# Fields and Field Extensions - Lovett, Chapter 7 .... ....

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MHB
I am reading Abstract Algebra: Structures and Applications" by Stephen Lovett ...

I am currently focused on Chapter 7: Field Extensions ... ...

I need help with Example 7.1.5 ...

https://www.physicsforums.com/attachments/6572
https://www.physicsforums.com/attachments/6573

In the above text from Lovett, we read the following:

" ... ... Then $$\displaystyle \mathbb{Q} [x] / ( p(x) ) = \mathbb{Q} [ \sqrt{5} ]$$ is a field. ... ... "

I understand that $$\displaystyle \mathbb{Q} [x] / ( p(x) ) = \mathbb{Q} [x] / ( x^2 - 5 )$$ is a field ... ... but why is it equal to $$\displaystyle \mathbb{Q} [ \sqrt{5} ]$$ ... ...?

Can someone please explain and demonstrate why the equality $$\displaystyle \mathbb{Q} [x] / ( x^2 - 5 ) = \mathbb{Q} [ \sqrt{5} ]$$ holds ... ?

Help will be appreciated ...

Peter

Last edited:

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Hi Peter,

Lovett means that $\Bbb Q[x]/(x^2 - 5)$ is isomorphic to $\Bbb Q[\sqrt{5}]$. Elements of $\Bbb Q[x]/(x^2 - 5)$ are of the form $a + bx + (x^2 - 5)$ where $a,b\in \Bbb Q$. Letting $x$ map to $\sqrt{5}$, we get a bijection $a + bx + (x^2 - 5)\mapsto a + b\sqrt{5}$ from $\Bbb Q[x]/(x^2 - 5)$ to $\Bbb Q[\sqrt{5}]$. This map is a homomorphism of rings, as you can check.

Gold Member
MHB
Hi Peter,

Lovett means that $\Bbb Q[x]/(x^2 - 5)$ is isomorphic to $\Bbb Q[\sqrt{5}]$. Elements of $\Bbb Q[x]/(x^2 - 5)$ are of the form $a + bx + (x^2 - 5)$ where $a,b\in \Bbb Q$. Letting $x$ map to $\sqrt{5}$, we get a bijection $a + bx + (x^2 - 5)\mapsto a + b\sqrt{5}$ from $\Bbb Q[x]/(x^2 - 5)$ to $\Bbb Q[\sqrt{5}]$. This map is a homomorphism of rings, as you can check.

Thanks for the help, Euge... appreciate it ...

Peter