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I Algebraically Closed Fields

  1. Jun 29, 2016 #1
    Dummit and Foote in their book Abstract Algebra give the following definition of an algebraically closed field ... ...


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    From the remarks following the definition it appears that the definition only applies to ##K[x]## ...

    Does it also apply to ##K[x_1, x_2], K[x_1, x_2, x_3], \ ... \ ... \ , K[x_1, x_2, \ ... \ ... \ , x_n] , \ ... \ ... \ ...## ?

    That is ... when we say K is an algebraically closed field does it imply that every polynomial in ##K[x_1, x_2, \ ... \ ... \ , x_n]## has a root in ##K## ... ... ?

    ... ... or maybe it is better if I say ... how does the definition of algebraically closed generalise to ##K[x_1, x_2, \ ... \ ... \ , x_n]## ... ... ?

    Hope someone can clarify this issue ... ...

    Peter
     

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    Last edited: Jun 29, 2016
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  3. Jun 30, 2016 #2

    mathwonk

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    take a polynomial in x and y and set x equal to anything. then what do you have?
     
  4. Jun 30, 2016 #3

    andrewkirk

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    It seems to me that extending the notion to multivariate polynomials makes no difference to the field.

    If every univariate polynomial over ##K## has a root, so must any multivariate polynomial ##P(x_1,...,x_m)##, since the univariate polynomial in ##x_1## that is obtained by setting all other ##x_j## to 0 will have a root ##a##, whence ##(a,0,....,0)## will be a root of ##P##.

    Conversely, if every multivariate polynomial has a root then univariate polynomials must all have roots, since they are just the case where ##m=1##.

    Edit: Oh, ah, er Jinx! I suppose.
     
  5. Jun 30, 2016 #4
    Thanks for the help ... I see that if you put x equal to something, then you have a polynomial in y, which, since K is algebraically closed has a root in K ...

    So we have a point ##(x, y) \in K^2## for which the polynomial is zero ...

    Is that what you are getting at? ...

    Peter
     
  6. Jun 30, 2016 #5

    Thanks for the help, Andrew ... that appears to clarify my problem/issue ...!!!

    Peter
     
  7. Jun 30, 2016 #6

    fresh_42

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    Algebraic closure is a property of the field ##K##. It means that all polynomial roots, which are per definition zeros, i.e. "numbers", are already in the field. E.g. ##\mathbb{C}## in which the additional element ##i## provides all possible polynomial zeros over the real numbers. It is not a property of the polynomial ring, so the question about multivariate polynomial rings doesn't make sense.
    No. ##K[x_1, x_2] = K[x_1][x_2]=(K[X_1])[X_2]##. But ##(K[X_1])## is a ring and thus fails the condition of the definition.
    One would have to consider its quotient field ##K(X_1)## instead, which isn't an algebraic extension over ##K## and thus again fails the definition. So to make sense at all, one has to consider ##K(X_1)## as the "starting field". But this reduces the entire question to a situation ##K' \, [X_2]## as described anyway.
    So the only question which might be left is, whether the quotient field of a polynomial ring over an algebraic closed field is algebraic closed again, which is a different issue. However, ##p(y) = y^2 -x \in \mathbb{C}(x)[y]## has a root ##y=\sqrt{x}## which is not an element of ##\mathbb{C}(x)##. Hence ##\mathbb{C}(x)## is not algebraic closed although ##\mathbb{C}## is.
    No. ##X_1 - X_2 = 0## as polynomial in ##X_2## over ##\mathbb{C}[X_1]## has no root in ##\mathbb{C}##.
    Only if you consider all variables as being replaced by elements of ##K## then you have a chance to end up in ##K##.
    But this situation is not distinguishable from ##K[a_1,...,a_{n-1}][X_n] = K[X_n] ##, which is exactly the situation in the definition.

    Edit : The reason for it doesn't make sense to speak of an algebraic closure of rings ##R## in general - such as polynomial rings ##R = K[X_1]## - is: Even the division, i.e. the zeros of ##a \cdot X_1 - 1 = 0## for an element ##a \in R - \{0\}## isn't a priori guaranteed. Therefore one would have to assume the existence of and pass to a quotient field of ##R## and ending up with a field again.
     
    Last edited: Jun 30, 2016
  8. Jun 30, 2016 #7

    micromass

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  9. Jul 1, 2016 #8


    Thanks for the extensive help fresh_42 ... really appreciate your assistance in clarifying this matter ...

    Still reflecting on what you have written ... ...

    Thanks again,

    Peter
     
  10. Jul 1, 2016 #9
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