[proofplan] Irreducibility in an integral domain means: every factorization is trivial, i.e., in any equation $\alpha = \beta\gamma$ one of $\beta, \gamma$ must be a unit. The proof exploits the multiplicativity of the norm to reduce this question to a statement in $\mathbb{Z}$: if $\alpha = \beta\gamma$ then $N(\alpha) = N(\beta) N(\gamma)$ in $\mathbb{Z}$. When $N(\alpha) = p$ is a rational prime, the only way to factor $p$ in $\mathbb{Z}$ is as $\pm 1 \cdot \pm p$, so one of $N(\beta), N(\gamma)$ equals $\pm 1$. By [Units via Norm](/theorems/1578), this forces the corresponding factor to be a unit in $\mathcal{O}_L$. [/proofplan] [step:Observe that $\alpha$ is nonzero and non-unit, so the irreducibility question is non-trivial] Since $N_{L/\mathbb{Q}}(\alpha)$ is a rational prime $p$, in particular $N(\alpha) \neq 0$ and $N(\alpha) \neq \pm 1$. From $N(\alpha) \neq 0$ we get $\alpha \neq 0$ (indeed the norm is multiplicative and $N(0) = 0$). From $N(\alpha) \neq \pm 1$, together with [Units via Norm](/theorems/1578), we conclude that $\alpha$ is **not** a unit in $\mathcal{O}_L$. Hence $\alpha$ is a nonzero non-unit, and the assertion that $\alpha$ is irreducible is the assertion that every factorization $\alpha = \beta\gamma$ in $\mathcal{O}_L$ has one factor a unit. [guided] Before analysing factorizations, we check the two standing hypotheses of irreducibility: an element of an integral domain is called **irreducible** when it is (i) nonzero, (ii) not a unit, and (iii) every factorization $\alpha = \beta\gamma$ has $\beta$ or $\gamma$ a unit. Both (i) and (ii) follow from $N(\alpha)$ being a rational prime $p$: - **Nonzero**: $N$ is multiplicative with $N(0) = 0$, so $N(\alpha) = p \neq 0$ forces $\alpha \neq 0$. - **Non-unit**: by [Units via Norm](/theorems/1578), $\alpha \in \mathcal{O}_L^\times$ iff $N(\alpha) = \pm 1$. Since $p$ is a rational prime (by convention, $|p| \geq 2$), $N(\alpha) = p \neq \pm 1$, so $\alpha$ is not a unit. Conditions (i) and (ii) secured, we now focus on (iii): showing every factorization is trivial. [/guided] [/step] [step:Factor the norm and apply primality in $\mathbb{Z}$] Let $\alpha = \beta\gamma$ with $\beta, \gamma \in \mathcal{O}_L$. Applying the norm map $N: L \to \mathbb{Q}$, which is a multiplicative homomorphism (by the determinant identity $\det(m_\beta m_\gamma) = \det(m_\beta) \det(m_\gamma)$), gives \begin{align*} N(\beta) \cdot N(\gamma) = N(\beta\gamma) = N(\alpha). \end{align*} Since $\beta, \gamma \in \mathcal{O}_L$, we have $N(\beta), N(\gamma) \in \mathbb{Z}$ by [Integrality and the Characteristic Polynomial](/theorems/1574). The right-hand side $N(\alpha) = p$ is a rational prime, so the above is a factorization of a prime in $\mathbb{Z}$. In $\mathbb{Z}$, a prime $p$ has only the factorizations \begin{align*} p = (\pm 1)(\pm p) \quad \text{or} \quad p = (\pm p)(\pm 1), \end{align*} so one of $N(\beta), N(\gamma)$ equals $\pm 1$. [guided] We have $\alpha = \beta\gamma$ with $\beta, \gamma \in \mathcal{O}_L$, and we want to show one factor is a unit. The strategy is to transport the equation from $\mathcal{O}_L$ to $\mathbb{Z}$ via the norm, where primality gives us leverage. The norm map $N_{L/\mathbb{Q}}: L^\times \to \mathbb{Q}^\times$ is multiplicative: $N$ is defined as $N(x) = \det(m_x)$ where $m_x: L \to L$ is the $\mathbb{Q}$-linear multiplication-by-$x$ map. Multiplicativity follows from $m_{\beta\gamma} = m_\beta \circ m_\gamma$ and $\det(AB) = \det(A)\det(B)$ for $\mathbb{Q}$-linear endomorphisms of $L$. Applying $N$ to $\alpha = \beta\gamma$: \begin{align*} N(\alpha) = N(\beta) \cdot N(\gamma). \end{align*} Two further facts: - The values $N(\beta), N(\gamma)$ lie in $\mathbb{Z}$ (not just $\mathbb{Q}$). This uses $\beta, \gamma \in \mathcal{O}_L$: by [Integrality and the Characteristic Polynomial](/theorems/1574), the characteristic polynomial of $m_x$ for $x \in \mathcal{O}_L$ has integer coefficients, so its constant term (which is $(-1)^n N(x)$) is in $\mathbb{Z}$. - $N(\alpha) = p$ is a rational prime by hypothesis. Substituting: \begin{align*} p = N(\beta) \cdot N(\gamma), \qquad N(\beta), N(\gamma) \in \mathbb{Z}. \end{align*} This is a factorization of $p$ in $\mathbb{Z}$. By definition of a prime integer, the only integer factorizations of $p$ are $(\pm 1)(\pm p)$ (in either order). Hence one of $N(\beta), N(\gamma)$ equals $\pm 1$ and the other equals $\pm p$. [/guided] [/step] [step:Translate $N(\beta) = \pm 1$ into $\beta \in \mathcal{O}_L^\times$ to complete the factorization analysis] By the previous step, at least one of $N(\beta), N(\gamma)$ equals $\pm 1$. Without loss of generality, say $N(\beta) = \pm 1$. Since $\beta \in \mathcal{O}_L$, [Units via Norm](/theorems/1578) gives \begin{align*} \beta \in \mathcal{O}_L^\times. \end{align*} Thus one factor of the arbitrary factorization $\alpha = \beta\gamma$ is a unit in $\mathcal{O}_L$. Combined with $\alpha \neq 0$ and $\alpha \notin \mathcal{O}_L^\times$ from the first step, $\alpha$ is irreducible in $\mathcal{O}_L$. [guided] From the previous step, either $N(\beta) = \pm 1$ or $N(\gamma) = \pm 1$. The two cases are symmetric (just relabel $\beta \leftrightarrow \gamma$), so assume WLOG $N(\beta) = \pm 1$. We invoke [Units via Norm](/theorems/1578): for $x \in \mathcal{O}_L$, \begin{align*} x \in \mathcal{O}_L^\times \iff N(x) = \pm 1. \end{align*} Hypotheses: $\beta \in \mathcal{O}_L$ (given) and $N(\beta) = \pm 1$ (just derived). Conclusion: $\beta \in \mathcal{O}_L^\times$. So in the factorization $\alpha = \beta\gamma$, the factor $\beta$ is a unit. Since the factorization was arbitrary and we have shown one factor is always a unit, together with the non-triviality hypotheses from Step 1 ($\alpha \neq 0$ and $\alpha$ not a unit), we conclude that $\alpha$ is irreducible in $\mathcal{O}_L$. A remark on why the converse fails: an element with composite norm may still be irreducible. For instance, in $\mathbb{Z}[\sqrt{-5}]$, the element $2$ has norm $4 = 2 \cdot 2$ (composite) but $2$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$ (no element has norm $2$). Irreducibility of norm-prime elements is a one-way implication: prime norm is a sufficient, not necessary, condition for irreducibility. [/guided] [/step]