prove tht sqrt(\alpha) is algebriac over K

**maroon_tiger** · April 10th 2008, 12:16 PM

Let K be a subfield of Q. Prove that if $\alpha$ is algebriac over K, then $\sqrt(\alpha)$ is algebriac over K

**ThePerfectHacker** · April 10th 2008, 12:20 PM

Originally Posted by maroon_tiger

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 $K$ .

**maroon_tiger** · April 10th 2008, 12:29 PM

Originally Posted by ThePerfectHacker

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?

**ThePerfectHacker** · April 10th 2008, 12:58 PM

Originally Posted by maroon_tiger

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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