# Thread: prove tht sqrt(\alpha) is algebriac over K

1. ## prove tht sqrt(\alpha) is algebriac over K

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

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

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

3. Originally Posted by ThePerfectHacker
There are some problems here. First, Q has no subfields other than itself. Second, $\displaystyle \sqrt{\alpha}$ is not well-defined. Third, we do not know even if $\displaystyle \sqrt{\alpha}$ exists.

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

There arent any subfields in q? isnt the integers a subfield in Q?

4. 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.