Exercise 2.6.11* (Multivariable generalization of Hensel's lemma)

Proof. By hypothesis $f(x)\in \mathbb{Z}[x_1,\dots ,x_n]$ is a polynomial with integral coefficients in the $n$ variables $x_1,x_2,\dots ,x_n$, and $a=(a_1,a_2,\dots ,a_n)\in {\mathbb{Z}}^n$. If $u_1,\dots ,u_n$ are variables, and $u=(u_1,\dots ,u_n)$, then $f(a+u)\in \mathbb{Z}[u_1,\dots u_n]$, so

$$f(a+u)=\underset{(i_1,\dots ,i_n)\in A}{\sum }{c}_{i_1,\dots ,i_k}{u}_1^{i_1}\cdots u_n^{i_n},$$

where $A\in {𝒫}_f({\mathbb{N}}^n)$ is a finite subset of ${\mathbb{N}}^n$, and $c_{i_1,\dots ,i_k}\in \mathbb{Z}$ for all indices $i_1,\dots ,i_k$.

The Taylor’s expansion for multivariate polynomials gives

$$\begin{array}{lll}\hfill f(a+u)=\underset{(i_1,\dots ,i_n)\in A}{\sum }\frac{1}{i_1!\cdots i_n!}\frac{{\partial }^{i_1+\cdots +i_n}f}{{\partial }^{i_1}{x}_1\cdots {\partial }^{i_n}{x}_n}(a)u_1^{i_1}\cdots u_n^{i_n}.& & \hfill \text{(1)}\end{array}$$

Note that $\frac{1}{i_1!\cdots i_n!}\frac{{\partial }^{i_1+\cdots +i_n}f}{{\partial }^{i_1}{x}_1\cdots {\partial }^{i_n}{x}_n}(a)=c_{i_1,\dots ,i_k}$ is an integer.

We show by induction that $f(x)\equiv 0(\mathit{mod}{p}^j)$ has a solution for every $j\ge 1$. This is true for $j=1$ since $a$ is a solution to $f(x)\equiv 0(\mathit{mod}p)$. Assume that there is a solution $a_j=(a_{1j},\dots ,a_{\mathit{nj}})$ to $f(x)\equiv 0(\mathit{mod}{p}^j)$, so that $f(a_j)\equiv 0(\mathit{mod}{p}^j)$.

If we apply the formula (1) with $a=a_j$ and $u=(t_1{p}^j,\cdots ,t_n{p}^j)$, we obtain, since $u_1^{i_1}\cdots u_n^{i_n}\equiv 0(\mathit{mod}{p}^{2j})$ if some $i_k>1$,

$$\begin{array}{lll}\hfill f(a_j+u)\equiv f(a_j)+\underset{i=1}{\overset{n}{\sum }}{t}_i{p}^j\frac{\mathit{\partial f}}{\partial x_i}(a_j)(mod{p}^{j+1}).& & \hfill \text{(2)}\end{array}$$ Write $d_i=\frac{\mathit{\partial f}}{\partial x_i}(a_j)\in \mathbb{Z}$. By hypothesis $d_i\not\equiv 0(\mathit{mod}p)$ for at least one $i$. Then $a_j+u=(a_{1j}+t_1{p}^j,\dots ,a_{\mathit{nj}}+t_n{p}^j)$ is a solution to the congruence $f(x)\equiv 0(\mathit{mod}{p}^{j+1})$ if and only if $$p^j(d_1{t}_1+\cdots +d_n{t}_n)\equiv -f(a_j)(\mathit{mod}{p}^{j+1}),$$

or equivalently,

$$d_1{t}_1+\cdots +d_n{t}_n\equiv -\frac{f(a_j)}{p^j}(\mathit{mod}p).$$

This last equation has a solution $(t_1,\dots ,t_n)$ modulo $p$. If we assume, without loss of generality, that $d_1\not\equiv 0,(\mathit{mod}p)$, then $(-\overline{d_1}\frac{f(a_j)}{p^j},0,\dots ,0)$ is a solution (where $\overline{d_1}$ is an integer chosen such that $d_1\overline{d_1}\equiv 1(\mathit{mod}p)$). The general solution is given by

$$t_1\equiv \overline{d_1}\left(-\frac{f(a_j)}{p^j}-d_2{t}_2-\cdots -d_n{t}_n\right)(\mathit{mod}p),$$

where $d_2,\dots ,d_n$ take arbitrary values.

This shows that $f(x)\equiv 0(\mathit{mod}{p}^{j+1})$ has a solution, given by

$$a_j+u=(a_{1j}+t_1{p}^j,\dots ,a_{\mathit{nj}}+t_n{p}^j),$$

where $(t_1,\dots ,t_n)$ is a solution of

$$d_1{t}_1+\cdots +d_n{t}_n\equiv -\frac{f(a_j)}{p^j}(\mathit{mod}p).$$

The induction is done, so the congruence $f(x)\equiv 0(\mathit{mod}{p}^j)$ has a solution for all $j\ge 1$. □