Apparently there is this general proof technique where one takes advantage of the fact that objects are usually specified by a finite amount of data. In commutative algebra one example is proving a result over $\mathbb{Z}[x_1,\dots,x_n]$ then transferring this result to an arbitrary ring $A$ via some morphism $\mathbb{Z}[x_1,\dots,x_n]\to A$.
I'm trying to understand the following application of this technique from Lombardi and Quitté's book on page 145. I think these are the relevant points:
- We have proven the formal Nullstellensatz over $\mathbb{Z}[x_1,\dots,x_n]$, which says that if $f_1,\dots,f_s\in\mathbb{Z}[x_1,\dots,x_n]$, then $\langle f_1,\dots,f_s\rangle=\mathbb{Z}[x_1,\dots,x_n]$ if and only if the $f_i$ do not admit a common zero over any finite field.
- We would like to show the following result: for a commutative ring $A$, an element $f$ of $A[x_1,\dots,x_n]$ is invertible if and only if its constant element is invertible in $A$ and all other coefficients are nilpotent.
The “backwards” implication is clear, but it is their proof of the forwards implication I am confused about. They argue by induction on the number of coefficients, so it suffices to work in $A[x]$. Suppose $f$ is invertible. Letting $g$ be its inverse, we may assume $f$ and $g$ have constant coefficient $1$ and write $f=1+xf_1$, $g=1+xg_1$, so that to say $fg=1$ is to say precisely that $f_1+g_1+xf_1g_1=0$.
At this point I don't see what the proof is trying to do, but I think it's this: we can write $f_1=a_0+a_1x+\dots+a_nx^n$ and $g_1=b_0+\dots+b_mx^m$. We want to show the $a_i$ are nilpotent in the ring $\mathbb{Z}[a_0,\dots,a_n,b_0,\dots,b_m,x]/\langle f_1+g_1+xf_1g_1\rangle$. Once this is done, we can take the expected morphism $\mathbb{Z}[a_0,\dots,a_n,b_0,\dots,b_m,x]/\langle f_1+g_1+xf_1g_1\rangle\to A[x]$ (well-defined because of the assumptions on $f$ and $g$) to conclude. In other words, "things true about indeterminates over $\mathbb{Z}$ must be true in every ring" as is the case with free objects.
The book argues "since the indeterminates are evaluated in a field, the coefficients of $f_1$ are annihilated at the zeros of the polynomial system in the indeterminates given by the equality $f_1+g_1+xf_1g_1=0$. We conclude with the formal Nullstellensatz." This is the part I don't understand at all.
Best Answer
The idea is this. Let $R=\mathbb{Z}[a_0,\dots,a_n,b_0,\dots,b_m]$, where the $a_i$ and $b_j$ are indeterminates. Let $I$ be the ideal generated by the $a_i$ and the $b_j$, and let $J$ be the smallest ideal of $R$ such that $f_1+g_1+Xf_1g_1=0$ in $R/J[X]$. Here $f_1$ and $g_1$ are the elements $\sum a_iX^i$ and $\sum b_jX^j$ of $R[X]$, and $J$-generators are obtained by equating the coefficients of the powers of $X$ on both sides of the equation. Both $I$ and $J$ are finitely generated.
Then $I=\sqrt{I}\subseteq\sqrt{J}$. This follows from the formal Nullstellensatz, in the form of corollary III.9.10 on p. 144. We merely have to check that, if $K$ is a field and $\varphi:R\to K$ a ring homomorphism, $\varphi(I)K\subseteq\varphi(J)K$. But if $\varphi(J)=0$, then, denoting the extension $R[X]\to K[X]$ by $\varphi$ again, $\varphi(f)\varphi(g)=1$ in $K[X]$, where $f=1+Xf_1$ and $g=1+Xg_1\in R[X]$. Since $K$ is a field, $\varphi(f_1)=0=\varphi(g_1)$ in $K[X]$, that is, $\varphi(I)=0$. And if $\varphi(J)$ contains a non-zero element of $K$, trivially $\varphi(I)K\subseteq\varphi(J)K$.
Now under the natural map $\psi:R\to A$, which sends the $a_i$ to the coefficients of $f_1$ and the $b_j$ to those of $g_1$, we have $\psi(J)=0$, so it follows that the elements of $\psi(I)$ are nilpotent in $A$.