Normal Closure of a Field Extension

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

Construct the Normal Closure from the given extension: $\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}$

So, I take the basis extension which is $\{1,\sqrt[5]{3},\sqrt[5]{3^2},\sqrt[5]{3^3},\sqrt[5]{3^4} \}$ 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 $\alpha^i=(\sqrt[5]{3})^i$ in $\mathbb{Q}$.

I know that the Normal Closure of $\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}$ 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 $m_{\alpha^i}$?