### CSIR JUNE 2011 PART B QUESTION 36 SOLUTION ($[\Bbb Q(\sqrt 2+\sqrt 2):\Bbb Q(\sqrt 2)$).

The degree of the extension $\Bbb Q(\sqrt 2+\sqrt 2)$ over $\Bbb Q(\sqrt 2)$ is
1) 1,
2)2,
3)3,
4)6.
Solution:
We have $\Bbb Q(\sqrt 2+\sqrt 2) = \Bbb Q(\sqrt 2, \sqrt 2)$. So we need to find the degree of the minimal polynomial satisfied by $\sqrt 2$ over $\Bbb Q(\sqrt 2)$.
Result: Minimal polynomial of an element $\alpha$ divides all the polynomials satisfied by $\alpha$.
Proof: Let $m(x)$ be the minimal polynomial of $\alpha$ and $f(x)$ be some polynomial satisfied by $\alpha$. Since $m(x)$ is the minimal polynomial, we have the degree of $f(x)$ is greater than or equal to the degree of $m(x)$. Now, by the division algorithm, we have $f(x) = q(x)m(x) + r(x)$ with degree of $r(x)$ strictly less than the degree of $m(x)$. Now, $f(\alpha) = q(\alpha)m(\alpha) + r(\alpha)$, this implies that $r(\alpha) = 0$. If $r(x)$ is not the zero polynomial, then it is polynomial of smaller degree than $m(x)$ satisfied by $\alpha$ which is not possible. Hence $r(x)$ is equal to zero. This shows that $m(x)$ divides $f(x)$.
Now, in our problem, consider the polynomial $x^3-2$ over $\Bbb Q(\sqrt 2)$. This polynomial satisfies $\sqrt 2$ and hence either this is the minimal polynomial of $\sqrt 2$
over $\Bbb Q(\sqrt 2)$ or the minimal polynomial of $\sqrt 2$ over $\Bbb Q(\sqrt 3)$ divides this polynomial. We know that a polynomial of degree $2$ or $3$ is irreducible over a field $\Bbb F$ if and only if it has no root in $\Bbb F$. Now the roots of $x^3-2$ are $\sqrt2,w \sqrt2, w^2 \sqrt2$ where $w$ is a primitive third root of unity and any of these roots of $x^3-2$ are not living in $\Bbb Q(\sqrt 2)$. Hence $x^3-2$ is irreducible over $\Bbb Q(\sqrt 2)$ and so it is the minimal polynomial of $\sqrt 2$ over $\Bbb Q(\sqrt 2)$. This shows that the required degree of the extension is $3$.
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