 f^{m-2} + \cdots + g_{m-1} ) = - g_m $$

However, we cannot apply 5.8 since $f$ is not apriori monic, and neither is $\pm g_m$. But we can force it to become monic as follows: Write the equality as

$$(x^N + f - x^N)^m + g_1( x^N + f - x^N)^{m-1} + \cdots + g_m = 0$$ or

$$(X^N + f)^m + \bar g_1 (x^N + f)^{m-1}+ \cdots + \bar g_m$$ where $\bar g_1$, $\ldots$, $\bar g_m \in A[x]$, and moreover $\pm \bar g_m$ monic if $N$ is large enough.

Now we Can apply 5.8 and Bob's your uncle