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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 \(\displaystyle \mathbb{Q} [x] / ( p(x) ) = \mathbb{Q} [ \sqrt{5} ] \) 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

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 \(\displaystyle \mathbb{Q} [x] / ( p(x) ) = \mathbb{Q} [ \sqrt{5} ] \) 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

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