Abstract Algebra First Isomorphsm Theorem

  • Thread starter CurtBuck
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Homework Statement


Use the First Isomorphism Theorem to show that Q[x]/(x^3-3) is isomorphic to {a+b*sqrt(3)}


Homework Equations


First Isomorphism Theorem:
If f: G-> H is a homomorphism then G/ker(f) is isomorphic to im(f)


The Attempt at a Solution


I understand that I need to show that Q[x] is homomorphic to {a+b*sqrt(3)} with a kernel of x^3-3. I am really struggling in finding that homomorphism. I see the connection between sqrt(3) and x^3-3, e.g. sqrt(3) is a root of x^3-3, but I don't understand how this helps.
 

Answers and Replies

  • #2
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It is clear that you should choose f(q)=q, for every [tex]q\in \mathbb{Q}[/tex]. So you only need to know what f(x) is. What you certainly want is [tex]f(x^3-3)=0[/tex]. Using that f should be a homomorphism yields that [tex]f(x)^3-3=0[/tex]. So what should be f(x)?
 
  • #3
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First, I apologize, I meant to right that the polynomial being modded out of Q[x] is x^2 - 3, not x^3 - 3.

So that being said, I understand that f(x)^2 - 3 = 0 implies that f(x) = sqrt(3).

But then how would one generalize this?
 
  • #4
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Well, f(q)=q for [tex]q\in \mathbb{Q}[/tex] and [tex]f(x)=\sqrt{3}[/tex].

So [tex]f(a+bx)=a+b\sqrt{3}[/tex] and [tex]f(a+bx+cx^2)=a+b\sqrt{3}+c\sqrt{3}^2=a+3c+b\sqrt{3}[/tex]. Now try to generalize what I'm doing here...
 
  • #5
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I understand what you're doing there.

Two more things:

How do I prove that this a homomorphism?

Also, wouldn't 0 also be in the kernel? Is this allowed for the First Isomorphism Theorem?
 
  • #6
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Hmm, showing that this is a homomorphism is indeed not so easy.

First take a general polymial and calculate

[tex]f\left(\sum_{k=0}^n{a_kx^k}\right)[/tex]

This will help you. Then with this information, you only need to calculate

[tex]f\left(\sum_{k=0}^n{a_kx^k}+\sum_{k=0}^n{b_kx^k}\right)[/tex]

and

[tex]f\left(\sum_{k=0}^n{a_kx^k}\sum_{k=0}^n{b_kx^k}\right)[/tex].



As for the other question. Of course 0 is going to be in our kernel, and this is not a problem. The first isomorphism theorem requires us that the kernel equals [tex](x^2-3)[/tex], that is: the ideal generated by [tex]x^2-3[/tex]. And 0 is in this ideal...
 
  • #7
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To show that "evaluation at [tex]\sqrt{3}[/tex]" is a homomorphism, it's actually a little easier to prove a general theorem that "evaluation maps are homomorphisms":

If [tex]A \subset B[/tex] are rings and [tex]\xi \in B[/tex], then the "evaluation at [tex]\xi[/tex]" map [tex]\mathrm{ev}_\xi: A[X] \to B[/tex] given by extending the inclusion map [tex]A \hookrightarrow B[/tex] by sending [tex]X \mapsto \xi[/tex] is a homomorphism.

(You can generalize this a little further by letting [tex]A[/tex] be embedded in [tex]B[/tex] via a homomorphism [tex]\varphi: A \to B[/tex] rather than directly a subring of [tex]B[/tex]; then the evaluation map extends [tex]\varphi[/tex] by sending [tex]X \mapsto \xi[/tex], rather than extending the inclusion map.)
 

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