Monic polynomial $= 0 \mod p$ for all $x$ For a monic polynomial with integer coefficients (leading coefficient of $1$) $f(x)$ where $f(x) \equiv 0$ mod $p$ for all $x$, where $p$ is a prime number how do I show that the degree of the polynomial must be greater than $p$? Frankly I don't even understand how such a polynomial is possible. If it's always $0$ for all choices of $x$, then shouldn't all coefficients be $0$? But then how is it monic?

I assume $p$ is prime. Since $F_p=\mathbb{Z}/p\mathbb{Z}$ is a field, a monic (or just nonzero) polynomial over $F_p$ having $p$ distinct roots must have degree at least $p$.
Of course such polynomials exist: $x(x-1)(x-2)\dots(x-p)g(x)$, where $g$ is any monic polynomial.
