Normal Closure of a Field Extension

I am studying Galois Theory and doing my homework I found this question:

Construct the Normal Closure from the given extension: $\mathbb{Q}(\sqrt{3}):\mathbb{Q}$

So, I take the basis extension which is $\{1,\sqrt{3},\sqrt{3^2},\sqrt{3^3},\sqrt{3^4} \}$ and from that make the polynomial $f=m_{\alpha^0}*m_{\alpha^1}*m_{\alpha^2}*m_{\alpha^3}*m_{\alpha^4}$, where $m_{\alpha^i}$ is the minimal polynomial of $\alpha^i=(\sqrt{3})^i$ in $\mathbb{Q}$.

I know that the Normal Closure of $\mathbb{Q}(\sqrt{3}):\mathbb{Q}$ must be the splitting field of $f$.

My questions are:

1) The splitting field of $f$ will be the field generated by the roots of $f$ right?

2) If I took the Normal Closure in $\mathbb{R}$ I must consider the complex roots of $m_{\alpha^i}$?

• Why don't just take the splitting field of the minimal polynomial of $\alpha$? – Student May 16 '18 at 22:44
• With $\sqrt3$ all the other roots of of the minimal polynomial $X^5-3\in \Bbb Q[X]$ have to be considered in the normal closure. Two such roots differ by a fifth root of unity, so one has to adjoin also $\zeta=\zeta_5$, a primitive $5$.th root of unity. The closure is thus $\Bbb Q(\zeta_5,\sqrt3)$. ($K=\Bbb Q(\zeta_5):\Bbb Q$ is a cyclotomic extension, it is Galois, then adjoining $\sqrt3$ is also Galois, $X^5-3$ spits a product, it has the factors $X-\zeta_5^k\sqrt3$, $k=0,1,2,3,4$. – dan_fulea May 16 '18 at 22:49
• I thought Dan doing that but I was not certain that will suffice. This only works because it is a simple extension? – Ettore Moura May 16 '18 at 22:52
• No, for the simple extension $\Bbb Q(\alpha)\supset\Bbb Q$, to get the normal closure, just adjoin all roots of the minimal polynomial for $\alpha$. That’s what happened here. – Lubin May 17 '18 at 1:07