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- #1

- Jun 22, 2012

- 2,937

Example 2 on page 526 reads as follows:

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(2) Consider the field [TEX] \mathbb{Q} ( \sqrt{2}, \sqrt{3} )[/TEX] generated over [TEX] \mathbb{Q} [/TEX] by [TEX] \sqrt{2} [/TEX] and [TEX] \sqrt{3} [/TEX].

Since [TEX] \sqrt{3} [/TEX] is of degree 2 over [TEX] \mathbb{Q} [/TEX] the degree of the extension [TEX] \mathbb{Q} ( \sqrt{2}, \sqrt{3} )[/TEX] is at most 2 and is precisely 2 if and only if [TEX] x^2 - 3 [/TEX] is irreducible over [TEX] \mathbb{Q} ( \sqrt{2} ) [/TEX]. ... ... etc etc

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My question is: why

__does it follow that the degree of the extension [TEX] \mathbb{Q} ( \sqrt{2}, \sqrt{3} )[/TEX] is at most 2 and is precisely 2 if and only if [TEX] x^2 - 3 [/TEX] is irreducible over [TEX] \mathbb{Q} ( \sqrt{2} ) [/TEX]?__

**exactly**Although I may be being pedantic I also have a concern about why exactly [TEX] \sqrt{3} [/TEX] is of degree 2 over [TEX] \mathbb{Q} [/TEX]. I know it is intuitively the case or it seems the case that the minimal polynomial is in this case [TEX] x^2 - 3 [/TEX] but how do we demonstrate this for sure - or is it obvious? (I may be overthinking this??)

Can someone help with the above issues/problems?

Peter

[Note: This has also been posted on MHF]