Is it meaningful to calculate ( x + 1 ) x + 2 in G F ( 3 2 ), e.g. using discrete logs?I constructed GF 3 2 using the polynomials degree &lt; 2 and arithmetic modulo the irreducible polynomial ( x 2 + 1 ). Then I built a table of logs to the base of the primitive element ( 2 x + 2 ).I can see that the log function maps only the non-zero elements of the field, so it seems sensible to use modulo 8 arithmetic between log values. This seems to work - for example: log 2 x + 2 ⁡ ( ( 2 x + 1 ) 2 m o d x 2 + 1 ) = ( 2 × log 2 x + 2 ⁡ ( 2 x + 1 ) ) m o d 8 log 2 x + 2 ⁡ ( x ) = ( 2 × ( 7 ) ) m o d 8 6 = 6 But how should I calculate, for example, ( x + 1 ) ( x + 2 ) ? Clearly modulo 8 arithmetic isn't going to do the trick. I guess that I need to use an irreducible polynomial but I am uncertain as to which one. Can I just use ( x 2 + 1 ) again?

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Is it meaningful to calculate   (  x  +  1      )          x      +      2       in   G  F  (      3    2    ), e.g. using discrete logs?I constructed GF      3    2   using the polynomials degree   &amp;lt;  2 and arithmetic modulo the irreducible polynomial   (      x    2    +  1  ). Then I built a table of logs to the base of the primitive element   (  2  x  +  2  ).I can see that the log function maps only the non-zero elements of the field, so it seems sensible to use modulo 8 arithmetic between log values. This seems to work - for example:                              log                      2            x            +            2                          ⁡                              (                                                (            2            x            +            1                          )              2                                            m            o            d                                      x              2                        +            1                                                )                                              =                              (                          2        ×                  log                      2            x            +            2                          ⁡        (        2        x        +        1        )                                            )                                            m            o            d                        8                                                        log                      2            x            +            2                          ⁡        (        x        )                            =                              (                          2        ×        (        7        )                                            )                                            m            o            d                        8                                              6                            =        6            But how should I calculate, for example,   (  x  +  1      )          (      x      +      2      )      ? Clearly modulo 8 arithmetic isn&#039;t going to do the trick. I guess that I need to use an irreducible polynomial but I am uncertain as to which one. Can I just use   (      x    2    +  1  ) again?

Brooklyn Mcintyre · Accepted Answer

Step 1Obviously,   (  1  +  x      )          x      +      2       should be   (  1  +  x      )    x    ⋅  (  1  +  x      )    2  , but there&#039;s no consistent way to define   (  1  +  x      )    x  .All the nonzero elements of the finite field are given by   1  ,  1  +  x  ,  (  1  +  x      )    2    ,  …  ,  (  1  +  x      )    7  , so whatever   (  1  +  x      )    x   is, we have   (  1  +  x      )    x    =  (  1  +  x      )    k   for some integer k.As a result,   (  1  +  x      )                  x        2              =  [  (  1  +  x      )    k        ]    x    =  [  (  1  +  x      )    x        ]    k    =  [  (  1  +  x      )    k        ]    k    =  (  1  +  x      )                  k        2            . But we also know that   (  1  +  x      )                  x        2             should be   (  1  +  x      )          −      1        =  (  1  +  x      )    7  , because   (  1  +  x      )                  x        2              (  1  +  x  )  =  (  1  +  x      )                  x        2            +      1        =  1. So whatever k is, it must satisfy       k    2    ≡  7    (  mod    8  ); but there is no such integer.Step 2(In other words,   f  (  a  )  =      a    x   should satisfy   f  (  a  b  )  =  f  (  a  )  f  (  b  ) and   f  (  f  (  a  )  )  =      a          −      1      , but the above proves that there is no such f.)

Is it meaningful to calculate (x+1)^{x+2} in GF(3^2), e.g. using discrete logs?

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