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Math Help - prove tht sqrt(\alpha) is algebriac over K

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    prove tht sqrt(\alpha) is algebriac over K

    Let K be a subfield of Q. Prove that if \alpha is algebriac over K, then \sqrt(\alpha) is algebriac over K
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    Quote Originally Posted by maroon_tiger View Post
    Let K be a subfield of Q. Prove that if \alpha is algebriac over K, then \sqrt(\alpha) is algebriac over K
    There are some problems here. First, Q has no subfields other than itself. Second, \sqrt{\alpha} is not well-defined. Third, we do not know even if \sqrt{\alpha} exists.

    Let me phrase the question differently: Let F be an extension field over K. Let \alpha \in F be algebraic over K. Suppose that \beta \in F has property that \beta^2 = \alpha. Show that \beta is algebraic over K.
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    Quote Originally Posted by ThePerfectHacker View Post
    There are some problems here. First, Q has no subfields other than itself. Second, \sqrt{\alpha} is not well-defined. Third, we do not know even if \sqrt{\alpha} exists.

    Let me phrase the question differently: Let F be an extension field over K. Let \alpha \in F be algebraic over K. Suppose that \beta \in F has property that \beta^2 = \alpha. Show that \beta is algebraic over K.

    There arent any subfields in q? isnt the integers a subfield in Q?
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    Quote Originally Posted by maroon_tiger View Post
    There arent any subfields in Q?
    Every field of charachteristic zero has Q (up to isomorphism) as a subfield. Thus, this is the smallest such sub-field.
    isnt the integers a subfield in Q?
    No. What is the inverse of 2 in the integers? None. So it cannot be a field.
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