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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 $\dfrac{\mathbb{C}[x_{1},\dots,x_{n}]}{x^{2}+y^{2}-z^{2}}$?

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

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

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

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

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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 $n$

In your example we get:

$1 - 1 = 0$ 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.

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