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I Separable Polynomials - Dummit and Foote - Proposition 37

  1. Jun 4, 2017 #1
    I am reading David S. Dummit and Richard M. Foote : Abstract Algebra ...

    I am trying to understand the proof of Proposition 37 in Section 13.5 Separable and Inseparable Extensions ...


    The Proposition 37 and its proof (note that the proof comes before the statement of the Proposition) read as follows:



    ?temp_hash=6dbe83814424bd9e74d92b99c61cd074.png



    In the above text by D&F we read the following:

    "If ##p(x)## were inseparable then we have seen that ##p(x) = q(x^p)## for some polynomial ##q(x) \in \mathbb{F} [x]##. ... ... "


    I cannot understand exactly why this is true ...


    Can someone please explain exactly why, as D&F assert, the following is true ... ...


    ##p(x)## inseparable ## \ \ \Longrightarrow \ \ p(x) = q(x^p)## for some polynomial ##q(x) \in \mathbb{F} [x]##


    I cannot find where D&F establish exactly this implication ... but maybe there is some idea in the proof of Corollary 34 which reads as follows:



    ?temp_hash=6dbe83814424bd9e74d92b99c61cd074.png
    ?temp_hash=6dbe83814424bd9e74d92b99c61cd074.png




    Corollary 36 may also be relevant ... so I am providing that as well ... as follows:




    ?temp_hash=6dbe83814424bd9e74d92b99c61cd074.png




    Hope someone can help explain the assertion by D&F mentioned above ..

    Peter
     
    Last edited: Jun 4, 2017
  2. jcsd
  3. Jun 4, 2017 #2

    andrewkirk

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    Call the irreducible polynomial ##P##, to distinguish it from ##p##, the characteristic of ##F##.

    Since ##P## is irreducible, it must either be co-prime with ##D_xP##, or else ##D_xP\equiv 0##

    If co-prime then by Proposition 9 (quoted in the proof of Corollary 34), ##P## has no roots in common with any of the irreducible factors of ##D_xP##. Hence ##P## has no roots in common with ##D_xP##. Hence ##P## is separable.

    So if ##P## is not separable, we must have ##D_xP\equiv 0##, which means that for each natural number ##k##, the ##k##th order coefficient in ##D_xP## must be zero. That coefficient is ##a_{k+1}(k+1)## where ##a_{k+1}## is the ##k##th order coefficient in ##P##. Since ##F## is a field there are no zero divisors, so either ##a_{k+1}=0_F## or ##(k+1)_F=0_F##. I have put subscripts in to make clear we are operating in ##F##.

    So for every term ##a_rx^r## of ##P## we must either have ##a_r=0## or else ##r=0\mod p##, in other words ##r=ps_r## for some natural number ##s_r##.

    But then each nonzero term of ##P## will be of the form ##a_{ps_r}x^{ps_r}## which can be written ##a_{ps_r}(x^p)^s_r##. So ##P## is a finite linear combination of powers of ##x^p##, ie a polynomial in ##x^p##. That new polynomial is the author's ##q##.
     
  4. Jun 5, 2017 #3
    Thanks Andrew ...

    Just reflecting on what you have said ...

    Peter
     
  5. Jun 5, 2017 #4

    andrewkirk

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    An interesting consequence of the above is that the counter-example I gave you a while back, of an irreducible, non-separable polynomial, was wrong!

    ##x^2+1## is indeed non-separable in ##\mathbb Z_2[x]##, but it is not irreducible, because it can be factorised as ##(x+1)^2##.

    Putting together Corollary 34 and Proposition 37, we conclude that the only fields containing irreducible, non-separable polynomials are infinite fields with non-zero characteristic.

    I think ##\mathbb Z_p(r)## for ##r## transcendental and ##p## prime, will be such fields, as they will have characteristic ##p>0## but will be infinite-dimensional vector spaces over ##\mathbb Z_p##, with basis ##\{r^k\ :\ r\in\mathbb Z,\ r\geq 0\}##.
     
  6. Jun 5, 2017 #5
    Oh God ... and I thought I understood that example!!!

    Thanks for pointing this out ...

    Must go through the example again tomorrow...

    Thanks again for correcting the record ...

    Peter
     
  7. Jun 5, 2017 #6

    fresh_42

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    I've found the following criterion (van der Waerden):

    A field of characteristic ##p## is perfect, i.e. every irreducible polynomial is separable, if and only if for each element there is a the ##p-##th root; or short ##F=F^p##.

    This also excludes finite fields (by consideration of ##a \mapsto a^p##). So the easiest example of an inseparable field is ##F=F_p(t)## the field of rational functions (polynomials) in ##t## and ##x^p-t \in F[x]##. Characteristic zero fields, algebraic extensions (esp. the algebraic closure) of separable fields and Galois fields are all separable, so many "interesting" fields are separable.
     
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