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?



Intuition behind Kronecker's congruence?
I'm assuming that the modular polynomial you are talking about is the same one as in [Wikipedia](https://en.wikipedia.org/wiki/Classical_modular_curve), 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](https://mathoverflow.net/questions/19390/intuition-behind-the-eichler-shimura-relation?rq=1) 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.


