Zackary Diaz

Answered

2022-11-23

Modular operation on polynomial rings

I originally started modular arithmetic by the following:

1mod2

1/2 is 0.5

0 times 2 is 0

$1-0=1$

1 equals 1mod2.

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

${x}^{3}+2x{y}^{2}-2x{z}^{2}+xmod\phantom{\rule{thickmathspace}{0ex}}{x}^{2}+{y}^{2}-{z}^{2}$

$\frac{{x}^{3}+2x{y}^{2}-2x{z}^{2}+x}{{x}^{2}+{y}^{2}-{z}^{2}}$

$\frac{({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)\times ({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)}{({x}^{2}+{y}^{2}-{z}^{2})}$

${x}^{3}+2x{y}^{2}-2x{z}^{2}+x-{\displaystyle \frac{({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)\times ({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)}{{x}^{2}+{y}^{2}-{z}^{2}}}$

I originally started modular arithmetic by the following:

1mod2

1/2 is 0.5

0 times 2 is 0

$1-0=1$

1 equals 1mod2.

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

${x}^{3}+2x{y}^{2}-2x{z}^{2}+xmod\phantom{\rule{thickmathspace}{0ex}}{x}^{2}+{y}^{2}-{z}^{2}$

$\frac{{x}^{3}+2x{y}^{2}-2x{z}^{2}+x}{{x}^{2}+{y}^{2}-{z}^{2}}$

$\frac{({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)\times ({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)}{({x}^{2}+{y}^{2}-{z}^{2})}$

${x}^{3}+2x{y}^{2}-2x{z}^{2}+x-{\displaystyle \frac{({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)\times ({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)}{{x}^{2}+{y}^{2}-{z}^{2}}}$

Answer & Explanation

Phiplyrhypelw0

Expert

2022-11-24Added 24 answers

Step 1

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

$a\equiv b\text{(mod}n)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}a-b$ 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:

$\frac{40}{17}\text{is}\mathrm{2.35294...}$ 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 $40=2(17)+6$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}40-6=2(17)$, which is a multiple of 17

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}40\equiv 6$ (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.

Step 2

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

$f(x)\equiv g(x)\phantom{\rule{thickmathspace}{0ex}}{\textstyle (}\text{mod}h(x){\textstyle )}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}f(x)-g(x)=p(x)h(x)$ 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

${x}^{3}+2x{y}^{2}-2x{z}^{2}+x$

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

We notice that ${x}^{3}+2x{y}^{2}-2x{z}^{2}+x=x(2{x}^{2}+2{y}^{2}-2{z}^{2})-{x}^{3}+x=2x({x}^{2}+{y}^{2}-{z}^{2})-{x}^{3}+x$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)-(-{x}^{3}+x)=2x({x}^{2}+{y}^{2}-{z}^{2})$ is a multiple of ${x}^{2}+{y}^{2}-{z}^{2}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{3}+2x{y}^{2}-2x{z}^{2}+x\equiv -{x}^{3}+x\text{(mod}{x}^{2}+{y}^{2}-{z}^{2})$

Step 2

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

${x}^{3}+2x{y}^{2}-2x{z}^{2}+x=x({x}^{2}+{y}^{2}-{z}^{2})+x{y}^{2}-x{z}^{2}+x$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{3}+2x{y}^{2}-2x{z}^{2}+x\equiv x{y}^{2}-x{z}^{2}+x\text{(mod}{x}^{2}+{y}^{2}-{z}^{2})$

and this is of course fine too.

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

$a\equiv b\text{(mod}n)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}a-b$ 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:

$\frac{40}{17}\text{is}\mathrm{2.35294...}$ 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 $40=2(17)+6$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}40-6=2(17)$, which is a multiple of 17

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}40\equiv 6$ (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.

Step 2

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

$f(x)\equiv g(x)\phantom{\rule{thickmathspace}{0ex}}{\textstyle (}\text{mod}h(x){\textstyle )}\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}f(x)-g(x)=p(x)h(x)$ 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

${x}^{3}+2x{y}^{2}-2x{z}^{2}+x$

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

We notice that ${x}^{3}+2x{y}^{2}-2x{z}^{2}+x=x(2{x}^{2}+2{y}^{2}-2{z}^{2})-{x}^{3}+x=2x({x}^{2}+{y}^{2}-{z}^{2})-{x}^{3}+x$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}({x}^{3}+2x{y}^{2}-2x{z}^{2}+x)-(-{x}^{3}+x)=2x({x}^{2}+{y}^{2}-{z}^{2})$ is a multiple of ${x}^{2}+{y}^{2}-{z}^{2}$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{3}+2x{y}^{2}-2x{z}^{2}+x\equiv -{x}^{3}+x\text{(mod}{x}^{2}+{y}^{2}-{z}^{2})$

Step 2

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

${x}^{3}+2x{y}^{2}-2x{z}^{2}+x=x({x}^{2}+{y}^{2}-{z}^{2})+x{y}^{2}-x{z}^{2}+x$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{3}+2x{y}^{2}-2x{z}^{2}+x\equiv x{y}^{2}-x{z}^{2}+x\text{(mod}{x}^{2}+{y}^{2}-{z}^{2})$

and this is of course fine too.

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