[proofplan] All three parts are direct corollaries of [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586), which provides an inverse $\mathfrak{c}^{-1}$ used to cancel $\mathfrak{c}$ on both sides of containments and divisibility relations. Part (1) is a symmetric pair of implications: one direction from monotonicity of ideal multiplication, the other from multiplying by $\mathfrak{c}^{-1}$. Part (2) is the same argument stated in the language of ideal divisibility. Part (3) — the slogan "to divide is to contain" — also has one direction immediate from the definition of divisibility, while the other direction constructs the quotient $\mathfrak{a}^{-1}\mathfrak{b}$ and shows it is an integral ideal, which produces the required witness for divisibility. [/proofplan] [step:Establish (1) — $\mathfrak{b} \subseteq \mathfrak{a}$ iff $\mathfrak{b}\mathfrak{c} \subseteq \mathfrak{a}\mathfrak{c}$, using invertibility of $\mathfrak{c}$] **($\Rightarrow$)** Suppose $\mathfrak{b} \subseteq \mathfrak{a}$. Ideal multiplication by a fixed ideal is monotonic: if $I \subseteq J$, then $IK \subseteq JK$ for any ideal $K$ (an element $\sum_i x_i y_i$ of $IK$ with $x_i \in I \subseteq J$ and $y_i \in K$ is also an element of $JK$). Applying with $I = \mathfrak{b}$, $J = \mathfrak{a}$, $K = \mathfrak{c}$: \begin{align*} \mathfrak{b}\mathfrak{c} \subseteq \mathfrak{a}\mathfrak{c}. \end{align*} **($\Leftarrow$)** Suppose $\mathfrak{b}\mathfrak{c} \subseteq \mathfrak{a}\mathfrak{c}$. Since $\mathfrak{c} \neq 0$ is a nonzero integral ideal — hence a nonzero fractional ideal — [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586) applies: $\mathfrak{c}$ has a fractional-ideal inverse $\mathfrak{c}^{-1}$ with $\mathfrak{c}\mathfrak{c}^{-1} = \mathcal{O}_L$. Multiplying both sides of $\mathfrak{b}\mathfrak{c} \subseteq \mathfrak{a}\mathfrak{c}$ by $\mathfrak{c}^{-1}$ (i.e., applying the same monotonicity of ideal multiplication — from $(\Rightarrow)$ — to the new pair with $K = \mathfrak{c}^{-1}$): \begin{align*} \mathfrak{b}\mathfrak{c}\mathfrak{c}^{-1} \subseteq \mathfrak{a}\mathfrak{c}\mathfrak{c}^{-1}. \end{align*} Using $\mathfrak{c}\mathfrak{c}^{-1} = \mathcal{O}_L$ and $\mathfrak{d} \cdot \mathcal{O}_L = \mathfrak{d}$ for any fractional ideal $\mathfrak{d}$: \begin{align*} \mathfrak{b} \subseteq \mathfrak{a}. \end{align*} [guided] We prove the biconditional $\mathfrak{b} \subseteq \mathfrak{a} \iff \mathfrak{b}\mathfrak{c} \subseteq \mathfrak{a}\mathfrak{c}$ for ideals $\mathfrak{a}, \mathfrak{b}, \mathfrak{c} \unlhd \mathcal{O}_L$ with $\mathfrak{c} \neq 0$. **Forward direction ($\Rightarrow$).** This is pure monotonicity, valid in any commutative ring. *Monotonicity of ideal multiplication.* If $I, J, K$ are ideals with $I \subseteq J$, then $IK \subseteq JK$. Proof: a general element of $IK$ is a finite sum $\sum_i x_i y_i$ with $x_i \in I$, $y_i \in K$; since $I \subseteq J$, each $x_i \in J$, so the same element is in $JK$. *Application.* With $I = \mathfrak{b}$, $J = \mathfrak{a}$, $K = \mathfrak{c}$, hypothesis $\mathfrak{b} \subseteq \mathfrak{a}$ gives $\mathfrak{b}\mathfrak{c} \subseteq \mathfrak{a}\mathfrak{c}$. **Reverse direction ($\Leftarrow$).** Here we use invertibility crucially: we cancel the common factor $\mathfrak{c}$ by multiplying by its inverse. *Invertibility of $\mathfrak{c}$.* By hypothesis $\mathfrak{c} \neq 0$. Integral ideals are fractional ideals (nonzero integral ideals $\mathfrak{c} \unlhd \mathcal{O}_L$ are finitely generated $\mathcal{O}_L$-submodules of $\mathcal{O}_L \subseteq L$ with the unit element $1 \in \mathcal{O}_L$ as common denominator). Hence [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586) applies: there exists a fractional ideal $\mathfrak{c}^{-1}$ with $\mathfrak{c}\mathfrak{c}^{-1} = \mathcal{O}_L$. *Cancellation.* Multiply both sides of $\mathfrak{b}\mathfrak{c} \subseteq \mathfrak{a}\mathfrak{c}$ by $\mathfrak{c}^{-1}$ — where "multiply" is the monoid operation on fractional ideals, and the monotonicity of this operation (same proof as in the forward direction, applied to fractional ideals rather than integral ones) preserves containment: \begin{align*} \mathfrak{b}\mathfrak{c} \cdot \mathfrak{c}^{-1} \subseteq \mathfrak{a}\mathfrak{c} \cdot \mathfrak{c}^{-1}. \end{align*} *Simplifying using $\mathfrak{c}\mathfrak{c}^{-1} = \mathcal{O}_L$ and $\mathfrak{d} \cdot \mathcal{O}_L = \mathfrak{d}$.* The monoid of fractional ideals is commutative, associative, with identity $\mathcal{O}_L$ (the full ring acts as identity: $\mathcal{O}_L \cdot \mathfrak{d}$ for any fractional $\mathfrak{d}$ consists of $\mathcal{O}_L$-combinations of $\mathfrak{d}$, which is $\mathfrak{d}$ since $\mathfrak{d}$ is already $\mathcal{O}_L$-closed). Hence \begin{align*} \mathfrak{b}\mathfrak{c}\mathfrak{c}^{-1} &= \mathfrak{b} \cdot (\mathfrak{c}\mathfrak{c}^{-1}) = \mathfrak{b} \cdot \mathcal{O}_L = \mathfrak{b}, \\ \mathfrak{a}\mathfrak{c}\mathfrak{c}^{-1} &= \mathfrak{a} \cdot (\mathfrak{c}\mathfrak{c}^{-1}) = \mathfrak{a} \cdot \mathcal{O}_L = \mathfrak{a}. \end{align*} *Conclusion.* $\mathfrak{b} \subseteq \mathfrak{a}$. **Structural remark.** The reverse direction is the only place invertibility is used. Without it — in a general commutative ring where not every ideal has an inverse — the implication $\mathfrak{b}\mathfrak{c} \subseteq \mathfrak{a}\mathfrak{c} \Rightarrow \mathfrak{b} \subseteq \mathfrak{a}$ *can fail*. In $\mathcal{O}_L$ it always works because of Theorem 1586. [/guided] [/step] [step:Establish (2) — $\mathfrak{a} \mid \mathfrak{b}$ iff $\mathfrak{a}\mathfrak{c} \mid \mathfrak{b}\mathfrak{c}$, using invertibility of $\mathfrak{c}$] Recall: $\mathfrak{a} \mid \mathfrak{b}$ means there exists an integral ideal $\mathfrak{d} \unlhd \mathcal{O}_L$ with $\mathfrak{b} = \mathfrak{a}\mathfrak{d}$. **($\Rightarrow$)** If $\mathfrak{a} \mid \mathfrak{b}$, write $\mathfrak{b} = \mathfrak{a}\mathfrak{d}$ with $\mathfrak{d}$ integral. Multiplying both sides by $\mathfrak{c}$: \begin{align*} \mathfrak{b}\mathfrak{c} &= \mathfrak{a}\mathfrak{d}\mathfrak{c} = (\mathfrak{a}\mathfrak{c})\mathfrak{d}. \end{align*} Since $\mathfrak{d}$ is integral, this witnesses $\mathfrak{a}\mathfrak{c} \mid \mathfrak{b}\mathfrak{c}$. **($\Leftarrow$)** If $\mathfrak{a}\mathfrak{c} \mid \mathfrak{b}\mathfrak{c}$, write $\mathfrak{b}\mathfrak{c} = (\mathfrak{a}\mathfrak{c})\mathfrak{d}$ with $\mathfrak{d}$ integral. Multiplying both sides by $\mathfrak{c}^{-1}$ (which exists by [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586), since $\mathfrak{c}$ is a nonzero fractional ideal): \begin{align*} \mathfrak{b}\mathfrak{c}\mathfrak{c}^{-1} &= \mathfrak{a}\mathfrak{c}\mathfrak{d}\mathfrak{c}^{-1} = \mathfrak{a}\mathfrak{d} \cdot (\mathfrak{c}\mathfrak{c}^{-1}) = \mathfrak{a}\mathfrak{d}. \end{align*} The left side simplifies similarly to $\mathfrak{b} \cdot (\mathfrak{c}\mathfrak{c}^{-1}) = \mathfrak{b}$. Hence $\mathfrak{b} = \mathfrak{a}\mathfrak{d}$ with $\mathfrak{d}$ integral, so $\mathfrak{a} \mid \mathfrak{b}$. [guided] We now lift the cancellation argument from containments (part 1) to divisibility (part 2). The difference is that divisibility requires the "witness" $\mathfrak{d}$ to be an *integral* ideal, not just a fractional one. **Definition refresher.** For integral ideals $\mathfrak{a}, \mathfrak{b} \unlhd \mathcal{O}_L$, "$\mathfrak{a}$ divides $\mathfrak{b}$" — written $\mathfrak{a} \mid \mathfrak{b}$ — means there exists an *integral* ideal $\mathfrak{d} \unlhd \mathcal{O}_L$ with $\mathfrak{b} = \mathfrak{a}\mathfrak{d}$. **Forward direction.** Given $\mathfrak{a} \mid \mathfrak{b}$, write $\mathfrak{b} = \mathfrak{a}\mathfrak{d}$ with $\mathfrak{d} \unlhd \mathcal{O}_L$ integral. Multiply on the right by $\mathfrak{c}$: \begin{align*} \mathfrak{b} \mathfrak{c} &= (\mathfrak{a}\mathfrak{d}) \mathfrak{c}. \end{align*} By associativity and commutativity of ideal multiplication: \begin{align*} (\mathfrak{a}\mathfrak{d})\mathfrak{c} &= \mathfrak{a}(\mathfrak{d}\mathfrak{c}) = \mathfrak{a}(\mathfrak{c}\mathfrak{d}) = (\mathfrak{a}\mathfrak{c})\mathfrak{d}. \end{align*} So $\mathfrak{b}\mathfrak{c} = (\mathfrak{a}\mathfrak{c})\mathfrak{d}$, and since $\mathfrak{d}$ is integral, this exhibits $\mathfrak{a}\mathfrak{c} \mid \mathfrak{b}\mathfrak{c}$. **Reverse direction.** Given $\mathfrak{a}\mathfrak{c} \mid \mathfrak{b}\mathfrak{c}$, write $\mathfrak{b}\mathfrak{c} = (\mathfrak{a}\mathfrak{c})\mathfrak{d}$ with $\mathfrak{d} \unlhd \mathcal{O}_L$ integral. We cancel $\mathfrak{c}$ as in part 1. *Verifying $\mathfrak{c}$ is invertible.* $\mathfrak{c} \neq 0$ is a nonzero integral ideal. Nonzero integral ideals are fractional ideals (they are finitely generated $\mathcal{O}_L$-submodules of $L$, having $1 \in \mathcal{O}_L$ as a common denominator). So [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586) applies: there is a fractional $\mathfrak{c}^{-1}$ with $\mathfrak{c}\mathfrak{c}^{-1} = \mathcal{O}_L$. *Multiplying by $\mathfrak{c}^{-1}$.* From $\mathfrak{b}\mathfrak{c} = \mathfrak{a}\mathfrak{c}\mathfrak{d}$, multiply both sides by $\mathfrak{c}^{-1}$: \begin{align*} \mathfrak{b}\mathfrak{c}\mathfrak{c}^{-1} = \mathfrak{a}\mathfrak{c}\mathfrak{d}\mathfrak{c}^{-1}. \end{align*} *Simplifying the left side.* $\mathfrak{b}\mathfrak{c}\mathfrak{c}^{-1} = \mathfrak{b}(\mathfrak{c}\mathfrak{c}^{-1}) = \mathfrak{b}\mathcal{O}_L = \mathfrak{b}$. *Simplifying the right side.* By commutativity and associativity: \begin{align*} \mathfrak{a}\mathfrak{c}\mathfrak{d}\mathfrak{c}^{-1} &= \mathfrak{a}\mathfrak{d}(\mathfrak{c}\mathfrak{c}^{-1}) = \mathfrak{a}\mathfrak{d}\mathcal{O}_L = \mathfrak{a}\mathfrak{d}. \end{align*} *Conclusion.* $\mathfrak{b} = \mathfrak{a}\mathfrak{d}$ with $\mathfrak{d}$ integral (unchanged from our hypothesis), so $\mathfrak{a} \mid \mathfrak{b}$. **Note on integrality of $\mathfrak{d}$.** The crucial subtlety is that we do not *construct* $\mathfrak{d}$ in the reverse direction — we re-use the $\mathfrak{d}$ witnessing $\mathfrak{a}\mathfrak{c} \mid \mathfrak{b}\mathfrak{c}$. That $\mathfrak{d}$ is integral by hypothesis, so integrality is preserved. If we had to construct $\mathfrak{d}$ as $\mathfrak{d} = \mathfrak{a}^{-1}\mathfrak{b}$ (a fractional ideal in general!), we would need an extra argument to show it is integral — and that is exactly the content of part (3) below. [/guided] [/step] [step:Establish ($\Rightarrow$) of (3) — $\mathfrak{a} \mid \mathfrak{b}$ implies $\mathfrak{b} \subseteq \mathfrak{a}$] Suppose $\mathfrak{a} \mid \mathfrak{b}$; write $\mathfrak{b} = \mathfrak{a}\mathfrak{d}$ with $\mathfrak{d} \unlhd \mathcal{O}_L$ integral. Since $\mathfrak{d} \subseteq \mathcal{O}_L$, monotonicity of ideal multiplication (same argument as in Step 1) gives \begin{align*} \mathfrak{b} = \mathfrak{a}\mathfrak{d} \subseteq \mathfrak{a}\mathcal{O}_L = \mathfrak{a}. \end{align*} Hence $\mathfrak{b} \subseteq \mathfrak{a}$. [guided] The forward direction of (3) — "$\mathfrak{a}$ divides $\mathfrak{b}$ $\Rightarrow$ $\mathfrak{b} \subseteq \mathfrak{a}$" — is a direct monotonicity argument, using the fact that the quotient $\mathfrak{d}$ is contained in $\mathcal{O}_L$. **Expanding the hypothesis.** $\mathfrak{a} \mid \mathfrak{b}$ means $\mathfrak{b} = \mathfrak{a}\mathfrak{d}$ for some integral ideal $\mathfrak{d} \unlhd \mathcal{O}_L$. Integral means $\mathfrak{d} \subseteq \mathcal{O}_L$ (this is the integrality of $\mathfrak{d}$). **Monotonicity of ideal multiplication.** From $\mathfrak{d} \subseteq \mathcal{O}_L$, multiplying on the left by $\mathfrak{a}$: \begin{align*} \mathfrak{a}\mathfrak{d} \subseteq \mathfrak{a}\mathcal{O}_L. \end{align*} **Identity property of $\mathcal{O}_L$.** $\mathfrak{a}\mathcal{O}_L = \mathfrak{a}$ since $\mathcal{O}_L$ is the multiplicative identity of the fractional-ideal monoid. **Combining.** $\mathfrak{b} = \mathfrak{a}\mathfrak{d} \subseteq \mathfrak{a}\mathcal{O}_L = \mathfrak{a}$. Hence $\mathfrak{b} \subseteq \mathfrak{a}$. **Intuition.** Every element of $\mathfrak{a}\mathfrak{d}$ is a finite sum of products $a_i d_i$ with $a_i \in \mathfrak{a}$ and $d_i \in \mathfrak{d} \subseteq \mathcal{O}_L$. Since $a_i \in \mathfrak{a}$ (an ideal) and $d_i \in \mathcal{O}_L$, the product $a_i d_i \in \mathfrak{a}$ (as $\mathfrak{a}$ is closed under $\mathcal{O}_L$-multiplication). Summing, $\mathfrak{b} = \mathfrak{a}\mathfrak{d} \subseteq \mathfrak{a}$. [/guided] [/step] [step:Establish ($\Leftarrow$) of (3) — construct $\mathfrak{c} = \mathfrak{a}^{-1}\mathfrak{b}$ and verify it is an integral ideal witnessing $\mathfrak{a} \mid \mathfrak{b}$] Suppose $\mathfrak{b} \subseteq \mathfrak{a}$ as integral ideals. If $\mathfrak{b} = 0$, then $\mathfrak{b} = \mathfrak{a} \cdot 0$ (where $0$ is the zero ideal), so $\mathfrak{a} \mid \mathfrak{b}$ with witness the zero ideal. Assume $\mathfrak{b} \neq 0$; we also have $\mathfrak{a} \neq 0$ since otherwise $\mathfrak{a} = 0 \supseteq \mathfrak{b}$ forces $\mathfrak{b} = 0$. Since $\mathfrak{a}$ is a nonzero fractional ideal, [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586) provides a fractional ideal $\mathfrak{a}^{-1}$ with $\mathfrak{a}\mathfrak{a}^{-1} = \mathcal{O}_L$. Define \begin{align*} \mathfrak{c} &:= \mathfrak{a}^{-1} \mathfrak{b}, \end{align*} a product of fractional ideals — hence a fractional ideal of $\mathcal{O}_L$. **Claim:** $\mathfrak{c}$ is integral, i.e., $\mathfrak{c} \subseteq \mathcal{O}_L$. From $\mathfrak{b} \subseteq \mathfrak{a}$, multiplying both sides on the left by $\mathfrak{a}^{-1}$ (using monotonicity of ideal multiplication for fractional ideals, exactly as in Step 1): \begin{align*} \mathfrak{a}^{-1}\mathfrak{b} \subseteq \mathfrak{a}^{-1}\mathfrak{a} = \mathcal{O}_L. \end{align*} Hence $\mathfrak{c} = \mathfrak{a}^{-1}\mathfrak{b} \subseteq \mathcal{O}_L$, so $\mathfrak{c}$ is an integral ideal. **$\mathfrak{b} = \mathfrak{a}\mathfrak{c}$:** multiply both sides of the definition $\mathfrak{c} = \mathfrak{a}^{-1}\mathfrak{b}$ by $\mathfrak{a}$: \begin{align*} \mathfrak{a}\mathfrak{c} &= \mathfrak{a} \cdot \mathfrak{a}^{-1}\mathfrak{b} = (\mathfrak{a}\mathfrak{a}^{-1})\mathfrak{b} = \mathcal{O}_L \mathfrak{b} = \mathfrak{b}. \end{align*} Hence $\mathfrak{a} \mid \mathfrak{b}$ with witness $\mathfrak{c}$. This completes the reverse direction and thereby part (3). Combined with Steps 1 and 2, all three parts of the theorem are established. [guided] The reverse direction of (3) — "$\mathfrak{b} \subseteq \mathfrak{a}$ $\Rightarrow$ $\mathfrak{a} \mid \mathfrak{b}$" — is the content-rich direction. We must construct an integral ideal $\mathfrak{c}$ with $\mathfrak{b} = \mathfrak{a}\mathfrak{c}$. **The natural candidate.** If such $\mathfrak{c}$ exists, multiplying $\mathfrak{b} = \mathfrak{a}\mathfrak{c}$ by $\mathfrak{a}^{-1}$ gives $\mathfrak{a}^{-1}\mathfrak{b} = \mathfrak{c}$. So $\mathfrak{c}$ is forced to be $\mathfrak{a}^{-1}\mathfrak{b}$. We just need to verify this actually is an integral ideal. **Degenerate case: $\mathfrak{b} = 0$.** Then $\mathfrak{b} = \mathfrak{a} \cdot 0$ with witness $0$ (the zero ideal, which is integral). So $\mathfrak{a} \mid \mathfrak{b}$ with the zero ideal as the required integral witness. (In some conventions the zero ideal is excluded from the divisibility relation; the slogan "divisibility equals containment" still holds vacuously since $\mathfrak{b} \subseteq \mathfrak{a}$ is automatic when $\mathfrak{b} = 0$.) **Main case: $\mathfrak{b} \neq 0$ and $\mathfrak{a} \neq 0$.** (The assumption $\mathfrak{a} \neq 0$ comes for free: if $\mathfrak{a} = 0$ then $\mathfrak{b} \subseteq \mathfrak{a} = 0$ gives $\mathfrak{b} = 0$, which is the degenerate case above.) *Invertibility of $\mathfrak{a}$.* $\mathfrak{a}$ is a nonzero integral ideal, hence a nonzero fractional ideal. By [Every Nonzero Fractional Ideal Is Invertible](/theorems/1586), there exists a fractional ideal $\mathfrak{a}^{-1}$ with $\mathfrak{a}\mathfrak{a}^{-1} = \mathcal{O}_L$. *Defining $\mathfrak{c}$.* Set \begin{align*} \mathfrak{c} := \mathfrak{a}^{-1}\mathfrak{b}. \end{align*} The product of two fractional ideals is a fractional ideal (the fractional-ideal monoid is closed under multiplication). So $\mathfrak{c}$ is a fractional ideal. *Integrality of $\mathfrak{c}$: $\mathfrak{c} \subseteq \mathcal{O}_L$.* Start from the hypothesis $\mathfrak{b} \subseteq \mathfrak{a}$. Multiply both sides on the left by $\mathfrak{a}^{-1}$. By monotonicity of ideal multiplication for fractional ideals — the same monotonicity as in Step 1, now applied in the fractional-ideal monoid rather than the integral-ideal monoid — containment is preserved: \begin{align*} \mathfrak{a}^{-1}\mathfrak{b} \subseteq \mathfrak{a}^{-1}\mathfrak{a}. \end{align*} But $\mathfrak{a}^{-1}\mathfrak{a} = \mathcal{O}_L$ (by definition of $\mathfrak{a}^{-1}$). Hence \begin{align*} \mathfrak{c} = \mathfrak{a}^{-1}\mathfrak{b} \subseteq \mathcal{O}_L, \end{align*} so $\mathfrak{c}$ is an integral ideal. *Verifying $\mathfrak{b} = \mathfrak{a}\mathfrak{c}$.* Multiply the definition $\mathfrak{c} = \mathfrak{a}^{-1}\mathfrak{b}$ on the left by $\mathfrak{a}$: \begin{align*} \mathfrak{a}\mathfrak{c} &= \mathfrak{a}(\mathfrak{a}^{-1}\mathfrak{b}) = (\mathfrak{a}\mathfrak{a}^{-1})\mathfrak{b} = \mathcal{O}_L \mathfrak{b} = \mathfrak{b}, \end{align*} using associativity, $\mathfrak{a}\mathfrak{a}^{-1} = \mathcal{O}_L$, and the identity property. *Conclusion.* $\mathfrak{b} = \mathfrak{a}\mathfrak{c}$ with $\mathfrak{c}$ integral, so $\mathfrak{a} \mid \mathfrak{b}$. **Structural remark — the slogan "to divide is to contain".** In general commutative rings, divisibility (as a constructive existence-of-witness statement) and containment (as a set-theoretic statement) can differ. For example, in $\mathbb{Z}[\sqrt{-5}]$ — a Dedekind domain but not a UFD — the ideal $\langle 2, 1 + \sqrt{-5} \rangle$ contains the element $2$ but does not divide the principal ideal $\langle 2 \rangle$ at the level of elements. However, at the level of *ideals*, divisibility and containment coincide exactly because of this theorem: the invertibility of ideals in a Dedekind domain lets us always construct the quotient $\mathfrak{a}^{-1}\mathfrak{b}$ and verify it is integral. This is the structural content that elevates the Dedekind-domain theory beyond elementary ring-theoretic divisibility. **Closing the whole theorem.** Combining: - Step 1 gave $\mathfrak{b} \subseteq \mathfrak{a} \iff \mathfrak{b}\mathfrak{c} \subseteq \mathfrak{a}\mathfrak{c}$ (part 1). - Step 2 gave $\mathfrak{a} \mid \mathfrak{b} \iff \mathfrak{a}\mathfrak{c} \mid \mathfrak{b}\mathfrak{c}$ (part 2). - Step 3 gave $\mathfrak{a} \mid \mathfrak{b} \Rightarrow \mathfrak{b} \subseteq \mathfrak{a}$; Step 4 gave the converse (part 3). All three parts of the theorem are established. [/guided] [/step]