Choose a $K$-basis $e_1,\ldots,e_n$ of $L$ and write
$$
\alpha=\sum_{i=1}^nx_ie_i,
$$
thereby identifying $L$ with $\mathbb A_K^n$. Let $M(x)$ be the $n\times n$ matrix whose $j$th column is the coordinate vector of $\alpha^{j-1}$ in this basis, for $1\leq j\leq n$.

Write the multiplication table as
$$
e_ie_j=\sum_{l=1}^nc_{ij}^{\,l}e_l
$$
with fixed <structure constants> $c_{ij}^{\,l}\in K$. Repeated multiplication shows that every entry of $M(x)$ is a polynomial in $x_1,\ldots,x_n$. Hence
$$
D(x)=\det M(x)\in K[x_1,\ldots,x_n].
$$
By the criterion supplied in the question,
$$
K[\alpha]=L
\quad\Longleftrightarrow\quad
1,\alpha,\ldots,\alpha^{n-1}\text{ are a basis}
\quad\Longleftrightarrow\quad
D(x)\ne0.
$$
Consequently
$$
\boxed{U=D(D)=\mathbb A_K^n\setminus V(D)},
$$
which is a <distinguished open set>, and therefore a <Zariski-open set>. This is the determinant construction showing that <primitive elements form a principal Zariski-open set>. No assertion that $D$ is nonzero is needed.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 25F / b / Solution

Fix a $K$-basis of a finite extension $L/K$ of degree $n$. The coordinates of $1,\alpha,\ldots,\alpha^{n-1}$ are polynomial functions of the coordinates of $\alpha$, because multiplication in $L$ has fixed structure constants. The determinant $D(\alpha)$ of those coordinate columns is therefore a polynomial, and
$$
K[\alpha]=L\quad\Longleftrightarrow\quad D(\alpha)\ne0.
$$
Thus the primitive-element locus is the distinguished <Zariski-open set> $D(D)\subseteq\mathbb A_K^n$, possibly empty for a non-simple extension.

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