= Solution
Let $\sigma_1,\ldots,\sigma_n:K\hookrightarrow\mathbb C$ be the embeddings of the <number field> $K$. The <discriminant of elements of a number field> is
$$
\operatorname{disc}(\alpha_1,\ldots,\alpha_n)
=\det\!\bigl(\sigma_i(\alpha_j)\bigr)^2
=\det\!\bigl(\operatorname{Tr}_{K/\mathbb Q}(\alpha_i\alpha_j)\bigr).
$$
If the $\alpha_j$ are $\mathbb Q$-linearly dependent, the embedding matrix has dependent columns and the discriminant vanishes. Conversely, if they are independent, they form a $\mathbb Q$-basis of the $n$-dimensional vector space $K$. The trace pairing of a characteristic-zero number field is nondegenerate, so its Gram matrix in this basis is nonsingular. Hence
$$
\boxed{\operatorname{disc}(\alpha_1,\ldots,\alpha_n)\ne0
\iff(\alpha_1,\ldots,\alpha_n)\text{ is a }\mathbb Q\text{-basis of }K}.
$$
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