[proofplan] We compare the prime subfield with the image of the canonical ring homomorphism from $\mathbb{Z}$ to $k$ sending $1$ to $1_k$. In characteristic $p$, this image is already a field and is isomorphic to $\mathbb{Z}/p\mathbb{Z}=\mathbb{F}_p$. In characteristic $0$, the image is a copy of $\mathbb{Z}$, and adjoining inverses of its nonzero elements gives a copy of $\mathbb{Q}$. In both cases, every subfield of $k$ contains the constructed copy, so it is exactly the prime subfield. [/proofplan] [step:Identify the prime subfield through the canonical integer map] By the theorem statement, $0_k$ denotes the zero element of $k$, $1_k$ denotes the multiplicative identity of $k$, $\mathcal{S}$ denotes the set of all subfields of $k$, and the prime subfield is \begin{align*} P:=\bigcap_{F\in \mathcal{S}} F. \end{align*} Since $k\in \mathcal{S}$, this collection is nonempty. Every subfield $F\in \mathcal{S}$ contains $0_k$, $1_k$, and is closed under addition and additive inverses. Hence, for every $n\in \mathbb{Z}$, the element $n\cdot 1_k$ belongs to every $F\in \mathcal{S}$. Define the canonical ring homomorphism \begin{align*} \iota:\mathbb{Z}\to k,\qquad n\mapsto n\cdot 1_k. \end{align*} The preceding paragraph shows that $\operatorname{im}\iota\subseteq P$. [guided] We first pin down exactly what the notation means. By the theorem statement, $0_k$ is the zero element of $k$, $1_k$ is the multiplicative identity of $k$, and $\mathcal{S}$ is the collection of all subfields of $k$. This collection is nonempty because $k$ itself is a subfield of $k$. The prime subfield is defined as the intersection \begin{align*} P:=\bigcap_{F\in \mathcal{S}} F. \end{align*} Thus an element of $k$ lies in $P$ precisely when it lies in every subfield of $k$. The element that every subfield must contain is $1_k$. Since a subfield is closed under addition and additive inverses, it must also contain every integer multiple of $1_k$. This motivates the canonical map \begin{align*} \iota:\mathbb{Z}\to k,\qquad n\mapsto n\cdot 1_k. \end{align*} This map is a ring homomorphism: addition of integers corresponds to addition of repeated copies of $1_k$, multiplication of integers corresponds to multiplication of repeated copies of $1_k$, and $\iota(1)=1_k$. Therefore every element of $\operatorname{im}\iota$ belongs to every subfield of $k$, so \begin{align*}\operatorname{im}\iota\subseteq P.\end{align*} This is the common starting point for both characteristic cases. [/guided] [/step] [step:In characteristic $p$, identify the image with $\mathbb{F}_p$] Assume $\operatorname{char}(k)=p$, where $p$ is prime. By the definition of characteristic, $p$ is the least positive integer such that $p\cdot 1_k=0_k$. If $n\in\ker\iota$, write \begin{align*} n = qp + r \end{align*} with integers $q,r$ and $0 \le r < p$. Then \begin{align*} 0_k=n\cdot 1_k. \end{align*} Using $n=qp+r$ and $p\cdot 1_k=0_k$, we obtain \begin{align*} 0_k=(qp+r)\cdot 1_k=r\cdot 1_k. \end{align*} So the minimality of $p$ forces $r=0$. Hence $n\in p\mathbb{Z}$. Conversely, every element of $p\mathbb{Z}$ maps to $0_k$ because $p\cdot 1_k=0_k$, and therefore $\ker\iota=p\mathbb{Z}$. The induced map \begin{align*} \bar{\iota}:\mathbb{Z}/p\mathbb{Z}\to\operatorname{im}\iota,\qquad \bar{n}\mapsto n\cdot 1_k \end{align*} is a well-defined ring isomorphism. It remains to check that $\operatorname{im}\iota$ is a subfield of $k$. Let $x\in \operatorname{im}\iota$ with $x\ne 0_k$. Then $x=n\cdot 1_k$ for some $n\in \mathbb{Z}$ whose class $\bar{n}$ is nonzero in $\mathbb{Z}/p\mathbb{Z}$. Thus $p\nmid n$. Since $p$ is prime, this gives $\gcd(n,p)=1$. By Bezout's identity, there exist integers $a,b\in \mathbb{Z}$ such that \begin{align*} an+bp=1. \end{align*} Multiplying by $1_k$ gives \begin{align*}(a\cdot 1_k)(n\cdot 1_k)+(b\cdot 1_k)(p\cdot 1_k)=1_k.\end{align*} Since $p\cdot 1_k=0_k$, this becomes \begin{align*}(a\cdot 1_k)x=1_k.\end{align*} Thus $x^{-1}=a\cdot 1_k\in \operatorname{im}\iota$, so $\operatorname{im}\iota$ is a subfield of $k$. Since $P$ is the intersection of all subfields of $k$ and $\operatorname{im}\iota$ is itself a subfield, we have $P\subseteq \operatorname{im}\iota$. Together with $\operatorname{im}\iota\subseteq P$, this gives $P=\operatorname{im}\iota\cong \mathbb{Z}/p\mathbb{Z}=\mathbb{F}_p$. [/step] [step:In characteristic $0$, construct the rational copy inside $k$] Assume $\operatorname{char}(k)=0$. Then no positive integer $n$ satisfies $n\cdot 1_k=0_k$, so $\ker\iota=\{0\}$ and $\iota$ is injective. Define the map \begin{align*}\psi:\mathbb{Q}\to k,\qquad \frac{a}{b}\mapsto \iota(a)\iota(b)^{-1}\end{align*} for $a,b\in\mathbb{Z}$ with $b\ne 0$. Since $\iota$ is injective, \begin{align*} \iota(b) \ne 0_k \end{align*} so $\iota(b)^{-1}$ exists in $k$. We verify that $\psi$ is well-defined. Suppose $a,b,c,d\in\mathbb{Z}$ with $b\ne 0$, $d\ne 0$, and \begin{align*}\frac{a}{b}=\frac{c}{d}.\end{align*} Then \begin{align*} ad = bc \end{align*} so applying $\iota$ gives \begin{align*} \iota(a)\iota(d)=\iota(b)\iota(c). \end{align*} Multiplying by $\iota(b)^{-1}\iota(d)^{-1}$ in the field $k$ gives \begin{align*} \iota(a)\iota(b)^{-1}=\iota(c)\iota(d)^{-1}. \end{align*} Thus $\psi$ is well-defined. The map $\psi$ respects multiplication because, for $a,b,c,d\in\mathbb{Z}$ with $b\ne0$ and $d\ne0$, \begin{align*} \psi\left(\frac{a}{b}\frac{c}{d}\right)=\psi\left(\frac{ac}{bd}\right)=\iota(ac)\iota(bd)^{-1}=\iota(a)\iota(b)^{-1}\iota(c)\iota(d)^{-1}. \end{align*} It respects addition because \begin{align*} \psi\left(\frac{a}{b}+\frac{c}{d}\right)=\psi\left(\frac{ad+bc}{bd}\right)=\iota(ad+bc)\iota(bd)^{-1}. \end{align*} Using that $\iota$ preserves addition and multiplication, the last expression equals \begin{align*} \iota(a)\iota(b)^{-1}+\iota(c)\iota(d)^{-1}. \end{align*} Also $\psi(1)=1_k$, so $\psi$ is a field homomorphism. Since every field homomorphism has kernel either $\{0\}$ or the whole field, and $\psi(1)=1_k\ne 0_k$, we have $\ker\psi=\{0\}$. Hence $\psi$ is injective. [guided] In characteristic $0$, the integer copy inside $k$ has no nonzero collapse. More precisely, if $\iota(n)=0_k$, then $n\cdot 1_k=0_k$. By the definition of characteristic $0$, this forces $n=0$. Hence \begin{align*}\ker \iota=\{0\}.\end{align*} So $\iota$ embeds $\mathbb{Z}$ into $k$. A subfield cannot contain only the integer multiples of $1_k$; it must also contain inverses of every nonzero such element. This is why the correct object in characteristic $0$ is $\mathbb{Q}$. Define \begin{align*}\psi:\mathbb{Q}\to k,\qquad \frac{a}{b}\mapsto \iota(a)\iota(b)^{-1}.\end{align*} where $a,b\in \mathbb{Z}$ and $b\ne 0$. The inverse $\iota(b)^{-1}$ exists because $\iota$ is injective and therefore $\iota(b)\ne 0_k$. The only delicate point is well-definedness, since a rational number may have many presentations. Suppose \begin{align*}\frac{a}{b}=\frac{c}{d}\end{align*} with $a,b,c,d\in \mathbb{Z}$ and $b,d\ne 0$. Cross-multiplication in $\mathbb{Z}$ gives $ad=bc$. Applying the ring homomorphism $\iota$ gives \begin{align*} \iota(a)\iota(d)=\iota(b)\iota(c). \end{align*} Since $\iota(b)$ and $\iota(d)$ are nonzero elements of the field $k$, both are invertible. Multiplying by $\iota(b)^{-1}\iota(d)^{-1}$ gives \begin{align*} \iota(a)\iota(b)^{-1}=\iota(c)\iota(d)^{-1}. \end{align*} Thus $\psi$ is well-defined. The addition and multiplication laws follow from the ordinary fraction laws in $\mathbb{Q}$ and the fact that $\iota$ preserves addition and multiplication. Also $\psi(1)=1_k$, so $\psi$ is a field homomorphism that is not the zero map. Its kernel is an ideal of the field $\mathbb{Q}$, and the only ideals of a field are $\{0\}$ and the whole field. Since $1\notin \ker\psi$, the kernel is $\{0\}$. Therefore $\psi$ is injective and embeds $\mathbb{Q}$ as a subfield of $k$. [/guided] [/step] [step:Show the rational copy is exactly the prime subfield] Let $Q_k:=\operatorname{im}\psi\subseteq k$. Since $\psi$ is an injective field homomorphism, $Q_k$ is a subfield of $k$ and $Q_k\cong \mathbb{Q}$. We show $Q_k\subseteq P$. Let $F\in \mathcal{S}$ be any subfield of $k$. From the first step, $\iota(n)\in F$ for every $n\in \mathbb{Z}$. If $b\in \mathbb{Z}$ and $b\ne 0$, then $\iota(b)\ne 0_k$, and since $F$ is a field, $\iota(b)^{-1}\in F$. Therefore, for all $a,b\in \mathbb{Z}$ with $b\ne 0$, \begin{align*} \psi\left(\frac{a}{b}\right)=\iota(a)\iota(b)^{-1}\in F. \end{align*} Since $F$ was arbitrary, $Q_k\subseteq P$. Because $Q_k$ is itself a subfield of $k$, the definition of $P$ as the intersection of all subfields gives $P\subseteq Q_k$. Hence $P=Q_k\cong \mathbb{Q}$. This proves the characteristic $0$ case and completes the classification. [/step]