# Abstract Algebra First Isomorphsm Theorem

## 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.

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It is clear that you should choose f(q)=q, for every $$q\in \mathbb{Q}$$. So you only need to know what f(x) is. What you certainly want is $$f(x^3-3)=0$$. Using that f should be a homomorphism yields that $$f(x)^3-3=0$$. So what should be f(x)?

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?

Well, f(q)=q for $$q\in \mathbb{Q}$$ and $$f(x)=\sqrt{3}$$.

So $$f(a+bx)=a+b\sqrt{3}$$ and $$f(a+bx+cx^2)=a+b\sqrt{3}+c\sqrt{3}^2=a+3c+b\sqrt{3}$$. Now try to generalize what I'm doing here...

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?

Hmm, showing that this is a homomorphism is indeed not so easy.

First take a general polymial and calculate

$$f\left(\sum_{k=0}^n{a_kx^k}\right)$$

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

$$f\left(\sum_{k=0}^n{a_kx^k}+\sum_{k=0}^n{b_kx^k}\right)$$

and

$$f\left(\sum_{k=0}^n{a_kx^k}\sum_{k=0}^n{b_kx^k}\right)$$.

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 $$(x^2-3)$$, that is: the ideal generated by $$x^2-3$$. And 0 is in this ideal...

To show that "evaluation at $$\sqrt{3}$$" is a homomorphism, it's actually a little easier to prove a general theorem that "evaluation maps are homomorphisms":

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

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