[proofplan] For part (1), the inclusion $A^\times \subset A \cap B^\times$ is immediate. For the reverse, if $a \in A$ has an inverse $b \in B$, we use the integral equation for $b$ over $A$ and multiply by a power of $a$ to show $b \in A$. Part (2) follows: if $B$ is a field, then $A \cap B^\times = A \setminus \{0\}$, so part (1) gives $A^\times = A \setminus \{0\}$, making $A$ a field. Conversely, if $A$ is a field and $0 \neq b \in B$, we use a minimal-degree integral equation to construct an inverse for $b$ in $B$. [/proofplan] [step:Prove part (1), easy inclusion: $A^\times \subset A \cap B^\times$] If $a \in A^\times$, then $a$ has an inverse $a^{-1} \in A \subset B$. In particular, $a$ is a unit in $B$, so $a \in A \cap B^\times$. This gives $A^\times \subset A \cap B^\times$. [/step] [step:Prove part (1), reverse inclusion: if $a \in A$ is a unit in $B$, then $a^{-1} \in A$] Let $a \in A \cap B^\times$. Then there exists $b \in B$ with $ba = 1_B$. Since $B$ is integral over $A$, the element $b$ satisfies a monic equation over $A$: there exist $a_1, \ldots, a_n \in A$ with \begin{align*} b^n + a_1 b^{n-1} + \cdots + a_{n-1} b + a_n = 0. \end{align*} Multiply both sides by $a^{n-1}$: \begin{align*} b^n a^{n-1} + a_1 b^{n-1} a^{n-1} + \cdots + a_{n-1} b a^{n-1} + a_n a^{n-1} = 0. \end{align*} Since $ba = 1$, we have $b^k a^k = (ba)^k = 1$ for all $k \geq 0$. Thus $b^n a^{n-1} = b \cdot (ba)^{n-1} \cdot \frac{1}{1} = b \cdot 1 = b$, and for each $1 \leq k \leq n-1$, $b^{n-k} a^{n-1} = a^{k-1} \cdot (ba)^{n-k} \cdot \frac{1}{1} = a^{k-1}$. More carefully: \begin{align*} b^n a^{n-1} &= b \cdot (b^{n-1} a^{n-1}) = b \cdot (ba)^{n-1} = b, \\ a_k b^{n-k} a^{n-1} &= a_k \cdot a^{k-1} \cdot (ba)^{n-k} = a_k a^{k-1}, \quad \text{for } 1 \leq k \leq n-1, \\ a_n a^{n-1} &= a_n a^{n-1}. \end{align*} Substituting: \begin{align*} b + a_1 + a_2 a + a_3 a^2 + \cdots + a_{n-1} a^{n-2} + a_n a^{n-1} = 0. \end{align*} Therefore $b = -(a_1 + a_2 a + a_3 a^2 + \cdots + a_n a^{n-1})$. Since $a \in A$ and $a_1, \ldots, a_n \in A$, the right-hand side is an element of $A$. Hence $b = a^{-1} \in A$, and so $a \in A^\times$. [guided] We want to show that if $a \in A$ has an inverse $b = a^{-1}$ in $B$, then $b$ actually lies in $A$. The inverse $b$ is integral over $A$, so it satisfies a monic equation $b^n + a_1 b^{n-1} + \cdots + a_n = 0$ with $a_i \in A$. The trick is to use the relation $ba = 1$ to eliminate powers of $b$. Multiplying the integral equation by $a^{n-1}$: \begin{align*} b^n a^{n-1} + a_1 b^{n-1} a^{n-1} + \cdots + a_{n-1} b a^{n-1} + a_n a^{n-1} = 0. \end{align*} Since $ba = 1$, we have $(ba)^k = 1$ for all $k \geq 1$. For the term $b^{n-k} a^{n-1}$, we factor out $k-1$ copies of $a$ and $n-k$ copies of $ba$: \begin{align*} b^{n-k} a^{n-1} = a^{k-1} (ba)^{n-k} = a^{k-1}. \end{align*} For $k = 0$: $b^n a^{n-1} = b \cdot b^{n-1} a^{n-1} = b \cdot (ba)^{n-1} = b$. Substituting all these simplifications: \begin{align*} b + a_1 \cdot 1 + a_2 \cdot a + a_3 \cdot a^2 + \cdots + a_n \cdot a^{n-1} = 0, \end{align*} so $b = -(a_1 + a_2 a + \cdots + a_n a^{n-1})$. Every term on the right is a product of elements of $A$, hence lies in $A$. Therefore $b \in A$, which means $a^{-1} \in A$ and $a \in A^\times$. [/guided] [/step] [step:Prove part (2), forward direction: if $B$ is a field, then $A$ is a field] Assume $B$ is a field, and assume $A$ and $B$ are integral domains. Since $B$ is a field, $B^\times = B \setminus \{0\}$. Since $A \subset B$ and $A$ is an integral domain, $A \setminus \{0\} \subset B \setminus \{0\} = B^\times$. Therefore \begin{align*} A \cap B^\times = A \cap (B \setminus \{0\}) = A \setminus \{0\}. \end{align*} By part (1), $A^\times = A \cap B^\times = A \setminus \{0\}$, which means every nonzero element of $A$ is a unit. Hence $A$ is a field. [/step] [step:Prove part (2), reverse direction: if $A$ is a field, then $B$ is a field] Assume $A$ is a field, and assume $A$ and $B$ are integral domains. Let $0 \neq b \in B$. We must show $b$ is a unit in $B$. Since $B$ is integral over $A$, there exist $a_1, \ldots, a_n \in A$ such that \begin{align*} b^n + a_1 b^{n-1} + \cdots + a_{n-1} b + a_n = 0. \end{align*} Choose such an equation with $n \geq 1$ minimal. Define $\Delta := b^{n-1} + a_1 b^{n-2} + \cdots + a_{n-1} \in B$, so that the integral equation factors as \begin{align*} b \Delta + a_n = 0, \quad \text{i.e.,} \quad b \Delta = -a_n. \end{align*} We claim $a_n \neq 0$. If $a_n = 0$, then $b \Delta = 0$. Since $B$ is an integral domain and $b \neq 0$, this forces $\Delta = 0$, i.e., $b^{n-1} + a_1 b^{n-2} + \cdots + a_{n-1} = 0$. This is a monic equation of degree $n - 1$ for $b$ over $A$, contradicting the minimality of $n$. Therefore $a_n \neq 0$. Since $A$ is a field and $a_n \in A \setminus \{0\}$, the element $a_n$ has an inverse $a_n^{-1} \in A$. From $b \Delta = -a_n$, we obtain \begin{align*} b \cdot (-a_n^{-1} \Delta) = 1_B. \end{align*} Therefore $b$ is a unit in $B$, with $b^{-1} = -a_n^{-1} \Delta \in B$. Since $0 \neq b \in B$ was arbitrary, $B$ is a field. [guided] Assume $A$ is a field and let $0 \neq b \in B$. We need to exhibit an inverse for $b$ in $B$. The integral equation for $b$ over $A$ provides the raw material. Take a monic equation $b^n + a_1 b^{n-1} + \cdots + a_n = 0$ with $n$ minimal. Why minimal? Because we need the constant term $a_n$ to be nonzero in order to construct the inverse. Rearranging: $b(b^{n-1} + a_1 b^{n-2} + \cdots + a_{n-1}) = -a_n$. Define $\Delta := b^{n-1} + a_1 b^{n-2} + \cdots + a_{n-1}$, so $b\Delta = -a_n$. If $a_n$ were zero, then $b\Delta = 0$. Since $B$ is an integral domain and $b \neq 0$, we would get $\Delta = 0$, i.e., $b^{n-1} + a_1 b^{n-2} + \cdots + a_{n-1} = 0$. But this is a monic equation of degree $n-1 < n$, contradicting the minimality of $n$. (This is also where we use $n \geq 1$: if $n = 1$, then $b + a_1 = 0$, and we need $a_1 = -b \neq 0$, which holds since $b \neq 0$ and $A \subset B$ is an inclusion of integral domains.) Therefore $a_n \neq 0$. Since $A$ is a field and $0 \neq a_n \in A$, the inverse $a_n^{-1}$ exists in $A$. Multiplying $b\Delta = -a_n$ by $-a_n^{-1}$: \begin{align*} b \cdot (-a_n^{-1} \Delta) = 1_B. \end{align*} So $b^{-1} = -a_n^{-1}\Delta \in B$, and $b$ is a unit. Since $b$ was an arbitrary nonzero element of $B$, every nonzero element of $B$ is invertible, and $B$ is a field. [/guided] [/step]