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A question about commutative rings and homomorphisms

  1. Jan 6, 2013 #1
    Let R be a commutative ring. Show that the function ε : R[x] → R, defined by
    [tex]\epsilon : a_0 + a_1x + a_2x +· · ·+a_n x^n \rightarrow a_0[/tex],
    is a homomorphism. Describe ker ε in terms of roots of polynomials.

    In order to show that it is a homomorphism, I need to show that ε(1)=1, right?

    But [tex]\epsilon(1) = a_0+a_1+...+a_n \not= 1 [/tex]

    So can anybody help me with this?

    Thanks in advance
     
  2. jcsd
  3. Jan 6, 2013 #2

    tiny-tim

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    Hi Artusartos! :smile:

    I'm confused :redface:

    isn't 1 = 1 + 0x + 0x2 + … + 0xn ? :confused:
     
  4. Jan 6, 2013 #3
    Oh...I thought we had to do it the other way around...

    But why do we set [tex]1=a_0 + a_1x + ... + a_nx^n [/tex] and then choose a_0 to be 1 and the rest of the coefficients to be zero? Do we have the freedom to choose the coefficients ourselves?

    I thought we were supposed to compute [tex]\epsilon(1)[/tex] and then find the multiplicative identity of [tex]a_0[/tex], and show that their equal? But the multiplicative identity for [tex]a_0[/tex] is 1 and [tex]\epsilon(1) = a_0 + a_1 + ... +a_n \not = 1 [/tex]. I must be doing something wrong...
     
  5. Jan 6, 2013 #4

    micromass

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    You seem to misunderstand the mapping [itex]\varepsilon[/itex]. The mapping is

    [tex]\varepsilon( a_0 + a_1X+...+a_nX^n)=a_0[/tex]

    For example:

    [tex]\varepsilon ( 2 + 3X + 4X^2)=2[/tex]

    because [itex]a_0=2,~a_1=3,~a_2=4[/itex].

    Other examples:

    [tex]\varepsilon ( 3 + 321X)=3[/tex]
    [tex]\varepsilon (421)=421[/tex]
    [tex]\varepsilon ( 1 + X^2 + 5X^4+6X^{77})=1[/tex]

    Does that clear things up for you?
     
  6. Jan 6, 2013 #5
    Thanks. I think I understand it now. :)
     
  7. Jan 6, 2013 #6

    tiny-tim

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    ah, the a's aren't constants in the function ε,

    they're variables (coordinates) in R …

    ε sends the variable point a0 + a11x + a2x +· · ·+a_n xn to the point a0 :wink:
     
  8. Jan 6, 2013 #7
    Thank you.
     
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