I am reading Dummit and Foote on algebraic extensions. I am having some issues understanding Example 2 on page 526 - see attachment.

Example 2 on page 526 reads as follows:

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(2) Consider the field $\displaystyle \mathbb{Q} ( \sqrt{2}, \sqrt{3} )$ generated over $\displaystyle \mathbb{Q} $ by $\displaystyle \sqrt{2} $ and $\displaystyle \sqrt{3} $.

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

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My question is: whydoes it follow that the degree of the extension $\displaystyle \mathbb{Q} ( \sqrt{2}, \sqrt{3} )$ is at most 2 and is precisely 2 if and only if $\displaystyle x^2 - 3 $ is irreducible over $\displaystyle \mathbb{Q} ( \sqrt{2} ) $?exactly

Although I may be being pedantic I also have a concern about why exactly $\displaystyle \sqrt{3} $ is of degree 2 over $\displaystyle \mathbb{Q} $. I know it is intuitively the case or it seems the case that the minimal polynomial is in this case $\displaystyle x^2 - 3 $ 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