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

[Math Amateur]

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 ...Example 7.1.5 reads as follows:
https://www.physicsforums.com/attachments/6572
https://www.physicsforums.com/attachments/6573
In the above text from Lovett, we read the following:

" ... ... Then $$\mathbb{Q} [x] / ( p(x) ) = \mathbb{Q} [ \sqrt{5} ] $$ is a field. ... ... "
I understand that $$\mathbb{Q} [x] / ( p(x) ) = \mathbb{Q} [x] / ( x^2 - 5 ) $$ is a field ... ... but why is it equal to $$\mathbb{Q} [ \sqrt{5} ]$$ ... ...?Can someone please explain and demonstrate why the equality $$\mathbb{Q} [x] / ( x^2 - 5 ) = \mathbb{Q} [ \sqrt{5} ]$$ holds ... ?Help will be appreciated ...

Peter

 

[Euge]

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.

 

[Math Amateur]

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