Do all monic polynomials factor into monic irreducibles?Do all monic polynomials in Z / n Z [ x ] where n is prime factor into monic irreducibles?Just by thinking about these monic polynomials and how they only act on a finite number of elements makes me think that the assertion is true, because if d is a root of a monic polynomial f(x) then f ( x ) = ( x − d ) g ( x ) where g(x) is monic, and you could apply the same rule iteratively until you exhaust all roots. But I am not sure how to really prove this, it sounds really similar to one part of the fundamental theorem of arithmetic, but I can't find how to quantify a polynomial the same way you would an integer for an inductive proof. How would I go about starting a proof of this?

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Do all monic polynomials factor into monic irreducibles?Do all monic polynomials in       Z        /    n      Z    [  x  ] where n is prime factor into monic irreducibles?Just by thinking about these monic polynomials and how they only act on a finite number of elements makes me think that the assertion is true, because if d is a root of a monic polynomial f(x) then   f  (  x  )  =  (  x  −  d  )  g  (  x  ) where g(x) is monic, and you could apply the same rule iteratively until you exhaust all roots. But I am not sure how to really prove this, it sounds really similar to one part of the fundamental theorem of arithmetic, but I can&#039;t find how to quantify a polynomial the same way you would an integer for an inductive proof. How would I go about starting a proof of this?

Haylie Park · Accepted Answer

Step 1If F is a field, then F[X] is a unique factorization domain, so every nonzero polynomial in F[X] can be factored into a unique product of irreducibles (up to scaling by units). If   f  ∈  F  [  X  ] is monic, then f can be factored as       f    1    ⋅      f    2    ⋯      f    n   for irreducible polynomials       f    1    ,  …  ,      f    n    ∈  F  [  X  ] whose leading coefficients       c          1        ,  …  ,      c          n        ∈  F must be units. Since       ∏          i      =      1              n            c          i        =  1, we can multiply each       f          i       by       c          i              −      1       to get                     f                  i                    ~       monic irreducible such that   f  =                    f                  1                    ~        ⋅                    f                  2                    ~        ⋯                    f                  n                    ~      , since this amounts to multiplying f by       ∏          i      =      1              n            c          i              −      1        =  (      ∏          i      =      1              n            c          i            )          −      1        =      1          −      1        =  1.Step 2This can be specialized to your case   F  =      Z        /    p      Z   for p prime.

kemecryncqe9 · Accepted Answer

Step 1If F is a field then any factorization of a monic polynomial in F[x] can be made monic. Inded, if f factors p and has leading coefficient   a  ≠  1, then       a          −      1        ⋅  f is monic and has the same roots as f, and hence also factos p.Step 2Now, if p is monic,   ∏      f    i   is any factorization of p and       a    i   is the leading coefficient of each       f    i  , then   ∏      a    i    =  1 so of course   ∏            (              a        i            )              −      1        =                    (        ∏                  a          i                )                    −      1        =  1. It follows that   ∏      (          a      i              −        1              ⋅          f      i        )   is a monic factorization of p.More generally, if p is a polynomial in F[x] with leading coefficient c then one can always factor p as c times a product of monic polynomials.

Do all monic polynomials in Z/nZ[x] where n is prime factor into monic irreducibles?

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