Modular operation on polynomial ringsI originally started modular arithmetic by the following:1mod21/2 is 0.50 times 2 is 0 1 − 0 = 11 equals 1mod2.Is it the same way to compute a quotient of a polynomial ring such as C [ x 1 , … , x n ] x 2 + y 2 − z 2 x 3 + 2 x y 2 − 2 x z 2 + x mod x 2 + y 2 − z 2 x 3 + 2 x y 2 − 2 x z 2 + x x 2 + y 2 − z 2 ( x 3 + 2 x y 2 − 2 x z 2 + x ) × ( x 3 + 2 x y 2 − 2 x z 2 + x ) ( x 2 + y 2 − z 2 ) x 3 + 2 x y 2 − 2 x z 2 + x − ( x 3 + 2 x y 2 − 2 x z 2 + x ) × ( x 3 + 2 x y 2 − 2 x z 2 + x ) x 2 + y 2 − z 2

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Modular operation on polynomial ringsI originally started modular arithmetic by the following:1mod21/2 is 0.50 times 2 is 0  1  −  0  =  11 equals 1mod2.Is it the same way to compute a quotient of a polynomial ring such as                               C                [                  x                      1                          ,        …        ,                  x                      n                          ]                              x                      2                          +                  y                      2                          −                  z                      2                                    x    3    +  2  x      y    2    −  2  x      z    2    +  x  mod        x    2    +      y    2    −      z    2                                x          3                +        2        x                  y          2                −        2        x                  z          2                +        x                              x          2                +                  y          2                −                  z          2                                        (                  x          3                +        2        x                  y          2                −        2        x                  z          2                +        x        )        ×        (                  x          3                +        2        x                  y          2                −        2        x                  z          2                +        x        )                    (                  x          2                +                  y          2                −                  z          2                )                  x    3    +  2  x      y    2    −  2  x      z    2    +  x  −                    (                  x          3                +        2        x                  y          2                −        2        x                  z          2                +        x        )        ×        (                  x          3                +        2        x                  y          2                −        2        x                  z          2                +        x        )                              x          2                +                  y          2                −                  z          2

Phiplyrhypelw0 · Accepted Answer

Step 1You can think of two numbers being congruent modulo n as follows:  a  ≡  b   (mod   n  )    ⟺    a  −  b is a multiple of nIn your example we get:  1  −  1  =  0 which is indeed divisible by 2, in fact it&#039;s divisible by any number n.For another example, suppose we wish to calculate 40 (mod 17), we could proceed as you did:      40    17     is   2.35294... is 2.35294...2 times 17 is 34  40  −  34  =  6.so   40  ≡  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    ⟹    40  −  6  =  2  (  17  ), which is a multiple of 17    ⟹    40  ≡  6 (mod 17).We have in essence done the same thing both times, but in the latter calculation it is clearer how we &quot;extract&quot; the multiples of 17 from 40.Step 2This latter way of thinking carries over nicely to other quotient rings, such as the polynomial ring you are interested in. That is  f  (  x  )  ≡  g  (  x  )              (        mod   h  (  x  )            )          ⟺    f  (  x  )  −  g  (  x  )  =  p  (  x  )  h  (  x  ) for some   p  (  x  )  ∈      C    [      x    1    ,  .  .  .      x    n    ]. ie.   f  (  x  )  −  g  (  x  ) is a multiple of h(x). We now want to calculate      x    3    +  2  x      y    2    −  2  x      z    2    +  xAs we did in our above example, we wish to &quot;extract&quot; any multiples of       x    2    +      y    2    −      z    2   from       x    3    +  2  x      y    2    −  2  x      z    2    +  xWe notice that       x    3    +  2  x      y    2    −  2  x      z    2    +  x  =  x  (  2      x    2    +  2      y    2    −  2      z    2    )  −      x    3    +  x  =  2  x  (      x    2    +      y    2    −      z    2    )  −      x    3    +  x    ⟹    (      x    3    +  2  x      y    2    −  2  x      z    2    +  x  )  −  (  −      x    3    +  x  )  =  2  x  (      x    2    +      y    2    −      z    2    ) is a multiple of       x    2    +      y    2    −      z    2      ⟹        x    3    +  2  x      y    2    −  2  x      z    2    +  x  ≡  −      x    3    +  x   (mod       x    2    +      y    2    −      z    2    )Step 2One final thing to note is that there are other equivalent ways to state this, for example, we could have instead said that      x    3    +  2  x      y    2    −  2  x      z    2    +  x  =  x  (      x    2    +      y    2    −      z    2    )  +  x      y    2    −  x      z    2    +  x    ⟹        x    3    +  2  x      y    2    −  2  x      z    2    +  x  ≡  x      y    2    −  x      z    2    +  x   (mod       x    2    +      y    2    −      z    2    )and this is of course fine too.

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

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