Find the degrees of these fields as extensions of $\mathbb{Q}$. Splitting field for $f(t)=t^3-1$: 
To find the splitting field for $f(x)$ over $\mathbb{Q}$, we must find all the roots of the polynomial. We can first use the difference of cubes formula to get $$t^3-1=(t-1)(t^2+t+1).$$
Hence one of the roots of $f(t)$ is $1$. To find the roots of $t^2+t+1$, we must use the quadratic formula.$$t=\frac{-1\pm \sqrt{1^2-4(1)(1)}}{2}=\frac{-1\pm i\sqrt{3}}{2}.$$
Hence all the roots of $f(t)$ are $$1\text{, and } \frac{-1\pm i\sqrt{3}}{2}.$$ So the splitting field of $f(t)$ over $\mathbb{Q}$ is $$\mathbb{Q}(1,\frac{-1\pm i\sqrt{3}}{2}).$$ However 
\begin{equation*}
\begin{aligned}
\mathbb{Q}(1,\frac{-1\pm i\sqrt{3}}{2}) & =\mathbb{Q}(\frac{-1\pm i\sqrt{3}}{2}), \text{ (since $1\in \mathbb{Q}$)} \\
& =\mathbb{Q}(-\frac{1}{2}, \pm \frac{i\sqrt{3}}{2}),\text{ (through a simple extension)} \\
& =\mathbb{Q}(\pm \frac{i\sqrt{3}}{2}),\text{ (since $\frac{-1}{2} \in \mathbb{Q}$)} \\
& =\mathbb{Q}(i\sqrt{3}), \text{ (since $\pm \frac{1}{2} \in \mathbb{Q}$ and is just a multiple of $i\sqrt{3}$)} \\
& =\mathbb{Q}(i,\sqrt{3}), \text{ (as seen in previous problems)} \\
\end{aligned}
\end{equation*}
Claim: The degree of the field extension of $\mathbb{Q}(i, \sqrt{3})$ is $4$.
Proof: First of all, $[\mathbb{Q}(3^{1/2}) : \mathbb{Q}] = 2$, because the basis of $\mathbb{Q}(3^{1/2}) : \mathbb{Q}$ is $\{ 1, 3^{1/2}\}$. Similarly, the basis of $\mathbb{Q}(i):\mathbb{Q}$ is $\{ 1, i\}$, then $[\mathbb{Q}(i):\mathbb{Q}]$ = 2.
Therefore,
\begin{align*}
[\mathbb{Q}(\sqrt{3},i):\mathbb{Q}] &= 2 \cdot 2 = 4
\end{align*}
My question is what I did to show the dimension of the splitting field is correct or not?
 A: The roots are $1, \mathrm j, \mathrm j^2$. Hence the splitting field is $\;\mathbf Q(1,\mathrm j, \mathrm j^2)=\mathbf Q(\mathrm j)$, which has degree $2$. The minimal polynomial of its generator is $x^2+x+1$.
This splitting field is contained in $\mathbf Q(\mathrm i,\sqrt 3)$, which is the splitting field of $(x^2-3)(x^2+1)$, but certainly not equal. The splitting field of $x^3-1$ is also the field $\;\mathbf Q(\mathrm i\sqrt3)\varsubsetneq\mathbf Q(\mathrm i,\sqrt3)$.
A: Let $f$ be a quadratic polynomial over a field $F$. If $f$ has a root in $F$, then also the other one is in $F$, by simple factorization. For the same reason, if $f$ has a root in an extension field $K$ of $F$, then both roots belong to $K$.
In your case, the root $1$ can be ignored, because it's rational. When you add either root of $t^2+t+1$ to $\mathbb{Q}$, you have obtained the splitting field, because both roots are in this extension.

Where did your computations get wrong? It's clear that adding $i\sqrt{3}$ or $(-1+\sqrt{3})/2$ is exactly the same, so your extension can be written $\mathbb{Q}(i\sqrt{3})$. Note that $\sqrt{3}$ doesn't belong to this extension: let's try solving $a+bi\sqrt{3}=\sqrt{3}$, with $a,b\in\mathbb{Q}$. This means
$$
\sqrt{3}(1-bi)=a
$$
or, squaring,
$$
3(1+b^2-2bi)=a^2
$$
so $b=0$ and $a=\sqrt{3}$: a contradiction.
A: You showed that the splitting field could be obtained by adjoining a single root of a quadratic polynomial irreducible over the rationals. So the splitting field has degree $2$ over the rationals.
The field $\mathbb{Q}(i,\sqrt{3})$ indeed has degree $4$ over the rationals. But it is not the splitting field of $t^3-1$. The degree argument shows that.
If you want to show it directly, you can observe that $\sqrt{3}$ cannot be expressed as $a+b\sqrt{-3}$, where $a$ and $b$ are rational.
