The degree of the extension $\Bbb Q(\sqrt 2+\sqrt[3] 2)$ over $\Bbb Q(\sqrt 2)$ is

1) 1,

2)2,

3)3,

4)6.

We have $\Bbb Q(\sqrt 2+\sqrt[3] 2) = \Bbb Q(\sqrt 2, \sqrt[3] 2)$. So we need to find the degree of the minimal polynomial satisfied by $\sqrt[3] 2$ over $\Bbb Q(\sqrt 2)$.

Now, in our problem, consider the polynomial $x^3-2$ over $\Bbb Q(\sqrt 2)$. This polynomial satisfies $\sqrt[3] 2$ and hence either this is the minimal polynomial of $\sqrt[3] 2$

over $\Bbb Q(\sqrt 2)$ or the minimal polynomial of $\sqrt[3] 2$ 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 $\sqrt[3]2,w \sqrt[3]2, w^2 \sqrt[3]2$ 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 $\Bbb Q(\sqrt[3] 2)$ and so it is the minimal polynomial of $\sqrt[3] 2$ over $\Bbb Q(\sqrt 2)$.

1) 1,

2)2,

3)3,

4)6.

**Solution**:We have $\Bbb Q(\sqrt 2+\sqrt[3] 2) = \Bbb Q(\sqrt 2, \sqrt[3] 2)$. So we need to find the degree of the minimal polynomial satisfied by $\sqrt[3] 2$ 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 $\sqrt[3] 2$ and hence either this is the minimal polynomial of $\sqrt[3] 2$

over $\Bbb Q(\sqrt 2)$ or the minimal polynomial of $\sqrt[3] 2$ 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 $\sqrt[3]2,w \sqrt[3]2, w^2 \sqrt[3]2$ 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 $\Bbb Q(\sqrt[3] 2)$ and so it is the minimal polynomial of $\sqrt[3] 2$ over $\Bbb Q(\sqrt 2)$.

**This shows that the required degree of the extension is $3$**.**Share to your groups:**

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