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Here is my proof for the statement that the sum of two algebraic numbers is algebraic:

Let $\alpha,\beta$ be algebraic numbers.

Then we know that $[\Bbb Q(\alpha):\Bbb Q]$ is finite. Similarly, $[\Bbb Q(\beta):\Bbb Q]$ is also finite. Let $f \in \Bbb Q[x]$ be the minimal polynomial of $\beta$ over $\Bbb Q.$ Then $f \in (\Bbb Q(\alpha))[x]$ statisfying $f(\beta)=0.$This means $[\Bbb Q(\alpha,\beta):\Bbb Q(\alpha)]$ is finite. Therefore $[\Bbb Q(\alpha,\beta):\Bbb Q]$ is finite. Therefore $\Bbb Q(\alpha,\beta)/\Bbb Q$ is a finite extension. Therefore $\Bbb Q(\alpha+\beta)/\Bbb Q$ is also a finite extension(hence an algebraic extension). Therefore $\alpha+\beta$ is algebraic.

Can someone check whether my proof for the statement is correct or not? Thanks.

Y.X.
  • 3,995
  • The proof is correct. And it works to show that $\alpha\beta$ is algebraic. It also works to show that the root of a polynomial with algebraic coefficients os algebraic. – Maxime Ramzi Apr 16 '17 at 10:01
  • The passages are seemingly correct. A person who is concerned with your understanding might ask you to explain the third-to-last "therefore": "$[\Bbb Q(\alpha,\beta):\Bbb Q(\alpha)]$ is finite therefore $[\Bbb Q(\alpha,\beta):\Bbb Q]$ is finite". –  Apr 16 '17 at 10:02
  • Looks great. I'd suggest possibly mentioning "...because $\mathbb{Q}( \alpha + \beta) \subset \mathbb{Q}(\alpha, \beta)$" as justification of the second-to-last sentence and possibly even citing the multiplicativity formula for degrees. I tend to like to be somewhat meticulous with my writing though (without sacrificing too much brevity ofc); I'm unsure if others would concur. – Kaj Hansen Apr 16 '17 at 10:03

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