Finite field adjunction - isomorphism that maps from the field into a polynomial ring factored over an idealA field ( K , + , 0 , ∗ , 1 ) is given, and it is a finite field with 25 elements. P is the prime field over K. Further it was given that a polynomial r ( x ) := x 2 + x + 1 ∈ P [ x ] is irreducible.Now, the polynomial y 2 + y + 1 ∈ K [ x ] has two non-equal zeros in K, say α 1 ≠ α 2 .The question is to describe all possible isomorphisms: φ : K → P [ x ] / ( r )by giving the values of φ ( α 1 ) and φ ( α 2 ).There are a few things that I don't understand about this question.First of all, what are the properties of K[y] apart from it being a polynomial ring? I think that the additive group ( K , + , 0 ) should be isomorph to a cyclic group C 5 × C 5 - can that help me in any way?Then - when I try to find the α 1 and α 2 I do my calculations in the Modulo 25 arithmetic - is that right (because I failed to find the zeros)?And since the φ ( k ) = k + ( r ) is one of the possibilites to describe this isomorphism, my question is - are there any more of them? Because my textbook doesn't state that this isomorphism is unique.Any kind of help - or a link to some other question is more than welcome.

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Finite field adjunction - isomorphism that maps from the field into a polynomial ring factored over an idealA field   (  K  ,  +  ,  0  ,  ∗  ,  1  ) is given, and it is a finite field with 25 elements. P is the prime field over K. Further it was given that a polynomial   r  (  x  )  :=      x    2    +  x  +  1  ∈  P  [  x  ] is irreducible.Now, the polynomial       y    2    +  y  +  1  ∈  K  [  x  ] has two non-equal zeros in K, say       α    1    ≠      α    2  .The question is to describe all possible isomorphisms:  φ  :  K  →  P  [  x  ]      /    (  r  )by giving the values of   φ  (      α    1    ) and   φ  (      α    2    ).There are a few things that I don&#039;t understand about this question.First of all, what are the properties of K[y] apart from it being a polynomial ring? I think that the additive group   (  K  ,  +  ,  0  ) should be isomorph to a cyclic group       C    5    ×      C    5   - can that help me in any way?Then - when I try to find the       α    1   and       α    2   I do my calculations in the Modulo 25 arithmetic - is that right (because I failed to find the zeros)?And since the   φ  (  k  )  =  k  +  (  r  ) is one of the possibilites to describe this isomorphism, my question is - are there any more of them? Because my textbook doesn&#039;t state that this isomorphism is unique.Any kind of help - or a link to some other question is more than welcome.

ustalovatfog · Accepted Answer

Step 1The polynomial r has two roots in K,       α    1   and       α    2  . Their images must also be roots of the polynomial r in P[x]/(r). The notation here can be confusing since   x  ∈  P  [  x  ]      /    (  r  ) is an element of the field.Let&#039;s compute the roots of       y    2    +  y  +  1 in P[x]/(r). They must be of the form   a  x  +  b for   a  ,  b  ∈  P, which is a root if and only if       a    2        x    2    +  2  a  b  x  +      b    2    +  a  x  +  b  +  1  =      a    2        x    2    +  (  2  a  b  +  a  )  x  +  (      b    2    +  b  +  1  )  =  0. This must be divisible by       x    2    +  x  +  1, and a is non-zero since r is irreducible over P, so       a    2    =  2  a  b  +  a  =      b    2    +  b  +  1. Hence   a  =  2  b  +  1 and   (  2  b  +  1      )    2    =      b    2    +  b  +  1, so   b  (  b  +  1  )  =  0. Hence the roots are x and   −  x  −  1.Step 2This is somewhat ad-hoc. This can be clarified and generalized by thinking about homomorphisms   P  [  x  ]      /    (  r  )  →  K, which are determined by where x goes. The image of x must also be a root of r, so the only possibilities are if the image of x is       α    1   or       α    2  . These actually define homomorphisms since r is irreducible, so the kernel of the evaluation map   P  [  x  ]  →  K,   x  ↦      α    i   is (r).

Laila Murphy · Accepted Answer

Step 1This is not in any way an answer to your question.Your elements of the field K are expressions of the form   a  +  b  x with   a  ,  b  ∈  P. You add them in the obvious way, always treating a and b as things modulo 5, and you multiply them by replacing       x    2   by   −  1  −  x  =  4  +  4  x every time that you see an       x    2  .Step 2Why don’t you take an element like   2  +  3  x and write down a few of its first powers? I mean   (  2  +  3  x      )    2  ,   (  2  +  3  x      )    3  , etc. If you’re enthusiastic, why don’t you write down all its powers, up to the twenty-fourth?I promise you that by this exercise you will gain much more insight into the situation than by contemplating head-on the question put to you.

A field (K,+,0,∗,1) is given, and it is a finite field with 25 elements. P is the prime field over K. Further it was given that a polynomial r(x):=x^2+x+1 in P[x] is irreducible.

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