Modular operation on polynomial rings

I originally started modular arithmetic by the following:

$$1 \bmod 2$$

$$1/2$$ is $$0.5$$

$$0$$ times $$2$$ is $$0$$

$$1-0=1$$

$$1$$ equals $$1 \bmod 2$$.

Is it the same way to compute a quotient of a polynomial ring such as $$\dfrac{\mathbb{C}[x_{1},\dots,x_{n}]}{x^{2}+y^{2}-z^{2}}$$?

$$x^3+2xy^2-2xz^2+x \bmod \; x^2+y^2-z^2$$

$$\dfrac{x^3+2xy^2-2xz^2+x}{x^2+y^2-z^2}$$

$$\dfrac {(x^3+2xy^2-2xz^2+x) \times (x^3+2xy^2-2xz^2+x)}{(x^2+y^2-z^2)}$$

$$x^3+2xy^2-2xz^2+x - \dfrac {(x^3+2xy^2-2xz^2+x) \times (x^3+2xy^2-2xz^2+x)}{x^2+y^2-z^2}$$

You can think of two numbers being congruent modulo $$n$$ as follows:

$$a \equiv b \text{ (mod } n) \iff a - b$$ is a multiple of $$n$$

$$1 - 1 = 0$$ which is indeed divisible by $$2$$, in fact it's divisible by any number $$n$$.

For another example, suppose we wish to calculate $$40$$ (mod $$17$$), we could proceed as you did:

$$\frac{40}{17} \text{ is } 2.35294...$$

$$2$$ times $$17$$ is $$34$$

$$40 -34 = 6$$.

so $$40 \equiv 6$$ (mod $$17$$).

This is correct but somehow it feels, for lack of a better word, clunky. Instead, we can approach this from a slightly different point of view.

We observe that $$40 = 2(17) + 6$$

$$\implies 40 - 6 = 2(17)$$, which is a multiple of 17

$$\implies 40 \equiv 6$$ (mod $$17)$$.

We have in essence done the same thing both times, but in the latter calculation it is clearer how we "extract" the multiples of $$17$$ from $$40$$.

This latter way of thinking carries over nicely to other quotient rings, such as the polynomial ring you are interested in. That is

$$f(x) \equiv g(x) \; \big(\text{mod }h(x)\big) \iff f(x) - g(x) = p(x)h(x)$$ for some $$p(x) \in \mathbb{C}[x_1,...x_n]$$. ie. $$f(x) - g(x)$$ is a multiple of $$h(x)$$. We now want to calculate

$$x^3+2xy^2-2xz^2+x$$ $$\big($$mod $$x^2+y^2-z^2 \big)$$.

As we did in our above example, we wish to "extract" any multiples of $$x^2+y^2-z^2$$ from $$x^3+2xy^2-2xz^2+x$$.

We notice that $$x^3+2xy^2-2xz^2 + x = x(2x^2+2y^2-2z^2) - x^3 + x = 2x(x^2+y^2-z^2) - x^3 + x$$

$$\implies (x^3+2xy^2-2xz^2+x) -(- x^3 + x) = 2x(x^2+y^2-z^2)$$ is a multiple of $$x^2+y^2-z^2$$

$$\implies x^3+2xy^2-2xz^2+x \equiv - x^3 + x \text{ (mod }x^2+y^2-z^2)$$.

One final thing to note is that there are other equivalent ways to state this, for example, we could have instead said that

$$x^3+2xy^2-2xz^2 + x = x(x^2+y^2-z^2) + xy^2 - xz^2 + x$$

$$\implies x^3+2xy^2-2xz^2 + x \equiv xy^2 - xz^2 + x \text{ (mod }x^2+y^2-z^2)$$

and this is of course fine too.