Question

Problem 3: Determine the splitting field of the polynomial (2 -2)(2-3)(2 -4) over Q. Find its degree over Q. Verify if all points of the splitting field are constructible.

Answer 1

We can factorize the given polynomial over $\mathbb{C}$ as, $(x-\sqrt{2})(x+\sqrt{2})(x-\sqrt{3})(x+\sqrt{3})(x-2)(x+2)$ . Hence the splitting field of the polynomial over $\mathbb{Q}$ is the field

Q(V2,V3)

We are asked to find out the degree of this extension. We have, $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]$

We know that $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$ , since it is the splitting field of a degree irreducible polynomial $x^2-2$. We claim that, $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})]=2$ . For this, consider the polynomial $x^2-3\in \mathbb{Q}(\sqrt{2})[x]$ . Clearly,
Q(V2,V3)
is the splitting field of this polynomial. Since $x^2-3$ is a degree polynomial, we have that $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})]\le 2$ . But clearly $\sqrt{3}\not\in\mathbb{Q}(\sqrt{2})$ . Hence, $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})]\ne 1$ . Then, $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})]=2$ . Thus we have the degree $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=2.2=4$

Now we need to check whether all the elements of this splitting field is constructible or not. Now we know that any $x\in\mathbb{C}$ is constructible if and only if $[\mathbb{Q}(z):\mathbb{Q}]$ is of degree some power of . Say, $z\in\mathbb{Q}(\sqrt{2},\sqrt{3})$ . Then we must have that $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(z)][\mathbb{Q}(z):\mathbb{Q}]$ , that is in particular $[\mathbb{Q}(z):\mathbb{Q}]$ divides $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=4$ . Hence, $[\mathbb{Q}(z):\mathbb{Q}]=1,2$ or . So is constructible. Thus every element of the splitting field is constructible.

Hope this helps. Feel free to comment if you need further clarifications. Cheers!