Intuition behind Kronecker's congruence? 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?
 A: I'm assuming that the modular polynomial you are talking about is the same one as in Wikipedia, although I don't see how to make that description match up with yours. For that one, if $X = j(\tau)$, then the roots of $\Phi_p(X, Y)$ are $j(\tau/p)$, $j((\tau+1)/p)$, $j((\tau+2)/p)$, ..., $j((\tau+p-1)/p)$ and $j(p \tau)$.
The motivation for this is as follows. Let $\Lambda$ be the free $\mathbb{Z}$-lattice spanned by $1$ and $\tau$. Then $\Lambda$ has $p+1$ index-$p$ sublattices: $\mathrm{Span}(p, \tau)$, $\mathrm{Span}(p, 1+\tau)$, ..., $\mathrm{Span}(p, \tau+p-1)$ and $\mathrm{Span}(1, p \tau)$. Call these lattices $\Lambda_j$. Thus, the elliptic curve $\mathbb{C}/\Lambda$ has a degree $p$ isogeny from the $p+1$ elliptic curves $\mathbb{C}/\Lambda_j$. Since rescaling $\Lambda_j$ by a scalar won't change the corresponding quotient curve, we may equally well work with $\mathrm{Span}(1, \tau/p)$, $\mathrm{Span}(1, (1+\tau)/p)$, ..., $\mathrm{Span}(1, (\tau+p-1)/p)$ and $\mathrm{Span}(1, p \tau)$.
Recall that the classical modular function $j(\tau)$ is the $j$-invariant of the elliptic curve $\mathbb{C}/\mathrm{Span}(1, \tau)$. So we see that, for elliptic curves $E_1$ and $E_2$ over $\mathbb{C}$, there is a degree $p$ isogeny $E_1 \to E_2$ if and only if $\Phi_p(j(E_1), j(E_2))=0$.
Now, the intuition for the question you ask about is that this result works not only over $\mathbb{C}$, but also over $\overline{\mathbb{F}_p}$. And so the key result is that, for $E_1$ and $E_2$ over $\overline{\mathbb{F}_p}$, there is a degree $p$ isogeny between $E_1$ and $E_2$ if and only if either $E_2$ is the Frobenius image of $E_1$ (in which case $j(E_2) = j(E_1)^p$) or vice versa (in which case $j(E_1) = j(E_2)^p$). See this answer for more on this. 
But I think that to make this intuition more precise, you are going to want to learn a lot more about how to talk about elliptic curves in the modern language of algebraic geometry. Have you worked through Silverman's books yet? That is the usual start.
A: 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.
