# Surjective Ring Homomorphism

• Mar 2nd 2010, 01:35 PM
Brattmat
Surjective Ring Homomorphism
Hi there!

This is my first time posting here so apologies if this is in the wrong section or there's already a post about this but I have looked everywhere to try and get a solution to this.

The question I'm stuck on is:

Let θ : Q[X] → Q(√3) be the map defined by θ(a0 + a1X + ... + anXn) = (a0 + a1√3 + ... + an(√3)n).
Show that θ is a surjective ring homomorphism.
Prove that Ker θ = (X^2 − 3)Q[X].
Deduce that the factor ring Q[X]/Ker θ is isomorphic to Q(√3).

The notes I have on this aren't particularly useful though I do believe that I need to use the first isomorphism theorem for the last part. The first part however I can't seem to find anything like in any of the books or websites I have looked at.

• Mar 2nd 2010, 06:23 PM
Brattmat
One example I have found seems to be relevant but unhelpful as it doesn't really explain each step properly.

φ: Q[x] -> Q(√2)
f(x) = ∑aiX^i - > φ(f(x)) = ∑ai(√2)^i = f(√2)

φ is a surjective homomorphism: if a + b√2 ∈ Q(√2), let f(x) = a + bX, then φf(x) = f(√2) = a + b√2.

Claim: Ker φ = (X^2 - 2)Q[x] = I

Which then goes on to find Ker φ.

Does anyone know the steps for Q(√3)?
• Mar 3rd 2010, 12:44 AM
aliceinwonderland
Quote:

Originally Posted by Brattmat
Hi there!

This is my first time posting here so apologies if this is in the wrong section or there's already a post about this but I have looked everywhere to try and get a solution to this.

The question I'm stuck on is:

Let θ : Q[X] → Q(√3) be the map defined by θ(a0 + a1X + ... + anXn) = (a0 + a1√3 + ... + an(√3)n).
Show that θ is a surjective ring homomorphism.
Prove that Ker θ = (X^2 − 3)Q[X].
Deduce that the factor ring Q[X]/Ker θ is isomorphic to Q(√3).

The notes I have on this aren't particularly useful though I do believe that I need to use the first isomorphism theorem for the last part. The first part however I can't seem to find anything like in any of the books or websites I have looked at.

Verify that $\theta$ is a ring homomorphism defined by $q \mapsto q$ for each $q \in \mathbb{Q}$ and $x \mapsto \sqrt{3}$, otherwise. An ideal ( $x^2-3$) is clearly kernel of $\theta$ since $\theta(I)=0$ for $I \in (x^2-3)$. Now, you apply the first isomorphism theorem and obtain the result.