I'm reading this introductory blog on the Fan-Vercauteren scheme and there are a few things I don't understand about polynomial moduli. The author uses practical examples:
Because we are considering remainders with respect to $x^{16}+1$, we only have to consider polynomials with powers from $x^0$ to $x^{15}$. Any larger powers will be reduced by division by the polynomial modulus. This can also be understood by considering that $x^{16}\equiv −1(\mathrm{mod} \ x^{16}+1)$, meaning that $x^{16}$ can be replaced by -1 to reduce the larger powers of x into the range 0 to 15.
I don't understand the third sentence at all, $x^{16} \equiv -1$?
Further down the tutorial, there is a practical example of polynomial multiplication followed by a division by polynomial modulus:
When we multiply two powers of x, say $2x^{14}$ and $x^4$, we add their exponents - making $2x^{18}$. One might assume that finding the remainder of this polynomial with respect to the polynomial modulus might involve just rotating the exponent back through 0 at $x^{16}$, to give $2x^{2}$, like it does for the integer coefficients shown above. This would be the case if the polynomial modulus was just $x^{16}$. However, our polynomial modulus is $x^{16}+1$ - as mentioned above, the extra plus one factor introduces a sign change which helps further scramble the result of the multiplication.
How does this extra plus one factor introduce a sign change?
and finally:
...multiplication of a term $2x^{14}$ by $x^4$ modulo $x^{16}+1$ takes this term (represented by the red dot above), rotates it forward 4 powers around the torus, and then reflects the value across the 0 point, making $22x^2$ (or $−2x^2$ if we consider numbers from -12 to 11 rather than 0 to 23).
$2x^{18}$ modulo $x^{16}+1 = 22x^2$, how?