[proofplan] The forward implication is the defining extension property of injective modules applied to the inclusion $J \hookrightarrow R$. For the converse, we prove that any $R$-[linear map](/page/Linear%20Map) from a submodule $A \subset B$ into $I$ extends to all of $B$. We use [Zorn's lemma](/theorems/1226) to choose a maximal partial extension, and if its domain is not all of $B$, we enlarge it by one cyclic submodule. The hypothesis on left ideals supplies exactly the compatibility needed to define the enlarged map. [/proofplan]

[step:Use injectivity to extend maps from left ideals to the regular module] Assume first that $I$ is an injective left $R$-module. Let $J \subset R$ be a left ideal, and let \begin{align*} f: J &\to I \end{align*} be an $R$-linear map. Let \begin{align*} \iota: J &\to R \\ x &\mapsto x \end{align*} be the inclusion of left $R$-modules. Since $\iota$ is injective and $I$ is injective, there exists an $R$-linear map \begin{align*} F: R &\to I \end{align*} such that $F \circ \iota = f$. Hence $F|_J = f$, proving the stated extension property. [/step]

[step:Set up the maximal partial extension problem]Conversely, assume that every $R$-linear map from a left ideal of $R$ to $I$ extends to an $R$-linear map from $R$ to $I$. To prove that $I$ is injective, let $A$ and $B$ be left $R$-modules, let \begin{align*} \alpha: A &\to B \end{align*} be an injective $R$-linear map, and let \begin{align*} f: A &\to I \end{align*} be an $R$-linear map. Replacing $A$ by the submodule $\alpha(A) \subset B$ and replacing $f$ by the map $\alpha(a) \mapsto f(a)$, which is well-defined because $\alpha$ is injective, it suffices to treat the case where $A \subset B$ and $\alpha$ is the inclusion. Define $\mathcal{P}$ to be the set of pairs $(C,g)$ such that $C$ is a left $R$-submodule of $B$ with $A \subset C$, and \begin{align*} g: C &\to I \end{align*} is an $R$-linear map satisfying $g|_A = f$. Partially order $\mathcal{P}$ by declaring \begin{align*} (C,g) \leq (C',g') \end{align*} if and only if $C \subset C'$ and $g'|_C = g$. The pair $(A,f)$ belongs to $\mathcal{P}$, so $\mathcal{P}$ is nonempty.[/step]

[guided]We want to prove injectivity of $I$, so we begin with the defining extension problem: a map $f: A \to I$ is given on a submodule of a larger module $B$, and we must extend it to all of $B$. The injective map $\alpha: A \to B$ need not literally be an inclusion. Since $\alpha$ is injective, it identifies $A$ with the submodule $\alpha(A) \subset B$. The map on this submodule is \begin{align*} \alpha(A) &\to I \\ \alpha(a) &\mapsto f(a). \end{align*} This is well-defined because if $\alpha(a)=\alpha(a')$, then injectivity gives $a=a'$, and hence $f(a)=f(a')$. Therefore it is enough to prove the extension property when $A$ is already a submodule of $B$. Now define $\mathcal{P}$ to be the collection of all partial extensions of $f$. An element of $\mathcal{P}$ is a pair $(C,g)$ where $C$ is a left $R$-submodule satisfying $A \subset C \subset B$, and \begin{align*} g: C &\to I \end{align*} is $R$-linear with $g|_A=f$. We order these pairs by extension: \begin{align*} (C,g) \leq (C',g') \end{align*} means $C \subset C'$ and $g'$ agrees with $g$ on $C$. The initial pair $(A,f)$ is in $\mathcal{P}$, so the poset is nonempty.[/guided]

[step:Use Zorn's lemma to obtain a maximal partial extension]Let $\mathcal{C}$ be a chain in $\mathcal{P}$. Define \begin{align*} C_* &:= \bigcup_{(C,g)\in \mathcal{C}} C. \end{align*} Since $\mathcal{C}$ is a chain under inclusion of domains, $C_*$ is a left $R$-submodule of $B$ containing $A$. Define \begin{align*} g_*: C_* &\to I \end{align*} as follows: for $x \in C_*$, choose $(C,g) \in \mathcal{C}$ with $x \in C$, and set $g_*(x):=g(x)$. This definition is independent of the choice of $(C,g)$: if $x \in C_1 \cap C_2$ for $(C_1,g_1),(C_2,g_2)\in\mathcal{C}$, then the chain condition gives either $(C_1,g_1)\leq (C_2,g_2)$ or $(C_2,g_2)\leq (C_1,g_1)$, and in both cases $g_1(x)=g_2(x)$. The map $g_*$ is $R$-linear. Indeed, if $x,y\in C_*$ and $r\in R$, choose chain elements containing $x$ and $y$; one of them contains both because $\mathcal{C}$ is linearly ordered. On that common domain, additivity and $R$-linearity follow from the corresponding partial extension. Also $g_*|_A=f$. Hence $(C_*,g_*)$ is an upper bound for $\mathcal{C}$ in $\mathcal{P}$. By Zorn's lemma (citing a result not yet in the wiki: Zorn's Lemma), there exists a maximal element $(M,h)$ of $\mathcal{P}$.[/step]

[guided]To use maximality, we need a maximal partial extension. We apply Zorn's lemma to the poset $\mathcal{P}$. Zorn's lemma requires that every chain have an upper bound in the same poset. Let $\mathcal{C}$ be a chain in $\mathcal{P}$. Its candidate upper domain is the union \begin{align*} C_* &:= \bigcup_{(C,g)\in \mathcal{C}} C. \end{align*} This is a left $R$-submodule of $B$: if $x,y\in C_*$, then $x\in C_1$ and $y\in C_2$ for some chain elements. Since $\mathcal{C}$ is linearly ordered, either $C_1\subset C_2$ or $C_2\subset C_1$, so both $x$ and $y$ lie in one common submodule. Therefore $x+y$ lies in that submodule, and for every $r\in R$, the element $rx$ lies in a chain submodule containing $x$. Also $A\subset C_*$ because every element of $\mathcal{P}$ contains $A$. Now define the union map \begin{align*} g_*: C_* &\to I. \end{align*} For $x\in C_*$, choose a pair $(C,g)\in\mathcal{C}$ with $x\in C$, and put $g_*(x)=g(x)$. The point needing verification is well-definedness. If $x$ also lies in another domain $C'$, the chain condition says that either $(C,g)\leq(C',g')$ or $(C',g')\leq(C,g)$. In either case, the larger map restricts to the smaller one, so both maps assign the same value to $x$. The same common-domain argument proves $R$-linearity. For $x,y\in C_*$ and $r\in R$, choose one chain element whose domain contains both $x$ and $y$. On that domain, the relevant partial extension is $R$-linear, so \begin{align*} g_*(x+y) &= g_*(x)+g_*(y), \\ g_*(rx) &= r g_*(x). \end{align*} Since each partial extension restricts to $f$ on $A$, the union map also satisfies $g_*|_A=f$. Thus $(C_*,g_*)$ is an upper bound for the chain. Zorn's lemma therefore gives a maximal partial extension $(M,h)$.[/guided]

[step:Construct the left ideal measuring whether a new element can be added]We claim that $M=B$. Suppose, toward a contradiction, that $M\neq B$. Choose an element $b\in B\setminus M$. Define \begin{align*} J_b &:= \{r\in R : rb\in M\}. \end{align*} Then $J_b$ is a left ideal of $R$: if $r,s\in J_b$ and $a\in R$, then $(r+s)b=rb+sb\in M$ and $(ar)b=a(rb)\in M$, since $M$ is a left $R$-submodule of $B$. Define \begin{align*} \varphi_b: J_b &\to I \\ r &\mapsto h(rb). \end{align*} This map is $R$-linear because $h$ is $R$-linear on $M$ and $rb\in M$ for all $r\in J_b$. By the assumed extension property for left ideals, there exists an $R$-linear map \begin{align*} \Phi_b: R &\to I \end{align*} such that $\Phi_b|_{J_b}=\varphi_b$.[/step]

[guided]Assume the maximal domain $M$ is not all of $B$. Choose an element $b\in B\setminus M$. We want to enlarge $M$ by adding all multiples of $b$, namely by passing from $M$ to $M+Rb$. The obstruction to defining an extension on $M+Rb$ is that an element of $M+Rb$ may have more than one representation as $m+rb$. The elements of $R$ that create this ambiguity form the set \begin{align*} J_b &:= \{r\in R : rb\in M\}. \end{align*} This is a left ideal. If $r,s\in J_b$, then $rb$ and $sb$ lie in $M$, so \begin{align*} (r+s)b=rb+sb\in M. \end{align*} If $a\in R$ and $r\in J_b$, then $rb\in M$, and because $M$ is a left $R$-submodule, \begin{align*} (ar)b=a(rb)\in M. \end{align*} Thus $ar\in J_b$. On this left ideal, define \begin{align*} \varphi_b: J_b &\to I \\ r &\mapsto h(rb). \end{align*} This is well-defined because $rb\in M$ for $r\in J_b$, so $h(rb)$ is defined. It is $R$-linear: for $r,s\in J_b$ and $a\in R$, \begin{align*} \varphi_b(r+s) &= h((r+s)b)=h(rb+sb)=h(rb)+h(sb)=\varphi_b(r)+\varphi_b(s), \\ \varphi_b(ar) &= h((ar)b)=h(a(rb))=a h(rb)=a\varphi_b(r). \end{align*} The hypothesis of Baer's criterion applies to this left ideal $J_b$ and this map $\varphi_b$. Therefore there exists an $R$-linear map \begin{align*} \Phi_b: R &\to I \end{align*} such that $\Phi_b|_{J_b}=\varphi_b$.[/guided]

[step:Extend the maximal map across the cyclic enlargement]Define $M' := M+Rb$, a left $R$-submodule of $B$ properly containing $M$ because $b\in M'$ and $b\notin M$. Define \begin{align*} h': M' &\to I \end{align*} by the rule \begin{align*} h'(m+rb) &:= h(m)+\Phi_b(r), \end{align*} where $m\in M$ and $r\in R$. This definition is independent of the representation of an element of $M'$. Indeed, suppose \begin{align*} m+rb &= m'+r'b \end{align*} with $m,m'\in M$ and $r,r'\in R$. Then \begin{align*} (r-r')b &= m'-m\in M, \end{align*} so $r-r'\in J_b$. Since $\Phi_b|_{J_b}=\varphi_b$, \begin{align*} \Phi_b(r-r')=\varphi_b(r-r')=h((r-r')b)=h(m'-m). \end{align*} Using additivity of $h$ and $\Phi_b$, we get \begin{align*} h(m)+\Phi_b(r)-h(m')-\Phi_b(r') &= h(m-m')+\Phi_b(r-r') \\ &= -h(m'-m)+h(m'-m) \\ &=0. \end{align*} Thus $h(m)+\Phi_b(r)=h(m')+\Phi_b(r')$. The map $h'$ is $R$-linear by direct computation from the displayed formula and the $R$-linearity of $h$ and $\Phi_b$. For $m\in M$, we have \begin{align*} h'(m+0b)=h(m)+\Phi_b(0)=h(m), \end{align*} so $h'|_M=h$. Hence $(M',h')\in\mathcal{P}$ and $(M,h)<(M',h')$, contradicting the maximality of $(M,h)$.[/step]

[guided]We now build the forbidden larger partial extension. Let \begin{align*} M' := M+Rb. \end{align*} This is a left $R$-submodule of $B$, it contains $M$, and it properly contains $M$ because $b=0+b\in M'$ while $b\notin M$. The natural formula for the enlarged map is \begin{align*} h'(m+rb) &:= h(m)+\Phi_b(r), \end{align*} where $m\in M$ and $r\in R$. The only serious issue is well-definedness. Suppose the same element of $M'$ has two representations: \begin{align*} m+rb &= m'+r'b. \end{align*} Then subtracting gives \begin{align*} (r-r')b &= m'-m. \end{align*} The right-hand side lies in $M$, so $r-r'\in J_b$ by the definition of $J_b$. On $J_b$, the map $\Phi_b$ agrees with $\varphi_b$, and $\varphi_b$ was defined by applying $h$ to the corresponding multiple of $b$. Hence \begin{align*} \Phi_b(r-r')=\varphi_b(r-r')=h((r-r')b)=h(m'-m). \end{align*} Now compute: \begin{align*} h(m)+\Phi_b(r)-h(m')-\Phi_b(r') &= h(m-m')+\Phi_b(r-r') \\ &= -h(m'-m)+h(m'-m) \\ &=0. \end{align*} So the value of $h'(m+rb)$ is independent of the chosen representation. The map $h'$ is $R$-linear. For additivity, take $m_1,m_2\in M$ and $r_1,r_2\in R$: \begin{align*} h'((m_1+r_1b)+(m_2+r_2b)) &= h'((m_1+m_2)+(r_1+r_2)b) \\ &= h(m_1+m_2)+\Phi_b(r_1+r_2) \\ &= h(m_1)+h(m_2)+\Phi_b(r_1)+\Phi_b(r_2) \\ &= h'(m_1+r_1b)+h'(m_2+r_2b). \end{align*} For scalar multiplication, take $a\in R$: \begin{align*} h'(a(m+rb)) &= h'(am+(ar)b) \\ &= h(am)+\Phi_b(ar) \\ &= a h(m)+a\Phi_b(r) \\ &= a h'(m+rb). \end{align*} Finally, if $m\in M$, then \begin{align*} h'(m)=h'(m+0b)=h(m)+\Phi_b(0)=h(m). \end{align*} Thus $h'$ extends $h$ to the strictly larger submodule $M'$, contradicting the maximality of $(M,h)$.[/guided]

[step:Conclude that every partial extension reaches the whole module] The contradiction shows that the maximal domain $M$ cannot be a proper submodule of $B$. Therefore $M=B$. The maximal map \begin{align*} h: B &\to I \end{align*} is an $R$-linear extension of $f: A\to I$. Thus for every inclusion $A\subset B$ of left $R$-modules and every $R$-linear map $f:A\to I$, there exists an $R$-linear extension $B\to I$. Hence $I$ is an injective left $R$-module. [/step]