Martin's Blog : Hensel's lemma and algebraic functions

Hensel's lemma and algebraic functions

Posted by martin on Monday, 05 April 2010 at 22:34

An algebraic function is a function which we obtain by solving a polynomial in two variables $x$ and $y$ to write $y$ as a function of $x$ . In general polynomials have more than one root, so (informally) we get a multi-valued function. In this post I will restrict attention to regions of the plane in which we can unambiguously pick a single “branch” of the function. What happens where branches meet will be the subject of a later post.

I shall give an algorithm for expressing an algebraic function (in such a nicely behaved region) as a power series, thereby proving that a power series solution to the original polynomial exists. The generalisation of this result to a complete discrete valuation ring is Hensel’s lemma, and is particularly important to number theorists in the case of p-adic integers (which were invented by Hensel). In this post I will focus on the case of algebraic functions, as it is easier to apply geometric intuition.

Algorithm for constructing power series

Suppose we have a polynomial $f(x, y) \in \mathbb{C}[x, y]$ , which defines a curve in the affine plane. We want to solve the equation $f(x, y) = 0$ to obtain $y$ as a power series in $x$ . (I am interested in formal solutions here, and don’t care about convergence of the power series.)

Of course, in general there will be several values of $y$ corresponding to a single $x$ , so there are several power series which solve the equation.

Consider a value $y_0$ at which $g(y) := f(0, y)$ has a simple zero (i.e. $g'(y_0) \neq 0$ ). Then as we move along the curve near $(0, y_0)$ , $y$ is a single-valued function of $x$ , so we can expect to look for a single power series solution with constant term $y_0$ .

Of course it is not obvious that the function $y(x)$ can actually be expressed by a power series. This will follow from giving an algorithm to construct a power series solution (ignoring convergence issues).

$y^3 - y^2 - x = 0$

For example, take $f(x, y) = y^3 - y^2 - x$ .

Then $f(0, y) = y^3 - y^2$ has a simple zero at $y_0 = 1$ .

So there should be a unique power series $y = 1 + a_1 x + a_2 x^2 + ...$ satisfying $f(x, y) = 0$ .

To find it, we can solve for $a_i$ one at a time:

Try $y_1 = 1 + a_1 x$ . Substitute this in $f(x, y)$ and we get a polynomial in $x$ with no constant term. The $x$ term is $(a_1 - 1) x$ so we set $a_1 = 1$ to make this term vanish.
Try $y_2 = 1 + x + a_2 x^2$ . Substituting this in $f(x, y)$ , we get a polynomial where the constant and $x$ terms vanish. The $x^2$ term is $(a_2 + 2) x^2$ so setting $a_2 = -2$ , this term vanishes.
Try $y_3 = 1 + x - 2x^2 + a_3 x^3$ . Substituting this in $f(x, y)$ , we get a polynomial where the constant, $x$ and $x^2$ terms vanish. The $x^3$ term is $(a_3 - 7) x^3$ so setting $a_3 = 7$ , this term vanishes.

Continuing in this way, we build up the power series $y_\infty = 1 + x - 2x^2 + 7x^3 - 30x^4 + 143x^5 - 728x^6 + \ldots$

The graph to the right shows that the first four terms of this power series approximate the curve $f(x, y) = 0$ near $(0, 1)$ .

Proof that the algorithm works

Now we shall prove that the algorithm does construct a power series $y_\infty$ satisfying $f(x, y_\infty) = 0$ .

At each step we are setting $y_k$ to be a polynomial of degree $k$ , with unknown $x^k$ coefficient and other coefficients known.

By induction, we have chosen $a_0, \ldots, a_{k-1}$ so that the terms of degree less than $k$ in the expansion of $f(x, y_{k-1})$ vanish. The $a_k x^k$ term in $y_k$ does not affect terms of lower degree in $f(x, y_k)$ , so terms of degree less than $k$ in $f(x, y_k)$ vanish.

We need to show that there is a unique value for $a_k$ such that the $x^k$ term of $f(x, y_k)$ vanishes. But this term has the form $(g'(y_0) a_k + c) x^k$ for some constant $c$ , so indeed the condition $g'(y_0) \neq 0$ ensures that there is a unique $a_k$ making it vanish.

If we let $y_\infty$ be the power series obtained as the limit of the $y_k$ , then the terms of $f(x, y_\infty)$ of degree at most $k$ are the same as the terms of $f(x, y_k)$ of degree at most $k$ , so are all 0. Since this holds for all $k$ , $f(x, y_\infty) = 0$ as formal power series.

(Note: it is perhaps wrong to call this an “algorithm” since it doesn’t terminate. It can’t terminate because it is constructing an infinite object, a power series. But in finite time it will give us any particular term of the power series.)

Hensel’s lemma

The above algorithm gives a constructive proof of:

Hensel’s lemma. Let $f(x, y)$ be a polynomial in $\mathbb{C}[x, y]$ , and suppose that $y_0$ is a simple root of $g(y) := f(0, y)$ . Then there is a unique $y(x) \in \mathbb{C}[[x]]$ (the ring of formal power series) such that $y(0) = y_0$ and $f(x, y(x)) = 0$ as formal power series.

Note that the lemma is entirely asymmetric between $x$ and $y$ . We can look at it as solving a polynomial in one variable $y$ , whose coefficients happen to be elements of the ring $\mathbb{C}[x]$ . Since the solutions end up in $\mathbb{C}[[x]]$ , it is more natural to consider polynomials in $y$ with coefficients in $\mathbb{C}[[x]]$ (the resulting functions $y(x)$ are sometimes called algebroid functions). Exactly the same algorithm still works.

If you know what a complete discrete valuation ring is, you can observe that the algorithm works for polynomials over any complete DVR (the function $g(y)$ , generalising $f(0, y)$ , being the reduction of $f$ mod the maximal ideal). Any DVR $A$ is a 1-dimensional ring, so $A[y]$ is 2-dimensional. You can look at Hensel’s lemma as being about giving a local expression for $y$ on the curve $f(y) = 0$ , near a point $(*, y_0)$ (where $*$ is the closed point of $\mathop{\mathrm{Spec}} A$ ). The condition that $g'(y_0) \neq 0$ is the condition that near this point, the curve has just one branch.