Degrees

Dr Kim,

A quick follow up question:

In certain solved examples of this type, we’ve assumed that Q(\sqrt{3},\sqrt{5}) has a degree 4, how do we know this already?

And in others we simply work out the degree of the minimum polynomial for Q[\sqrt{3}+\sqrt{5}] and prove it’s irreducible. Then it says by the tower theorem the degree of Q(\sqrt{3},\sqrt{5}) is less than or equal to 4?

It does seem that the degree of Q(\sqrt{3},\sqrt{5}) would be 4, but i’m not sure about the theory behind it.

————————————————

Reply:

Firstly, the fact that the degree of Q(\sqrt{3},\sqrt{5}) is less than or equal to 4 requires only the tower theorem since

[Q(\sqrt{3},\sqrt{5}):Q(\sqrt{3})] \leq 2

and

[Q(\sqrt{3}):Q]\leq 2.

So when you’ve proved that the minimal polynomial of \sqrt{3}+\sqrt{5} has degree 4, then you know that the subfield Q[\sqrt{3}+\sqrt{5}] \subset Q(\sqrt{3},\sqrt{5}) has degree *equal* to 4, showing that both have degree four and are the same field. This kind of argument is common.

For some more examples of the requisite argument, you can look at the document `Some irrationality’ on the course webpage.

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