Calculations on GF(16) find 0111/1111It's my first time doing finite field arithmetics. As an exercise, I want to find 0111 / 1111 ∈ G F ( 16 ) generated by Π ( α ) = 1 + α + α 4 that is an irreducible polynomial.In polynomial form we have: 0111 → α + α 2 + α 3 1111 → 1 + α + α 2 + α 3 If I perform the polynomial division, I obtain -1 (that is the same result obtained writing 0111 ≡ − 1 ( mod 1111 ).How can I compute this result -1 in the right element of the field? Or perhaps this some kind of sign that the result ∉ G F ( 16 )?

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Calculations on GF(16) find 0111/1111It&#039;s my first time doing finite field arithmetics. As an exercise, I want to find   0111      /    1111  ∈  G  F  (  16  ) generated by   Π  (  α  )  =  1  +  α  +      α    4   that is an irreducible polynomial.In polynomial form we have:  0111  →  α  +      α    2    +      α    3    1111  →  1  +  α  +      α    2    +      α    3  If I perform the polynomial division, I obtain -1 (that is the same result obtained writing   0111  ≡  −  1    (  mod    1111  ).How can I compute this result -1 in the right element of the field? Or perhaps this some kind of sign that the result   ∉  G  F  (  16  )?

canhaulatlt · Accepted Answer

Step 1As it happens, the discrete logarithm table for GF(16) that I prepared for referrals like this, uses the same minimal polynomial of the primitive element. The only difference is that in the linked thread I denote by γ the element that you refer to as α.Step 2Anyway, consulting that table, we see that  1111  =  1  +  α  +      α    2    +      α    3    =      α          12        ,and  0111  =  α  +      α    2    +      α    3    =      α          11        .Therefore                              0111          1111                                    =                              α                          11                                            α                          12                                      =                              α                          11                                            α                          12                                      ⋅                              α            3                                α            3                                                            =                              α                          14                                            α                          15                                      =                              α                          14                                1                                                  =                  α                      14                          =                  α          3                +        1        =        1001.

Kendrick Finley · Accepted Answer

Step 1GF(16)  has characteristic 2. That is, each coefficient of   α is either 0 or 1. And   −  1  =  1.However, the polynomial division does not just result in -1 or 1. Instead we have:                    α        3            +              α        2            +      α                      α        3            +              α        2            +      α      +      1        =  1  +      1                  α        3            +              α        2            +      α      +      1      As an alternative to the answers in the comments, we can write each element (except 0) as a power of   α. After all, GF(16) has a cyclic multiplicative group and       α          15        =  1.Step 2Over the polynomial       X    4    +  X  +  1 we have   0111  =      α          11       and   1111  =      α          12      . Therefore:0111/1111=α11/α12=α11+15−12=α14=1001This is consistent with your result:  1  +      1                  α        3            +              α        2            +      α      +      1        =  1  +      1          α              12              =  1  +            α              15                    α              12              =  1  +      α    3

Calculations on GF(16) find 0111/1111. It's my first time doing finite field arithmetics. As an exercise, I want to find 0111/1111 in GF(16) generated by Pi(alpha)=1+alpha+alpha^4 that is an irreducible polynomial.

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