Results 1 to 5 of 5

Math Help - Extension Fields - Gallian Ch 20 - Example 1

  1. #1
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    558
    Thanks
    2

    Extension Fields - Gallian Ch 20 - Example 1

    I am working through Gallian's book "Contemporary Abstract Algebra" (Fifth Edition) and I am on Chapter 20 - Extension Fields.

    I need help with Example 1 on page 345 - see attached pdf. [My text below is on the pdf - I am trying to understand the equations]

    Example 1 states:

    Let f(x) = x^2 + 1 \in Q[x]

    Then in E = Q[x]/< x^2 + 1> we have

     f(x + <x^2 +1> ) = (x + <x^2 +1>)^2 +1

    = x^2 + <x^2 +1> + 1

    Can someone please help me understand how the above two lines are equal?

    Could you also explain how the last two eqations of the 4 are equal - see attached sheet.

    Bernhard
    Attached Files Attached Files
    Last edited by Bernhard; June 4th 2011 at 05:43 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by Bernhard View Post
    I am working through Gallian's book "Contemporary Abstract Algebra" (Fifth Edition) and I am on Chapter 20 - Extension Fields.

    I need help with Example 1 on page 345 - see attached pdf. [My text below is on the pdf - I am trying to understand the equations]

    Example 1 states:

    Let f(x) = x^2 + 1 \in Q[x]

    Then in E = Q[x]/< x^2 + 1> we have

     f(x + <x^2 +1> ) = (x + <x^2 +1>)^2 +1

    = x^2 + <x^2 +1> + 1

    Can someone please help me understand how the above two lines are equal?

    Could you also explain how the last two eqations of the 4 are equal - see attached sheet.

    Bernhard


    You´re supposed to know elementary ring theory BEFORE getting into fields extensions, and thus you must know that

    if R is a ring and I and ideal in it, then for any element a\in R\,,\,\,(a+I)^2:=a^2+I\in R/I , since we define (a+I)(b+I):=ab+I ...

    I don't understand what "the last equations of the 4..." can possibly mean, but remember

    that a+I=I\iff a\in I , with I being, of course, the zero element of

    the quotient ring R/I

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,370
    Thanks
    739
    one way some authors approach this is the following:

    let α = x + <x^2 + 1> in Q[x]/<x^2 + 1>. note that the mapping q ---> q + <x^2 + 1> is an monomorphism (when q is in Q), so we can identify

    Q with its image under this mapping.

    then α^2 + 1 = (x + <x^2 + 1>)(x + <x^2 + 1>) + 1 + <x^2 + 1> = x^2 + 1 + <x^2 + 1> = <x^2 + 1> = 0 + <x^2 + 1> = 0.

    (where this last 0 is the additive identity of Q[x]/<x^2 + 1>, which we identify with 0 (in Q)).

    in other words: in the larger ring (which is actually a field) Q[x]/<x^2+1>, α^2 = -1, that is, α is a square root of -1 (actually, -1 + <x^2 + 1>,

    but this "acts just like the rational number -1").
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    558
    Thanks
    2
    Thanks Deveno

    Just a clarifying question.

    If we let a = x + < x^2 +1> in Q[x]/ < x^2 +1>

    then surely

     a^2 + 1 = (x + < x^2 +1>) (x + < x^2 +1>) +1

    But you write:

     a^2 + 1 = (x + < x^2 +1>) (x + < x^2 +1>) +1 + < x^2 +1>

    Why is this? Is it to formally acknowledge or indicate that the product is a coset?

    Bernhard
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,370
    Thanks
    739
    because the "1" in Q[x]/<x^2 + 1> is actually the coset 1 + <x^2 + 1>. however, because <x^2 + 1> is an ideal of Q[x], we can regard an element

    q + <x^2 + 1>, where q is a rational number (that is, a constant polynomial in Q[x]), as being just "the number q".

    this is similar to how we regard "real numbers" as being "complex numbers". we have the isomorphism a --> a+0i = (a,0).

    note that these aren't quite the same thing. a real number a, is just one number. the complex number a+0i = (a,0) is a PAIR of numbers,

    but the second coordinate "doesn't do anything", it's always 0.

    the same thing happens in a quotient ring R/I. the ideal I in R/I is the additive identity, the "0" of that ring.

    think about what happens in the ring Z/nZ. we take all the multiples of n in Z, and "call them all 0" (that is, they are 0 mod n).

    in the case of Q[x]/<x^2 + 1>, we call any polynomial having x^2 + 1 as a factor, "0".

    in other words, we are creating a structure where x^2 + 1 has a solution (by setting it "equal to 0").

    that what ideals DO. an ideal is a subring that can "take the place of 0". specifically, these are the properties of 0 we keep:

    0 + 0 = 0
    a0 = 0a = 0. note that if we replace "0" with "I", we get the rules an ideal obeys.

    the short answer to your question is "yes". in abstract algebra, there is a tendency to replace "equality" with the broader equivalence of "isomorphism".

    why? because we are concerned with algebraic properties of structures, not the specific "names" or definitions.

    instead of Q, one might say: form the field of fractions of Z, Frac(Z). and certainly ZxZ*/~ where (a,b) ~ (c,d) iff ad = bc isn't quite

    the same thing as Q (we think of the rationals as magnitudes on a line, generally, whereas we think of the fraction a/b as a ratio of two numbers).

    but they are isomorphic, so the differences are mainly a matter of convention. the algebra works out the same.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Extension fields
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 16th 2011, 07:47 AM
  2. Extension Fields
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 22nd 2011, 06:41 PM
  3. Extension fields
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: September 19th 2010, 07:15 PM
  4. Need help with Extension Fields
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 2nd 2010, 06:49 PM
  5. Extension fields / splitting fields proof...
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: December 19th 2007, 07:29 AM

Search Tags


/mathhelpforum @mathhelpforum