# Algebraic Extension Question [closed]

If $$K$$ is an algebraic extension of $$\mathbb{Q}(\sqrt{2})$$, prove $$K$$ is also an algebraic extension of $$\mathbb{Q}$$.

Can someone kindly provide some ideas to approach this question?

Thanks a lot.

## closed as off-topic by Jyrki Lahtonen, астон вілла олоф мэллбэрг, Greg Martin, Daniele Tampieri, José Carlos SantosAug 21 at 7:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Jyrki Lahtonen, астон вілла олоф мэллбэрг, Greg Martin, Daniele Tampieri, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.

• You need to a better job describing the tools at your disposal. Such as results characterizig algebraic extensions. Starting from the definition! Just listing those may allow you to answer the question, because they point at some directions! Anyway, I recommend that you check out our guide for new askers. – Jyrki Lahtonen Aug 21 at 4:58
• Finite extensions are algebraic. If $L\rightarrow K$ is algebraic, and $K\rightarrow E$ is algebraic, then $L\rightarrow E$ is algebraic. – take008 Aug 21 at 5:15