Chowla's theorem on class number of real quadratic field

Question

Let $p\equiv1\bmod 4$ be a prime number and $h$ the class number of real quadratic field $\mathbb Q(\sqrt{p})$, $\epsilon=\frac{t+u\sqrt{p}}{2}$ its fundamental unit. In this paper https://www.pnas.org/doi/abs/10.1073/pnas.47.6.878(Proc. Nat. Acad. Sci. U.S.A. 47 (1961), 878.) Chowla proves the following result $$h\equiv1\bmod 4\Longleftrightarrow\left(\frac{p-1}{2}\right)!\equiv-\frac{t}{2}\bmod p$$ or $$\left(\frac{p-1}{2}\right)!\equiv(-1)^{\frac{h+1}{2}}\cdot\frac{t}{2}\bmod p.$$

In the proof, his first show that $$\sqrt{p}\cdot\epsilon^{h}=\prod\limits_{\left(\frac{n}{p}\right)=-1}(1-\zeta^n)$$ where the product is over all the quadratic non-residue modulo $p$. This identity can be deduced by Dirichlet's class number formula. Let $\zeta=e^{\frac{2\pi i}{p}}$ be the $p$-th primitive root of unit. We know that $$p\mathbb Z[\zeta]=(1-\zeta)^{p-1},\quad \mathbb Z/p\mathbb Z\cong \mathbb Z[\zeta]/(1-\zeta).$$ Then Chowla Claim that in the ring $\mathbb Z[\zeta]$, $$\prod\limits_{\left(\frac{n}{p}\right)=-1}(1-\zeta^n)\equiv \left(\prod\limits_{\left(\frac{n}{p}\right)=-1}n\right)(1-\zeta)^{\frac{p-1}{2}}\equiv(1-\zeta)^{\frac{p-1}{2}}\bmod (1-\zeta)^{\frac{p+1}{2}}$$ and $$\left(\frac{p-1}{2}\right)!(1-\zeta)^{\frac{p-1}{2}}\equiv\epsilon^{-h}\cdot(1-\zeta)^{\frac{p-1}{2}}\bmod(1-\zeta)^{\frac{p+1}{2}}.$$ From this, he can deduce his theorem. I kown how to prove the first congruence identity. In fact we have that $$\frac{1-\zeta^k}{1-\zeta}=1+\zeta+\zeta^2+\cdots+\zeta^{k-1}\equiv k\bmod (1-\zeta)\Longrightarrow 1-\zeta^k\equiv k(1-\zeta)\bmod (1-\zeta)^2.$$ My Question: How to prove the second congruence identity, I mean how to prove $$\sqrt{p}\equiv\left(\frac{p-1}{2}\right)!\cdot(1-\zeta)^{\frac{p-1}{2}}\bmod(1-\zeta)^{\frac{p+1}{2}}.$$

Answer 1

Note that $$ \prod_{n=1}^{p-1}(1-\zeta^n)=\Phi_p(1)=p, $$ where $\Phi_p(x)=\frac{x^p-1}{x-1}$ is the cyclotomic polynomial. Next, $1-\zeta^{p-n}=-\zeta^{-n}(1-\zeta^n)$, hence we have $$ p=\prod_{n=1}^{(p-1)/2} (1-\zeta^n)^2(-\zeta^{-n}), $$ therefore $$ \left(\prod_{n=1}^{(p-1)/2}(1-\zeta^n)\right)^2=p\zeta^{(p^2-1)/ 8}, $$ because $1+2+\cdots+(p-1)/2=\frac{p^2-1}{8}$. Also, $1-\zeta^n=-2i \zeta^{n/2}\sin\frac{\pi n}{p}$, hence the product in LHS above is equal to $$ (-i)^{(p-1)/2}\zeta^{(p^2-1)/16}\sqrt{p}. $$ Let $\eta=\sqrt{\zeta}$ be a primitive $2p$-th root of unity. Then $(-i)^{(p-1)/2}=(-1)^{(p-1)/4}=\eta^{p(p-1)/4}$. Note now that $p(p-1)/4+(p^2-1)/8=(p-1)(3p+1)/8$ is an even number, so $$ (-i)^{(p-1)/2}\zeta^{(p^2-1)/16}=\eta^{(p-1)(3p+1)/8}=\zeta^k\equiv 1\pmod {1-\zeta} $$ Hence $$ \prod_{n=1}^{(p-1)/2}(1-\zeta^n)=\zeta^k \sqrt{p}. $$ Dividing both sides by $\zeta^k(1-\zeta)^{(p-1)/2}$ and using the observation $\frac{1-\zeta^n}{1-\zeta}\equiv n\pmod {1-\zeta}$ stated in the question, we see that $$ \frac{\sqrt{p}}{(1-\zeta)^{(p-1)/2}}\equiv \left(\frac{p-1}{2}\right)! \pmod {1-\zeta}, $$ as needed.