diff --git a/book/src/background/fields.md b/book/src/background/fields.md
index d793356a..779ef080 100644
--- a/book/src/background/fields.md
+++ b/book/src/background/fields.md
@@ -254,11 +254,97 @@ describes the general strategy. (There is another square root algorithm that use
 quadratic extension fields, but it doesn't pay off in efficiency until the prime becomes
 quite large.)
 
-> TODO: describe more recent algorithms, e.g. Bernstein's table-based method and
-> [eprint 2020/1407](https://eprint.iacr.org/2020/1407).
-
 [ts-sqrt]: https://en.wikipedia.org/wiki/Tonelli%E2%80%93Shanks_algorithm
 
+### Sarkar square-root algorithm (table-based variant)
+We use a technique from [Sarkar2020](https://eprint.iacr.org/2020/1407.pdf) to compute
+square roots in `halo2`. The intuition behind the algorithm is that we can split the task
+into computing square roots in each multiplicative subgroup.
+
+Suppose we want to find the square root of $u$ modulo an odd prime $p$, where $u$ is a
+non-zero square in $\mathbb{Z}_p^\times$. We write $p - 1 \equiv 2^{n}m$ with $n \geq 1$
+and $m$ odd; $g = z^m$ where $z$ is a non-square in $\mathbb{Z}_p^\times$.
+
+Let $x_3 = uv^2, x_2 = x_3^{2^8}, x_1 = x_2^{2^8}, x_0 = x_1^{2^8}.$
+
+#### Precompute the following tables:
+$$
+gtab = \begin{bmatrix}
+g^0 & g^1 & ... & g^{255} \\
+(g^{2^8})^0 & (g^{2^8})^1 & ... & (g^{2^8})^{255} \\
+(g^{2^{16}})^0 & (g^{2^{16}})^1 & ... & (g^{2^{16}})^{255} \\
+(g^{2^{24}})^0 & (g^{2^{24}})^1 & ... & (g^{2^{24}})^{255}
+\end{bmatrix}
+$$
+
+$$
+invtab = \begin{bmatrix}
+(g^{2^{-24}})^0 & (g^{2^{-24}})^1 & ... & (g^{2^{-24}})^{255}
+\end{bmatrix}
+$$
+
+### i = 0, 1
+Using $invtab$, we lookup $t_0$ s.t. $x_0 = (g^{2^{-24}})^{t_0} \implies x_0 \cdot g^{t_0 \cdot 2^{24}} = 1.$
+
+Update global variable: $t = t_0.$
+
+Define $\alpha_1 = x_1 \cdot (g^{2^{16}})^{t}.$
+
+### i = 2
+Lookup $t_1$ s.t. 
+$$
+\begin{array}{l}
+\alpha_1 = (g^{2^{-24}})^{t_1} &\implies x_1 \cdot (g^{2^{16}})^{t_0} = (g^{2^{-24}})^{t_1} \\
+&\implies
+x_1 \cdot g^{(t_0 + 2^8 \cdot t_1) \cdot 2^{16}} = 1.
+\end{array}
+$$
+
+Update global variable:
+$t = t_0 + 2^8 \cdot t_1$
+
+Define $\alpha_2 = x_2 \cdot (g^{2^8})^{t}.$
+         
+### i = 3
+Lookup $t_2$ s.t. 
+
+$$
+\begin{array}{l}
+\alpha_2 = (g^{2^{-24}})^{t_2} &\implies x_2 \cdot (g^{2^8})^{t_0 + 2^8\cdot {t_1}} = (g^{2^{-24}})^{t_2} \\
+&\implies x_2 \cdot (g^{2^8})^{t_0 + 2^8 \cdot t_1 + 2^{16} \cdot t_2} = 1.
+\end{array}
+$$
+
+Update global variable:
+$t = t_0 + 2^8 \cdot t_1 + 2^{16} \cdot t_2$
+
+Define $\alpha_3 = x_3 \cdot g^{t}.$
+
+### Final result
+Lookup $t_3$ s.t.
+
+$$
+\begin{array}{l}
+\alpha_3 = (g^{2^{-24}})^{t_3} &\implies x_3 \cdot g^{t_0 + 2^8\cdot {t_1} + 2^{16} \cdot t_2} = (g^{2^{-24}})^{t_3} \\
+&\implies x_3 \cdot g^{t_0 + 2^8 \cdot t_1 + 2^{16} \cdot t_2 + 2^{24} \cdot t_3} = 1.
+\end{array}
+$$
+
+Update global variable:
+$t = t_0 + 2^8 \cdot t_1 + 2^{16} \cdot t_2 + 2^{24} \cdot t_3$
+
+We can now write
+$$
+\begin{array}{l}
+x_3 \cdot g^{t} = 1 &\implies x_3 \cdot g^{t + 1} = g \\
+&\implies uv^2 \cdot g^{t + 1} = g \\
+&\implies uv^2 = g^{-t} \\
+&\implies uv \cdot g^{t / 2} = v^{-1} g^{-t / 2}.
+\end{array}
+$$
+
+Squaring the RHS, we observe that $(v^{-1} g^{-t / 2})^2 = v^{-2}g^{-t} = u.$ Therefore, the square root of $u$ is $v^{-1} g^{-t / 2}.$
+
 ## Roots of unity
 
 In the previous sections we wrote $p - 1 = 2^k \cdot t$ with $t$ odd, and stated that an