[proofplan] The forward direction is multiplicativity of the norm: if $\alpha\beta = 1$ in $\mathcal{O}_L$, then $N(\alpha)N(\beta) = 1$ in $\mathbb{Z}$, forcing $N(\alpha) \in \{\pm 1\}$. The reverse direction uses the embedding description of the norm (from [Norm and Trace via Embeddings](/theorems/1577)): fixing one embedding as the identity, the identity $N(\alpha) = \alpha \cdot \prod_{i \geq 2} \sigma_i(\alpha)$ exhibits an explicit candidate for $\alpha^{-1}$, namely $\pm \prod_{i \geq 2} \sigma_i(\alpha)$. We then show this candidate lies in $\mathcal{O}_L$ by noting that each $\sigma_i(\alpha)$ is an algebraic integer and $\mathcal{O}_L$ is closed under sums and products (from [Ring of Integers is a Ring](/theorems/1567)). [/proofplan] [step:Assume $\alpha$ is a unit and deduce $N(\alpha) \in \{\pm 1\}$ from multiplicativity] Suppose $\alpha \in \mathcal{O}_L^\times$. Then there exists $\beta \in \mathcal{O}_L$ with $\alpha\beta = 1$. The norm $N_{L/\mathbb{Q}}: L^\times \to \mathbb{Q}^\times$ is a group homomorphism (multiplicativity of the determinant applied to the multiplication-by-$x$ map), so \begin{align*} N(\alpha) \cdot N(\beta) = N(\alpha\beta) = N(1) = 1. \end{align*} Moreover, $N$ sends $\mathcal{O}_L$ into $\mathbb{Z}$: for any $x \in \mathcal{O}_L$, the characteristic polynomial of multiplication by $x$ has integer coefficients (by [Integrality and the Characteristic Polynomial](/theorems/1574)), and $N(x)$ is up to sign its constant term. Therefore $N(\alpha), N(\beta) \in \mathbb{Z}$ and their product is $1$, which forces $N(\alpha), N(\beta) \in \mathbb{Z}^\times = \{\pm 1\}$. In particular $N(\alpha) = \pm 1$. [guided] Assume $\alpha$ is a unit in $\mathcal{O}_L$ — by definition this means there is some $\beta \in \mathcal{O}_L$ with $\alpha\beta = 1$. We want to deduce that $N(\alpha) = \pm 1$. The key input is multiplicativity of the norm. Concretely, for $x \in L$, the norm $N_{L/\mathbb{Q}}(x)$ is defined as the determinant of the $\mathbb{Q}$-linear map $m_x: L \to L$, $y \mapsto xy$. Multiplicativity of the determinant (for the composition of linear maps) immediately gives $N(xy) = N(x) N(y)$ for any $x, y \in L$. In particular: \begin{align*} N(\alpha) \cdot N(\beta) = N(\alpha\beta) = N(1) = 1. \end{align*} (The last equality, $N(1) = 1$, comes from $m_1 = \operatorname{id}_L$, whose determinant is $1$.) Why does this pin down $N(\alpha)$ to $\pm 1$? Because $N(\alpha), N(\beta)$ are not just rationals but **integers**: by [Integrality and the Characteristic Polynomial](/theorems/1574), for $x \in \mathcal{O}_L$ the characteristic polynomial of $m_x$ lies in $\mathbb{Z}[t]$, so every coefficient — including the constant term $(-1)^n N(x)$ — is an integer. Applying this to $x = \alpha$ and $x = \beta$: both $N(\alpha), N(\beta) \in \mathbb{Z}$ with $N(\alpha) N(\beta) = 1$. The only units of $\mathbb{Z}$ are $\pm 1$, so $N(\alpha) = \pm 1$. [/guided] [/step] [step:Write $N(\alpha)$ as $\alpha$ times a product of conjugates using the embedding formula] Conversely, assume $N(\alpha) = \pm 1$. Let $\sigma_1, \ldots, \sigma_n: L \to \mathbb{C}$ be the $n$ distinct $\mathbb{Q}$-embeddings of $L$ into $\mathbb{C}$. Fix a choice of embedding $L \hookrightarrow \mathbb{C}$ and label it $\sigma_1$, so that $\sigma_1$ is the identity on the image of $L$ (we identify $L$ with its image under $\sigma_1$, so $\sigma_1(\alpha) = \alpha$). By [Norm and Trace via Embeddings](/theorems/1577), \begin{align*} N(\alpha) = \prod_{i=1}^n \sigma_i(\alpha) = \alpha \cdot \prod_{i=2}^n \sigma_i(\alpha). \end{align*} Since $N(\alpha) = \pm 1 \neq 0$, the element $\alpha$ is nonzero, and we may divide to obtain \begin{align*} \alpha^{-1} = \frac{1}{N(\alpha)} \prod_{i=2}^n \sigma_i(\alpha) = \pm \prod_{i=2}^n \sigma_i(\alpha), \tag{$\ast$} \end{align*} where the sign is that of $N(\alpha)$. The right-hand side is a priori an element of $\mathbb{C}$; a priori $\alpha^{-1}$ exists in $L \subseteq \mathbb{C}$ (as $L$ is a field and $\alpha \neq 0$), and $(\ast)$ is an equality inside $\mathbb{C}$. [guided] To prove $\alpha \in \mathcal{O}_L^\times$, we must exhibit an inverse $\alpha^{-1}$ and show it lies in $\mathcal{O}_L$. Since $L$ is a field and $\alpha \neq 0$ (because $N(\alpha) = \pm 1$ is nonzero, whereas $N(0) = 0$), the inverse $\alpha^{-1}$ exists inside $L$. The question is whether it lies in the subring $\mathcal{O}_L \subseteq L$. We look for a formula for $\alpha^{-1}$ that makes its integrality manifest. The [Norm and Trace via Embeddings](/theorems/1577) theorem gives us \begin{align*} N(\alpha) = \prod_{i=1}^n \sigma_i(\alpha), \end{align*} where $\sigma_1, \ldots, \sigma_n$ are the $\mathbb{Q}$-embeddings of $L$ into $\mathbb{C}$. To extract $\alpha$ from the product, fix an embedding $L \hookrightarrow \mathbb{C}$ and label it $\sigma_1$; identifying $L$ with its image, $\sigma_1$ is the identity, so $\sigma_1(\alpha) = \alpha$. Splitting off this factor: \begin{align*} N(\alpha) = \alpha \cdot \prod_{i=2}^n \sigma_i(\alpha). \end{align*} We are given $N(\alpha) = \pm 1$, so dividing both sides by $\alpha$ (valid because $\alpha \neq 0$): \begin{align*} \frac{N(\alpha)}{\alpha} = \prod_{i=2}^n \sigma_i(\alpha), \end{align*} and rearranging, \begin{align*} \alpha^{-1} = \frac{1}{N(\alpha)} \prod_{i=2}^n \sigma_i(\alpha) = \pm \prod_{i=2}^n \sigma_i(\alpha). \end{align*} We have written $\alpha^{-1}$ as (up to sign) a product of conjugates of $\alpha$. The next step is to verify that this product lands in $\mathcal{O}_L$. [/guided] [/step] [step:Show each conjugate $\sigma_i(\alpha)$ is an algebraic integer] Each embedding $\sigma_i: L \to \mathbb{C}$ sends $\alpha$ to a root of the minimal polynomial $p_\alpha(t) \in \mathbb{Q}[t]$ of $\alpha$ over $\mathbb{Q}$: indeed $\sigma_i$ fixes $\mathbb{Q}$, so $p_\alpha(\sigma_i(\alpha)) = \sigma_i(p_\alpha(\alpha)) = \sigma_i(0) = 0$. Since $\alpha \in \mathcal{O}_L$, [Integrality Criterion via Minimal Polynomial](/theorems/1570) gives $p_\alpha \in \mathbb{Z}[t]$. Thus $\sigma_i(\alpha)$ is a root of a monic polynomial with integer coefficients, i.e., an algebraic integer. [guided] Each $\sigma_i(\alpha)$ is a complex number; we must show it is an algebraic integer (a root of some monic polynomial in $\mathbb{Z}[t]$). The idea is that **$\sigma_i$ maps $\alpha$ to a root of its minimal polynomial**, and this minimal polynomial has integer coefficients because $\alpha$ is itself an algebraic integer. Concretely, let $p_\alpha(t) \in \mathbb{Q}[t]$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Two facts: - **$p_\alpha \in \mathbb{Z}[t]$**: this is [Integrality Criterion via Minimal Polynomial](/theorems/1570) — an element of $L$ is in $\mathcal{O}_L$ if and only if its minimal polynomial over $\mathbb{Q}$ has integer coefficients. Applied to $\alpha \in \mathcal{O}_L$, we get $p_\alpha \in \mathbb{Z}[t]$. - **$\sigma_i(\alpha)$ is a root of $p_\alpha$**: because $\sigma_i: L \to \mathbb{C}$ is a $\mathbb{Q}$-algebra homomorphism (it fixes $\mathbb{Q}$ pointwise — every embedding of a number field fixes $\mathbb{Q}$), applying $\sigma_i$ to the equation $p_\alpha(\alpha) = 0$ gives $p_\alpha(\sigma_i(\alpha)) = \sigma_i(p_\alpha(\alpha)) = \sigma_i(0) = 0$. Combining: $\sigma_i(\alpha) \in \mathbb{C}$ is a root of the monic polynomial $p_\alpha \in \mathbb{Z}[t]$, so by definition $\sigma_i(\alpha)$ is an algebraic integer. [/guided] [/step] [step:Conclude $\alpha^{-1} \in \mathcal{O}_L$ and hence $\alpha \in \mathcal{O}_L^\times$] Let $\bar{\mathbb{Z}} \subset \mathbb{C}$ denote the ring of all algebraic integers. By [Ring of Integers is a Ring](/theorems/1567), $\bar{\mathbb{Z}}$ is closed under sums, differences, and products. By the previous step, each $\sigma_i(\alpha) \in \bar{\mathbb{Z}}$, so \begin{align*} \prod_{i=2}^n \sigma_i(\alpha) \in \bar{\mathbb{Z}}, \end{align*} and therefore $\alpha^{-1} = \pm \prod_{i=2}^n \sigma_i(\alpha) \in \bar{\mathbb{Z}}$ by $(\ast)$. Since $\alpha^{-1} \in L$ (as $L$ is a field containing $\alpha \neq 0$), we have \begin{align*} \alpha^{-1} \in L \cap \bar{\mathbb{Z}} = \mathcal{O}_L, \end{align*} by the definition of $\mathcal{O}_L$ as the elements of $L$ that are algebraic integers over $\mathbb{Z}$. Thus $\alpha\beta = 1$ with $\beta = \alpha^{-1} \in \mathcal{O}_L$, so $\alpha \in \mathcal{O}_L^\times$. [guided] The formula $\alpha^{-1} = \pm \prod_{i=2}^n \sigma_i(\alpha)$ has expressed the inverse as a product of algebraic integers. We now need two facts to conclude: 1. **The product of algebraic integers is an algebraic integer.** Let $\bar{\mathbb{Z}} \subset \mathbb{C}$ be the set of all algebraic integers. By [Ring of Integers is a Ring](/theorems/1567), $\bar{\mathbb{Z}}$ is a subring of $\mathbb{C}$ — closed under sums, differences, and products. (This is nontrivial: even the sum of two algebraic integers is not a priori an algebraic integer, but it can be shown via the characterization in terms of finitely generated $\mathbb{Z}$-submodules.) Each $\sigma_i(\alpha)$ is in $\bar{\mathbb{Z}}$ by the previous step, hence their product is too: \begin{align*} \prod_{i=2}^n \sigma_i(\alpha) \in \bar{\mathbb{Z}}. \end{align*} Therefore $\alpha^{-1} = \pm \prod_{i=2}^n \sigma_i(\alpha) \in \bar{\mathbb{Z}}$. 2. **$L \cap \bar{\mathbb{Z}} = \mathcal{O}_L$.** This is the definition of $\mathcal{O}_L$: the ring of integers of $L$ is the set of elements of $L$ which are integral over $\mathbb{Z}$, i.e., which are algebraic integers. Combining: $\alpha^{-1}$ lies in $L$ (since $L$ is a field containing the nonzero $\alpha$) and in $\bar{\mathbb{Z}}$, hence in $\mathcal{O}_L$. Taking $\beta := \alpha^{-1} \in \mathcal{O}_L$ gives $\alpha\beta = 1$ with $\beta \in \mathcal{O}_L$, which is exactly the definition of $\alpha$ being a unit in $\mathcal{O}_L$. The result is a concrete test for unit-ness: to decide if $\alpha \in \mathcal{O}_L$ is a unit, compute $N(\alpha) \in \mathbb{Z}$ and check whether it is $\pm 1$. This reduces an infinite question (does an inverse exist?) to a finite computation (is a specific integer $\pm 1$?). [/guided] [/step]