Using the First Isomorphism Theorem

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Discussion Overview

The discussion focuses on understanding the first isomorphism theorem for groups, specifically in the context of the quotient ring Q[x]/(x^2-3) and its isomorphism to the set of elements of the form {a+b*sqrt(3)}. Participants explore the construction of a homomorphism and the identification of elements within the quotient ring.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant seeks assistance in finding a homomorphism from Q[x] to {a+b*sqrt(3)} with the kernel x^2-3.
  • Another participant questions the initial claim about the kernel being x^3-3, suggesting it should be x^2-3 instead.
  • A later reply confirms the correction regarding the kernel being x^2-3.
  • One participant proposes that to find the isomorphism, one must identify an element in Q[x]/(x^2-3) whose square equals 3, indicating a two-step process involving finding such an element and constructing the isomorphism.
  • Another participant states that the homomorphism can be defined by sending x to sqrt(3), explaining that modding out by f(x) results in x being treated as a root of f.

Areas of Agreement / Disagreement

Participants generally agree on the kernel being x^2-3, but there is ongoing exploration regarding the construction of the homomorphism and the identification of the appropriate elements within the quotient ring.

Contextual Notes

Participants have not fully resolved the steps necessary to construct the isomorphism, and there may be additional assumptions or definitions that are not explicitly stated.

CurtBuck
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I'm trying to understand the first isomorphism theorem for groups.

Part of the examples given in the book is showing that Q[x]/(x^3-3) is isomorphic to {a+b*sqrt(3)}

As I understand it, by finding a homomorphism from Q[x] to {a+b*sqrt(3)} in which the kernel is x^3-3, the two are isomorphic.

I am struggling in finding the homomorphism from Q[x] to {a+b*sqrt(3)}

Any help would be great.
 
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Are you sure it's not supposed to be Q[x]/(x2-3)?
 
Yeah, it was supposed to be Q[x]/(x^2-3)
 
You need to find an element in Q[x]/(x2-3) whose square is equal to 3, i.e. you want to find some y such that y2+(x2-3) = 3+(x2-3) (y is a polynomial here). So there's really two steps here

1) Find y.
2) Use y to construct an isomorphism
 
the homomorphism sends x to sqrt(3).

when you mod out by f(x), you make f(x) = 0, so x becomes a root of f. since sqrt(3) is the root of x^2-3, x becomes sqrt(3).
 

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