[proofplan] We prove both inclusions. Ground-model ordinals remain ordinals because $M \subseteq M[G]$ and forcing extensions of transitive models do not change the membership relation on old sets. For the converse, we take an ordinal $\alpha \in M[G]$, choose a $\mathbb{P}$-name in $M$ whose interpretation by $G$ is $\alpha$, and use a rank induction to bound the interpreted set by the syntactic rank of the name. Since an ordinal has rank equal to itself, this places $\alpha$ below a ground-model ordinal, hence inside $M$ by transitivity. [/proofplan] [step:Show that every ground-model ordinal remains an ordinal in the extension] Let $\xi \in \operatorname{Ord}^{M}$. Since $M \subseteq M[G]$, the set $\xi$ belongs to $M[G]$. We use the forcing-extension construction theorem for transitive models, namely the standard result asserting transitivity of the generic extension and preservation of the old membership relation. Its hypotheses apply here because $M$ is transitive, $\mathbb P \in M$ is the forcing notion, and $G \subseteq \mathbb P$ is $M$-generic. The theorem states that $M[G]$ is transitive and that the membership relation on old sets is the ambient membership relation restricted to those sets. Thus, for all $a,b \in M$, one has $a \in b$ in $M$ iff $a \in b$ in $M[G]$. Because $M$ is transitive, every element of $\xi$ belongs to $M$, and the same membership structure on $\xi$ is seen in the transitive model $M[G]$. Thus $\xi$ is still transitive and well-ordered by $\in$ in $M[G]$. Therefore $\xi \in \operatorname{Ord}^{M[G]}$, so \begin{align*} \operatorname{Ord}^{M} \subseteq \operatorname{Ord}^{M[G]}. \end{align*} [/step] [step:Bound the rank of an interpreted name by the rank of the name] Let $\tau \in M$ be a $\mathbb{P}$-name, meaning a hereditary collection of pairs $(\sigma,p)$ with $\sigma$ again a $\mathbb{P}$-name and $p \in \mathbb P$. Define the $M$-rank of $\tau$, denoted $\operatorname{rank}^{M}(\tau)$, to be the ordinal computed in $M$ by the recursive rule \begin{align*} \operatorname{rank}^{M}(\tau)=\sup \{\operatorname{rank}^{M}(\sigma)+1 : \exists p \in \mathbb{P}\, ((\sigma,p) \in \tau)\}. \end{align*} Since $M \models \mathrm{ZFC}$, this rank is an ordinal of $M$. Because $M$ is transitive and the recursion defining name rank uses only membership among names already in $M$, this ordinal agrees with the ambient recursive rank of the same name. For any set $y \in M[G]$, write $\operatorname{rank}(y)$ for the ordinary von Neumann rank of $y$ computed in the transitive model $M[G]$, equivalently the ambient rank of $y$ in the surrounding universe. Here $\tau_G$ denotes the name evaluation of $\tau$ by $G$, namely the set obtained recursively by keeping exactly the interpretations $\sigma_G$ for pairs $(\sigma,p) \in \tau$ with $p \in G$. We claim that for every $\mathbb{P}$-name $\tau \in M$, \begin{align*} \operatorname{rank}(\tau_G) \leq \operatorname{rank}^{M}(\tau). \end{align*} [claim:Evaluation does not increase rank beyond the name rank] For every $\mathbb{P}$-name $\tau \in M$, the interpreted set $\tau_G$ satisfies \begin{align*} \operatorname{rank}(\tau_G) \leq \operatorname{rank}^{M}(\tau). \end{align*} [/claim] [proof] We prove the claim by induction on the ordinal $\operatorname{rank}^{M}(\tau)$. Fix a $\mathbb{P}$-name $\tau \in M$, and assume the assertion has already been proved for every $\mathbb{P}$-name $\sigma \in M$ with \begin{align*} \operatorname{rank}^{M}(\sigma) < \operatorname{rank}^{M}(\tau). \end{align*} By definition of name evaluation, \begin{align*} \tau_G=\{\sigma_G : \exists p \in G\, ((\sigma,p) \in \tau)\}. \end{align*} Let $x \in \tau_G$. Then there exist a $\mathbb{P}$-name $\sigma \in M$ and a condition $p \in G$ such that $(\sigma,p) \in \tau$ and $x=\sigma_G$. From the definition of $\operatorname{rank}^{M}(\tau)$, this implies \begin{align*} \operatorname{rank}^{M}(\sigma) + 1 \leq \operatorname{rank}^{M}(\tau), \end{align*} and hence \begin{align*} \operatorname{rank}^{M}(\sigma) < \operatorname{rank}^{M}(\tau). \end{align*} The induction hypothesis gives \begin{align*} \operatorname{rank}(x)=\operatorname{rank}(\sigma_G)\leq\operatorname{rank}^{M}(\sigma)<\operatorname{rank}^{M}(\tau). \end{align*} Since this holds for every $x \in \tau_G$, the definition of rank gives \begin{align*} \operatorname{rank}(\tau_G)=\sup\{\operatorname{rank}(x)+1 : x \in \tau_G\}\leq\operatorname{rank}^{M}(\tau). \end{align*} This proves the claim. [/proof] [guided] The point of this step is to make precise the idea that evaluating a name cannot create sets of rank higher than the syntactic rank already present in the name. Let $\tau \in M$ be a $\mathbb{P}$-name. A $\mathbb P$-name is built hereditarily from pairs $(\sigma,p)$, where $\sigma$ is again a $\mathbb P$-name and $p \in \mathbb P$. Its $M$-rank is the ordinal \begin{align*} \operatorname{rank}^{M}(\tau)=\sup \{\operatorname{rank}^{M}(\sigma)+1 : \exists p \in \mathbb{P}\, ((\sigma,p) \in \tau)\}. \end{align*} This definition is made inside $M$, and since $M$ satisfies $\mathrm{ZFC}$, the value is an ordinal of $M$. Because $M$ is transitive and the recursion only inspects the membership structure of names belonging to $M$, the ordinal computed by $M$ is the same ordinal obtained by carrying out the same recursion in the ambient universe. For any set $y \in M[G]$, the notation $\operatorname{rank}(y)$ denotes the ordinary von Neumann rank of $y$ computed in the transitive model $M[G]$, equivalently the ambient rank of $y$ in the surrounding universe. We prove by induction on $\operatorname{rank}^{M}(\tau)$ that \begin{align*} \operatorname{rank}(\tau_G) \leq \operatorname{rank}^{M}(\tau). \end{align*} Assume the assertion is known for all names $\sigma \in M$ of strictly smaller $M$-rank than $\tau$. The name evaluation of $\tau$ by $G$ is the recursively defined interpretation \begin{align*} \tau_G=\{\sigma_G : \exists p \in G\, ((\sigma,p) \in \tau)\}. \end{align*} Thus every element $x \in \tau_G$ has the form $x=\sigma_G$ for some pair $(\sigma,p) \in \tau$ with $p \in G$. For such a $\sigma$, the definition of the name rank gives \begin{align*} \operatorname{rank}^{M}(\sigma)+1 \leq \operatorname{rank}^{M}(\tau), \end{align*} so $\sigma$ has strictly smaller $M$-rank than $\tau$. The induction hypothesis applies to $\sigma$ and yields \begin{align*} \operatorname{rank}(x)=\operatorname{rank}(\sigma_G)\leq\operatorname{rank}^{M}(\sigma)<\operatorname{rank}^{M}(\tau). \end{align*} Since every element of $\tau_G$ has rank below $\operatorname{rank}^{M}(\tau)$, taking the supremum in the definition of rank gives \begin{align*} \operatorname{rank}(\tau_G)=\sup\{\operatorname{rank}(x)+1 : x \in \tau_G\}\leq\operatorname{rank}^{M}(\tau). \end{align*} This proves the rank bound. [/guided] [/step] [step:Use the rank bound to show that no new ordinal appears] Let $\alpha \in \operatorname{Ord}^{M[G]}$. By definition of the forcing extension, there exists a $\mathbb{P}$-name $\tau \in M$ such that \begin{align*} \tau_G = \alpha. \end{align*} Let $\beta := \operatorname{rank}^{M}(\tau)$. By the previous step, \begin{align*} \operatorname{rank}(\alpha) \leq \beta. \end{align*} Since $\alpha$ is an ordinal, its rank is itself: \begin{align*} \operatorname{rank}(\alpha)=\alpha. \end{align*} Therefore $\alpha \leq \beta$. Because $\beta \in \operatorname{Ord}^{M}$ and $M \models \mathrm{ZFC}$, the successor ordinal $\beta+1$ belongs to $M$. Since $M$ is transitive, every element of $\beta+1$ belongs to $M$. The inequality $\alpha \leq \beta$ means exactly that $\alpha \in \beta+1$, and hence $\alpha \in M$. Since $M$ is transitive, ordinalhood is absolute between $M$ and the ambient universe for elements of $M$; therefore $M$ also regards $\alpha$ as an ordinal. Thus \begin{align*} \operatorname{Ord}^{M[G]} \subseteq \operatorname{Ord}^{M}. \end{align*} [/step] [step:Combine the two inclusions] The first step proved \begin{align*} \operatorname{Ord}^{M} \subseteq \operatorname{Ord}^{M[G]}. \end{align*} The previous step proved \begin{align*} \operatorname{Ord}^{M[G]} \subseteq \operatorname{Ord}^{M}. \end{align*} Therefore \begin{align*} \operatorname{Ord}^{M[G]}=\operatorname{Ord}^{M}. \end{align*} This completes the proof. [/step]