Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 136 / 2

One form of Hensel lemma is: if $R$ is a complete discrete valuation ring, $f\in R[X]$, and $f(a_1)\equiv 0(\mathrm{mod}\pi )$ while $f^{\prime }(a_1)\not\equiv 0(\mathrm{mod}\pi )$, then there is a unique $\alpha \in R$ with $f(\alpha )=0$ and $\alpha \equiv a_1(\mathrm{mod}\pi )$.

Inductively, if $f(a_n)\equiv 0(\mathrm{mod}{\pi }^n)$, choose $t$ modulo $\pi$ so thatand put $a_{n+1}=a_n+{\pi }^{n}t$. The unit $f^{\prime }(a_n)$ makes $t$ unique. The resulting sequence is Cauchy, so completeness gives a root $\alpha$. Applying the same first-order congruence to two roots proves uniqueness.

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For $f(X)=X^2-X-2m$, $f(0)\equiv 0(\mathrm{mod}2)$ and $f^{\prime }(0)\equiv 1(\mathrm{mod}2)$. Hensel's lemma gives $x\in {\mathbb{Z}}_2$ with $x^2-x=2m$, hence

For $v=\infty$, signs give two square classes and $4/|2{|}_{\infty }=2$. For odd $p$, parity of $v_p$ gives two classes and ${\mathbb{F}}_p^{\ast }/({\mathbb{F}}_p^{\ast }{)}^2$ gives two more, while Hensel makes every unit congruent to $1$ modulo $p$ a square; hence there are four. For $p=2$, valuation parity gives two classes and odd units modulo squares are represented by $1,3,5,7\mathrm{mod}8$, giving eight. Thus in every case

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The Laurent series field $k((t))$ consists of series ${\sum }_{n\ge N}{a}_n{t}^n$. The mapis a discrete valuation. A Cauchy sequence has each coefficient eventually constant, and these stabilized coefficients define its limit, proving completeness.

For $K={\mathbb{F}}_p((t))$, write $K^{\ast }=t^{\mathbb{Z}}\times {\mathbb{F}}_p^{\ast }\times (1+t{\mathbb{F}}_p[[t]])$. If $p$ is odd, Hensel's lemma makes squaring an automorphism of the last factor, so the square-class group has order four. If $p=2$, Frobenius sends $\sum a_n{t}^n$ to $\sum a_n^2{t}^{2n}$, and the classes of units with arbitrarily placed odd-degree terms give infinitely many square classes. Hence the group is finite exactly for odd $p$.

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