TITLE: Intuition behind Kronecker's congruence?
QUESTION [6 upvotes]: The modular polynomial is defined by$$\Phi_n(X, \tau) = \prod_{\tau} (X - j(\tau)),$$where $j$ is the elliptic modular function and $\tau$ is running through classes of imaginary quadratic integers of discriminant $n$. What is an easy way to see that$$\Phi_p(X, Y) \equiv (X - Y^p)(X^p - Y) \text{ (mod }p \text{)},$$i.e. what is the intuition behind Kronecker's congruence?

REPLY [8 votes]: David Speyer posted his answer while I was typing, and it's great, but here's a slightly different take.
Consider a generic curve $E/\mathbb C_p$ having multiplicative reduction. Then $E/\mathbb C_p$ has a Tate model,
$$ E(\mathbb C_p) \cong \mathbb C_p^*/q^{\mathbb Z}, $$
where $q\in\mathbb C_p$ satisfies $0<|q|<1$. What are the curves $p$-isogenous to $E$. Letting $Q=q^{1/p}$ ($Q$ is any of the $p$'th roots of $q$), it's a nice exercise to check that they are the curves
$$ E_\zeta := \mathbb C_p^*/(\zeta Q)^{\mathbb Z} \quad\hbox{and}\quad E_0:=\mathbb C_p^*/(q^p)^{\mathbb Z},$$
where  $\zeta$ is any primitive $p$'th root of unity.
(These correspond to the subgroups $\boldsymbol\mu_p$ and $\langle\zeta Q\rangle$.)
From the $q$-expansion of the $j$-invariant, we have
$$ j(E) = q^{-1} + 744 + 196884q + \cdots, $$
So
$$ \Phi_p(X) = (X-j(q^p))\prod_\zeta (X-j(\zeta Q)) .$$
We have $j(q^p)\equiv j(q)^p\pmod{p}$, while the product looks like
$$
  \prod_\zeta (X-j(\zeta Q)) = \prod_\zeta (X-(\zeta Q)^{-1} - \cdots)
  = (X^p - Q^{-p} + \cdots) = (X^p-q^{-1} + \cdots) = (X^p-j(q) + \cdots).
$$
(I'm being a bit imprecise here, but we're looking at neighborhood of $q=0$, and
the dots represent lower order terms as $q\to0$.) Putting it together, we get
$$ \Phi_p(X) \equiv (X-j(q)^p)(X^p-j(q))\pmod{p}. $$
N.B. I'm not suggesting that this is a rigorous proof. But you asked for some intuition. For me, (1) it's always been writing $E/K$ as $K^*/q^{\mathbb Z}$ that's explained Kronecker; (2) you can turn this into a proof over $\mathbb C$ by being a bit more careful with the $q$ expansions; (3) you can also make the proof I've sketched over $\mathbb C_p$ rigorous with more work.
Of course, the fancy way to express it is that when you reduce the Hecke correspondence $T_p$ modulo $p$, you get $\Phi_p+\Phi_p^t$, the sum of the graph of Frobenius and it's transpose.