I've been given this problem. I think I must proceed in the following way:
Given $a=\sqrt[4]{3}$ and $b=\cos\frac{2\pi}{4}+ \sin\frac{2\pi}{4}i$, the roots of $x^4-3$ are $a$, $ab$, $ab^2$, $ab^3$.
We have $\mathbb{Q}\subset \mathbb{Q}[a]\subset\mathbb{Q}[ab]$ so, is $\mathbb{Q}[ab]$ the desired spliting field?
Given a previous example in the book, it seems it is this way it should be done. However, we extend a previous field in various steps, it's not clear to me when this process "ends", for example: I could extend further and find $\mathbb{Q}[ab^2]$ but in a previous example, it stops before adjoining all the roots, I think the same applies here.
Just addressing your concerns about finiteness and stopping.
$p(x) = x^4 - 3$ has at most four roots in any field. Call them $r_1$, $r_2$, $r_3$, and $r_4$. Then, at the very least $\Bbb{Q}(r_1, r_2, r_3, r_4)$, whatever field that is, is a field that splits $p$. More generally, for a polynomial of degree $d$, extending $\Bbb{Q}$ by its (up to (see "separable")) $d$ roots yields a splitting field of the polynomial. Proceeding in this way, the process is evidently finite.
It may turn out that $\{r_1, r_2, r_3, r_4\}$ are not independent. For instance, it may occur that $r_4 \in \Bbb{Q}(r_1, r_2, r_3)$. So the field we get may be more directly described using a shorter sequence of extensions. To detect this, we ask "Is $r_2 \in \Bbb{Q}(r_1)$? To really ask this, we have to be explicit about which roots are which. That can be difficult, so another way to go is to see if you can partially factor the polynomial in some (smaller) extension field.
We see that $x^4 - 3 = (x^2 + \sqrt{3})(x^2 - \sqrt{3})$. This is not a factorization in $\Bbb{Q}$, but is a factorization in $\Bbb{Q}(-\sqrt{3}, \sqrt{3})$.We get to decide which of these we label as the first root and which the second. In order to potentially reduce the numbers of minus signs, let's try to determine whether $-\sqrt{3} \in \Bbb{Q}(\sqrt{3})$. To do so, we look at the $\Bbb{Q}(\sqrt{3})$ vector space over $\Bbb{Q}$. Generally, this is $$ \{q_0 + q_1 \sqrt{3} + q_2 (\sqrt{3})^2 + \cdots q_n (\sqrt{3})^n \mid n \in \Bbb{Z}_{\geq 0}, q_i \in \Bbb{Q} \} \text{.} $$ This is a $\Bbb{Q}$-linear vector space with spanning set $\{1, \sqrt{3}, \sqrt{3}^2, \dots \}$. However, this spanning set is redundant. We know this because $\sqrt{3}$ is an algebraic number, of some degree, $m$. So $\sqrt{3}$ is a root of a polynomial of degree $m$. So there is a $\Bbb{Q}$-linear combination of the first $m$ powers of $\sqrt{3}$ that is zero -- that is, those $m$ powers are not $\Bbb{Q}$-linearly independent. For $\sqrt{3}$, its minimal polynomial over $\Bbb{Q}$ is $x^2 - 3 = 0$, so $\sqrt{3}^2 \in \Bbb{Q}$. Therefore, a basis for $\Bbb{Q}(\sqrt{3})$ is $\{1, \sqrt{3}\}$ and (as a set) $$ \Bbb{Q}(\sqrt{3}) = \{q_0 + q_1\sqrt{3} \mid q_0, q_1 \in \Bbb{Q} \} \text{.} $$
All of that setup lets us observe $-\sqrt{3} = 0 + (-1)\sqrt{3} \in \Bbb{Q}(\sqrt{3})$, so $\Bbb{Q}(-\sqrt{3}, \sqrt(3)) = \Bbb{Q}(\sqrt{3})$.
Now what? We need a further extension. The factor $x^2 - \sqrt{3} = (x - \sqrt[4]{3})(x + \sqrt[4]{3})$ As above, any field containing $\sqrt[4]{3}$ contains $-\sqrt[4]{3}$, so we will extend to $\Bbb{Q}(\sqrt{3}, \sqrt[4]{3})$. But notice that $\sqrt{3} = (\sqrt[4]{3})^2$ is in the $\Bbb{Q}$-linear span of the powers of $\sqrt[4]{3}$, so $\sqrt{3} \in \Bbb{Q}(\sqrt[4]{3})$. So our new field extension is still principal. (We first chopped $3$ into finer factors to get the first field extension. If we chop it even finer, there is a second extension that contains all the slices. For instance, $\Bbb{Q}(\sqrt{3}, \sqrt[3]{3})$ is contained in $\Bbb{Q}(\sqrt[6]{3})$ -- all the generators of the first extension are powers of the generator of the second.)
That leaves the other factor: $x^2 + \sqrt{3} = (x - \mathrm{i}\sqrt[4]{3})(x + \mathrm{i}\sqrt[4]{3})$. Since the field $\Bbb{Q}(\sqrt[4]{3})$ is a bunch of real numbers, $\Bbb{Q}$, extended by a real number, $\sqrt[4]{3}$, it contains neither of these complex roots. We showed this slowly, above, but more quickly here: any field containing $\mathrm{i}\sqrt[4]{3}$ contains $(-1)\mathrm{i}\sqrt[4]{3}$. So we study the extension $\Bbb{Q}(\sqrt[4]{3})(\mathrm{i}\sqrt[4]{3})$. Just as we can use field operations to go from the element $\mathrm{i}\sqrt[4]{3}$ to $(-1)\mathrm{i}\sqrt[4]{3}$, we can use field elements to replace $\mathrm{i}\sqrt[4]{3}$ with $\frac{\mathrm{i}\sqrt[4]{3}}{\sqrt[4]{3}} = \mathrm{i}$. (That is, $\sqrt[4]{3}$ is a (nonzero) coefficient in the base field $\Bbb{Q}(\sqrt[4]{3})$ of $\Bbb{Q}(\sqrt[4]{3})(\mathrm{i}\sqrt[4]{3})$, so we are free to add, subtract, multiply or divide any generator of the further extension by it.) So a more compact way to write our extension is $\Bbb{Q}(\sqrt[4]{3}, \mathrm{i})$. We know the second generator is not in the span of the first by comparing real to nonreal numbers. Do we know the first is not in the span of the second?
So we study the question: $\sqrt[4]{3} \in \Bbb{Q}(\mathrm{i})$? The basis here is $\{1, \mathrm{i}\}$. We write $$ \sqrt[4]{3} \in q_0 + q_1 \mathrm{i} $$ and solve for $q_0$ and $q_1$. From our long experience with complex numbers, we know the only way for this to happen is $q_0 = \sqrt[4]{3}$ and $q_1 = 0$, which is not a pair of rational numbers. Therefore, $\sqrt[4]{3} \not\in \Bbb{Q}(\mathrm{i})$ and our splitting field is $\Bbb{Q}(\sqrt[4]{3}, \mathrm{i})$.
How does that respond to your question? First, we extend by the (degree of polynomial) many roots. Everything after that is reduction -- at each step we determine whether each generator is in the field extended by the other generators, so in at most (degree of polynomial) steps we have a subset of the roots that are independent extension generators. So this is a finite process.
We have used the fact that the extension generators are algebraic to get finite lists of powers to use in writing generic elements.
What does this have to do with your $a$ and $b$? In either $\Bbb{Q}(\mathrm{i})(\sqrt[4]{3})$ or $\Bbb{Q}(\sqrt[4]{3})(\mathrm{i})$ (which is two ways to describe the same field extension of $\Bbb{Q}$), we have $a = \mathrm{i}$ and $b = \sqrt[4]{3}$ and a complete set of generators for the field extension is $\{1,b, b^2, b^3, a, ab, ab^2, ab^3\}$. We are free to see this as "powers of $b$ with coefficients from powers of $a$" or "powers of $a$ with coefficients from powers of $b$", precisely corresponding to the two ways we wrote the extension. However, this is also true for the other versions of the field extension that we wrote down or passed by without explicitly writing down. For instance, it also holds for $\Bbb{Q}(-\sqrt[4]{3})(-\mathrm{i}\sqrt[4]{3})$. We do get a different basis, but it spans the same extension field.
Again, this is finite because we start with a finite number of roots, $r_1, \dots, r_n$, each of finite degree, $d_1, \dots d_n$, and the resulting set of $\prod_{i=1}^n d_i$ generators is finite. We then eliminate redundant generators (which induces a partial order on the field extensions) to obtain a minimal set of generators, a basis. There are only finitely many generators to eliminate and there are only finitely many partial orders of intermediate extensions, so the entire process is finite.