If $Q \vartriangleleft \Bbb Z[i]$ lies over $(p) \vartriangleleft \Bbb Z$ where $p\in \Bbb Z\setminus\{2\}$, then $e(Q|p) = 1$.

Question

Let $R = \Bbb Z$, $S = \Bbb Z[i]$, and $p \in R\setminus\{2\}$. If the ideal $Q \vartriangleleft S$ lies over $(p) \vartriangleleft R$, then $e(Q|p) = 1$.

The primes (non-zero prime ideals) in $S$ lying over a given prime $P$ in $R$ are the ones which occur in the prime decomposition of $PS := \{\alpha_1\beta_1 + \ldots + \alpha_r\beta_r: \alpha_i \in P, \beta_i \in S, r\in \Bbb N\}.$ The exponents with which they occur are called ramification indices. Thus, if $Q^e$ is the exact power of $Q$ dividing $PS$, then $e$ is called the ramification index of $Q$ over $P$, denoted by $e(Q|P)$.

Interestingly, in the context of the above problem, if $p = 2$ and $Q = (1-i)$, then $e(Q|p) = 2$ as $(1-i)$ lies over $(2)$ and $2S = (1-i)^2$.

My attempt: Suppose $p\in R$, $p\ne 2$, and the ideal $Q$ in $S$ lies above $(p)$. By definition, $Q \mid (p)S$. Let $Q^e$ be the exact power of $Q$ dividing $(p)S$, i.e., $(p)S = Q^e P_1^{n_1} \ldots P_k^{n_k}$ for some prime ideals $P_1, \ldots, P_k$ in $S$, such that $Q \ne P_i$ for any $1\le i\le r$. I'm pretty much stuck here, but the goal is to conclude $e = 1$.

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Reference: Marcus' Number Fields, Chapter $3$, Pg. $45$.

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Note: To be technically correct, we should write $e(Q\mid (p))$, but we use $p$ instead of $(p)$ to avoid ugly notation.

Answer 1

For $p\ne 2$,

If $x^2+1\in \Bbb{F}_p[x]$ is irreducible then $\Bbb{Z}[i]/(p) \cong \Bbb{F}_p[x]/(x^2+1)$ is a field and $(p)\subset \Bbb{Z}[i]$ is a prime ideal.

If $x^2+1\in \Bbb{F}_p[x]$ is not irreducible then it has a root $c$, and a second distinct one $-c$, then $(p,c+i)$, $(p,c-i)$ are two distinct prime ideals and their product is $(p,c+i)(p,c-i)=(p^2,c^2+1,p(c+i),p(c-i))$ $=(p(p,c+i,c-i),c^2+1) = (p(p,2c,2i),c^2+1) = (p)$.

Answer 2

I will say that @reuns's answer above is quite nice. However, since you requested something more along the lines of the narrative in Marcus, here's another argument. A prime $(p) \in \mathbb{Z}$ ramifies in a number ring $\mathcal{O}_K$ if and only if it divides the discriminant of $\mathcal{O}_K$ (Marcus proves only the forward implication in chapter 3, but this is all you need). In chapter $2$, Marcus shows that for $K = \mathbb{Q}(\sqrt{m})$ with $m$ squarefree, disc$(\mathcal{O}_K)$ is $m$ when $m \equiv 1 \pmod{4}$ and is $4m$ when $m \equiv 2, 3 \pmod{4}$. So the discriminant of the Gaussian integers is -4, hence $2$ is the only prime that ramifies in this number ring.