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I Separable Polynomials - Paul E Bland's definition and exampl

  1. Jun 2, 2017 #1
    I am reading Paul E Bland's book: The Basics of Abstract Algebra and I am trying to understand his definition of "separable polynomial" and his second example ...

    Bland defines a separable polynomial as follows:


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    ... and Bland's second example is as follows:



    ?temp_hash=3fe8cdf2beb62400e3e56319cd158346.png




    I am uncomfortable with, and do not fully understand the above definition and am uncomfortable with the example as well ... I hope someone can clarify my difficulties and problems ...

    The first and second sentences of the definition above seem to lead to different notions of separability to me ...

    The first sentence of the definition:

    " ... ... For a field ##F##, an irreducible polynomial ##f(x) \in F[x]## of degree ##n## is said to be separable if ##f(x)## has ##n## distinct roots in its splitting field ... ... "

    Under this definition of a separable polynomial, the polynomial in Bland's example:

    ##f(x) = (x^2 + 2)^2 (x^2 - 3)##

    is of degree ##n = 6## and splits in (among other fields) in ##\mathbb{C}## and does NOT have ##6## distinct roots (as the roots ##\pm \sqrt{2} i## are repeated) ... ...


    (EDIT ... hmm ... last minute thought! ... but i guess you could argue that ##f(x)## is not irreducible ... is that the key to my confusion ...?)


    ... BUT ...


    The second sentence of Bland's definition reads:

    " ... ... A polynomial in ##F[x]## is said to be separable if each of its irreducible factors is separable if each of its irreducible factors is separable ... "

    Well ... under this definition I (uncomfortably) go along with Bland's analysis of

    ##f(x) = (x^2 + 2) (x^2 + 2) (x^2 - 3)##

    in his example ...


    BUT i remain uncomfortable with this ... I cannot think of an example of a case where this definition gives rise to a repeated root ...can someone please give a simple example of a polynomial with a repeated root under Bland's definition ...



    Help will be much appreciated ...

    Peter
     

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  3. Jun 2, 2017 #2

    andrewkirk

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    That's because they are not easy to find.

    If the field has zero characteristic, all irreducible polynomials have distinct roots in the splitting field.

    But that is not the case for fields with nonzero characteristic. An example is the polynomial ##x^2+1## over the field ##\mathbb Z_2##. The splitting field is ##\mathbb Z_2+i\mathbb Z_2##, and the root ##i## has multiplicity two.

    There's more detail here.
     
  4. Jun 2, 2017 #3

    fresh_42

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    Me too. I wonder why he needs separability in such a broad sense, will say for reducible polynomials as well as for irreducible. Maybe it makes further considerations easier. But he is consistent with this definition as he requires all irreducible parts to be separable which is the point that counts. It counts because the whole theory is about field extensions, and for them irreducibility is important. Composed polynomials can be factored and field extensions therefore reduced to steps in which each is with an irreducible polynomial. In the example above we have the two possible towers: ##\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{2}i) \subseteq \mathbb{Q}(\sqrt{2}i,\sqrt{3})## and ##\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}i,\sqrt{3})##. So that's why I'm a bit confused, too. I cannot see the need for his definition.

    But it isn't easy to find an inseparable field extension anyway.

    In the characteristic ##0## case, all irreducible polynomials are separable. And with his definition: all polynomials are separable.

    In the characteristic ##p## case, a field extension is perfect (= all elements are separable) if and only if the field contains to each element also its ##p-##th root.

    That narrows the inseparable cases a lot. So the main problem for you is probably to keep the definitions in the various books apart and not so much the definition itself. What counts are the irreducible polynomials and I guess that other authors simply don't define separability for reducible polynomials. Apparently Bland does. So what? Btw. that was my first thought as I read his definition: What about the doubles in the reducible case? I had expected that he would have ruled them out and was surprised he didn't. Good that he gave this example to destroy any doubts how he means it. So as a summary: separable is if all irreducible factors are separable, regardless how often they occur.
     
  5. Jun 2, 2017 #4
    Thanks for the help Andrew, fresh_42 ...

    Reflecting on what you have said ...

    Peter
     
  6. Jun 2, 2017 #5

    Thanks Andrew ... appreciate the help ...

    But just clarifying your example ...

    If ##i## is a root of ##f(x)## with multiplicity ##2## then

    ##f(x) = (x -i ) ( x - i ) = x^2 - 2xi - 1 = x^2 - 1## ...


    Did you mean ##f(x) = x^2 - 1## ... or have I made an error ...


    Peter


    EDIT ... oh ... I think that in ##\mathbb{F}_2## we have ##-1 \equiv 1## ...

    ... so ... ##x^2 - 1 \equiv x^2 + 1## ... is that right?
     
  7. Jun 2, 2017 #6

    andrewkirk

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    Yes, that's right.
     
  8. Jun 3, 2017 #7

    Hi Andrew ...

    Just another clarification ...

    You write:

    " ... ... An example is the polynomial ##x^2+1## over the field ##\mathbb Z_2##. The splitting field is ##\mathbb Z_2+i\mathbb Z_2##, and the root ##i## has multiplicity two. ... "

    Given that the roots of the polynomial ##x^2+1## over the field ##\mathbb Z_2## are ##i## of multiplicity ##2## isn't the splitting field just ##i\mathbb Z_2##?

    Peter

    EDIT ... now I realise that I'm a bit unsure how to determine a splitting field ..
     
  9. Jun 3, 2017 #8

    andrewkirk

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    The splitting field is the smallest field containing ##\mathbb Z_2## and the roots of the polynomial, ie containing ##\mathbb Z_2\cup\{i\}##. That field is ##\mathbb Z_2(i)##.

    ##i\mathbb Z_2## is not a field, as it's not closed under multiplication, so it can't be the splitting field. But the splitting field is the field generated by ##i\mathbb Z_2##.
     
  10. Jun 3, 2017 #9

    Thanks Andrew ... that certainly clarified that issue ...

    Thanks for all your help and support with the above series of posts ..

    Peter
     
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