[proofplan]
This is a direct corollary of the [Norm and Discriminant](/theorems/1594) identity $\Delta(\alpha_1, \ldots, \alpha_n) = N(\mathfrak{a})^2 D_L$. If the left-hand side is square-free, then the square factor $N(\mathfrak{a})^2$ on the right must equal $\pm 1$, forcing $N(\mathfrak{a}) = 1$, which means $\mathfrak{a} = \mathcal{O}_L$. The remaining factor on the right is $D_L$, which must therefore equal $\Delta(\alpha_1, \ldots, \alpha_n)$ up to sign — and hence be square-free as well.
[/proofplan]
[step:Apply the norm-discriminant relation]
Since $\alpha_1, \ldots, \alpha_n$ is a $\mathbb{Z}$-basis of $\mathfrak{a}$, by [Norm and Discriminant](/theorems/1594),
\begin{align*}
\Delta(\alpha_1, \ldots, \alpha_n) = N(\mathfrak{a})^2\, D_L.
\end{align*}
Both $N(\mathfrak{a}) \in \mathbb{Z}_{\geq 1}$ (positive integer, by definition of the ideal norm) and $D_L \in \mathbb{Z}$ (the field discriminant is an integer, being the determinant-squared of an integer matrix obtained from $\Delta(\alpha_1', \ldots, \alpha_n')$ after expanding in terms of embeddings; equivalently, $D_L \in \mathbb{Z}$ because $\operatorname{Tr}_{L/\mathbb{Q}}(\alpha_i' \alpha_j') \in \mathbb{Z}$ and the trace-form discriminant equals $\det(\operatorname{Tr}(\alpha_i' \alpha_j'))$).
[guided]
The hypothesis tells us that $\alpha_1, \ldots, \alpha_n$ is a $\mathbb{Z}$-basis of $\mathfrak{a}$. The [Norm and Discriminant](/theorems/1594) theorem applies immediately, giving
\begin{align*}
\Delta(\alpha_1, \ldots, \alpha_n) = N(\mathfrak{a})^2 D_L.
\end{align*}
This is the only machinery we need; the rest of the argument is arithmetic in $\mathbb{Z}$.
**Integrality of the factors.**
- $N(\mathfrak{a}) = |\mathcal{O}_L/\mathfrak{a}|$ is the cardinality of a finite abelian group, hence a positive integer.
- $D_L \in \mathbb{Z}$: the discriminant $\Delta(\alpha_1', \ldots, \alpha_n')$ of an integral basis equals $\det(\operatorname{Tr}_{L/\mathbb{Q}}(\alpha_i' \alpha_j'))$, which is the determinant of an integer matrix (integrality of trace on $\mathcal{O}_L$), hence an integer.
So the factorisation takes place entirely inside $\mathbb{Z}$.
[/guided]
[/step]
[step:Force $N(\mathfrak{a}) = 1$ by the square-free hypothesis]
Suppose for contradiction that $N(\mathfrak{a}) \geq 2$. Then $N(\mathfrak{a})$ has a prime factor $p \in \mathbb{Z}$, and so $p^2 \mid N(\mathfrak{a})^2$. By the identity of Step 1, $p^2 \mid \Delta(\alpha_1, \ldots, \alpha_n)$. This contradicts the hypothesis that $\Delta(\alpha_1, \ldots, \alpha_n)$ is square-free.
Therefore $N(\mathfrak{a}) = 1$.
[guided]
**Square-free.** An integer $m \in \mathbb{Z}$ is **square-free** if $m \neq 0$ and for every prime $p$, $p^2 \nmid m$. Equivalently, in the prime factorisation $|m| = \prod_p p^{v_p}$, every exponent $v_p$ equals $0$ or $1$.
**Applying the definition.** Suppose $N(\mathfrak{a}) \geq 2$. Then $N(\mathfrak{a})$ has a prime factor $p$, so $p \mid N(\mathfrak{a})$. Squaring, $p^2 \mid N(\mathfrak{a})^2$. From the identity $\Delta(\alpha_1, \ldots, \alpha_n) = N(\mathfrak{a})^2 D_L$ of Step 1 and $D_L \in \mathbb{Z}$, we get $p^2 \mid \Delta(\alpha_1, \ldots, \alpha_n)$, contradicting square-freeness.
**Conclusion.** The only way out is $N(\mathfrak{a}) < 2$. Since $N(\mathfrak{a}) \geq 1$ always (nonzero ideal), this forces $N(\mathfrak{a}) = 1$.
**Why "square-free" is the right hypothesis.** The identity has $N(\mathfrak{a})^2$ — a square — as a factor. Square-freeness of the discriminant means no prime $p$ enters with exponent $\geq 2$, which is exactly the obstruction that prevents $N(\mathfrak{a})^2$ from having any nontrivial prime factor.
[/guided]
[/step]
[step:Conclude $\mathfrak{a} = \mathcal{O}_L$ from $N(\mathfrak{a}) = 1$]
By definition, $N(\mathfrak{a}) = |\mathcal{O}_L/\mathfrak{a}|$. If this index is $1$, then the quotient $\mathcal{O}_L/\mathfrak{a}$ has a single element, namely the zero coset. Since $\mathfrak{a} \subseteq \mathcal{O}_L$, this means $\mathfrak{a} = \mathcal{O}_L$.
[guided]
**Unpacking $|\mathcal{O}_L/\mathfrak{a}| = 1$.** The quotient $\mathcal{O}_L/\mathfrak{a}$ is the set of cosets $\{\beta + \mathfrak{a} : \beta \in \mathcal{O}_L\}$. "Cardinality $1$" means there is exactly one coset — necessarily the zero coset $0 + \mathfrak{a} = \mathfrak{a}$. So every $\beta \in \mathcal{O}_L$ represents the zero coset, i.e., $\beta + \mathfrak{a} = \mathfrak{a}$, which is equivalent to $\beta \in \mathfrak{a}$.
Thus $\mathcal{O}_L \subseteq \mathfrak{a}$. Combined with the standing $\mathfrak{a} \subseteq \mathcal{O}_L$,
\begin{align*}
\mathfrak{a} = \mathcal{O}_L.
\end{align*}
[/guided]
[/step]
[step:Deduce that $D_L$ is square-free]
Substituting $N(\mathfrak{a}) = 1$ into the identity of Step 1,
\begin{align*}
\Delta(\alpha_1, \ldots, \alpha_n) = D_L.
\end{align*}
By hypothesis, the left-hand side is square-free, hence so is $D_L$.
Combining, $\mathfrak{a} = \mathcal{O}_L$ and $D_L$ is square-free, as claimed.
[guided]
Now that $N(\mathfrak{a}) = 1$, the identity
\begin{align*}
\Delta(\alpha_1, \ldots, \alpha_n) = N(\mathfrak{a})^2 D_L
\end{align*}
simplifies to
\begin{align*}
\Delta(\alpha_1, \ldots, \alpha_n) = D_L.
\end{align*}
Since the left-hand side is a square-free integer by hypothesis, and the right-hand side equals it, $D_L$ is square-free.
**Consistency check.** With $\mathfrak{a} = \mathcal{O}_L$, any $\mathbb{Z}$-basis of $\mathfrak{a}$ is actually an integral basis of $\mathcal{O}_L$, so $\Delta(\alpha_1, \ldots, \alpha_n)$ is by definition a value of $D_L$ — it should equal $D_L$. The computation above confirms this.
**Why this is useful.** Given an algebraic number $\alpha$ generating $L$, the discriminant $\Delta(1, \alpha, \ldots, \alpha^{n-1})$ is the polynomial discriminant of the minimal polynomial of $\alpha$ — a finite, explicit integer computation. If this integer happens to be square-free, the criterion immediately yields $\mathcal{O}_L = \mathbb{Z}[\alpha]$ and identifies $D_L$ exactly — sidestepping the general problem of determining the ring of integers.
[/guided]
[/step]
