The Stacks project

Proof. Special case of Homology, Lemma 12.27.3. $\square$

Proof. This is true because $\mathop{\mathrm{Hom}}\nolimits _ R(M, E) = \mathop{\mathrm{Hom}}\nolimits _ S(M \otimes _ R S, E)$ by Algebra, Lemma 10.14.3 and the fact that tensoring with $S$ is exact. $\square$

Proof. This is true because $\mathop{\mathrm{Hom}}\nolimits _ R(N, E) = \mathop{\mathrm{Hom}}\nolimits _ S(N, E)$ for any $S$-module $N$, see Algebra, Lemma 10.107.14. $\square$

Proof. This is true because $\mathop{\mathrm{Hom}}\nolimits _ S(N, \mathop{\mathrm{Hom}}\nolimits _ R(S, E)) = \mathop{\mathrm{Hom}}\nolimits _ R(N, E)$ by Algebra, Lemma 10.14.4. $\square$

Proof. The final assertion follows from the first and the fact that the category of $R$-modules has enough injectives (More on Algebra, Section 15.55).

Assume (1). Let $E \subset E' \subset I$ as in (2). Then the map $\text{id}_ E : E \to E$ can be extended to a map $\alpha : E' \to E$. The kernel of $\alpha$ has to be zero because it intersects $E$ trivially and $E'$ is an essential extension. Hence $E = E'$.

Assume (2). Let $M \subset N$ be $R$-modules and let $\varphi : M \to E$ be an $R$-module map. In order to prove (1) we have to show that $\varphi$ extends to a morphism $N \to E$. Consider the set $\mathcal{S}$ of pairs $(M', \varphi ')$ where $M \subset M' \subset N$ and $\varphi ' : M' \to E$ is an $R$-module map agreeing with $\varphi$ on $M$. We define an ordering on $\mathcal{S}$ by the rule $(M', \varphi ') \leq (M'', \varphi '')$ if and only if $M' \subset M''$ and $\varphi ''|_{M'} = \varphi '$. It is clear that we can take the maximum of a totally ordered subset of $\mathcal{S}$. Hence by Zorn's lemma we may assume $(M, \varphi )$ is a maximal element.

Choose an extension $\psi : N \to I$ of $\varphi$ composed with the inclusion $E \to I$. This is possible as $I$ is injective. If $\psi (N) \subset E$, then $\psi$ is the desired extension. If $\psi (N)$ is not contained in $E$, then by (2) the inclusion $E \subset E + \psi (N)$ is not essential. hence we can find a nonzero submodule $K \subset E + \psi (N)$ meeting $E$ in $0$. This means that $M' = \psi ^{-1}(E + K)$ strictly contains $M$. Thus we can extend $\varphi$ to $M'$ using

This contradicts the maximality of $(M, \varphi )$. $\square$

Proof. Let $E_ i$ be a family of injective modules parametrized by a set $I$. Set $E = \bigoplus E_ i$. To show that $E$ is injective we use Injectives, Lemma 19.2.6. Thus let $\varphi : I \to E$ be a module map from an ideal of $R$ into $E$. As $I$ is a finite $R$-module (because $R$ is Noetherian) we can find finitely many elements $i_1, \ldots , i_ r \in I$ such that $\varphi$ maps into $\bigoplus _{j = 1, \ldots , r} E_{i_ j}$. Then we can extend $\varphi$ into $\bigoplus _{j = 1, \ldots , r} E_{i_ j}$ using the injectivity of the modules $E_{i_ j}$. $\square$

Proof. Since $R \to S^{-1}R$ is an epimorphism of rings, it suffices to show that $S^{-1}E$ is injective as an $R$-module, see Lemma 47.3.3. To show this we use Injectives, Lemma 19.2.6. Thus let $I \subset R$ be an ideal and let $\varphi : I \to S^{-1} E$ be an $R$-module map. As $I$ is a finitely presented $R$-module (because $R$ is Noetherian) we can find an $f \in S$ and an $R$-module map $I \to E$ such that $f\varphi$ is the composition $I \to E \to S^{-1}E$ (Algebra, Lemma 10.10.2). Then we can extend $I \to E$ to a homomorphism $R \to E$. Then the composition

is the desired extension of $\varphi$ to $R$. $\square$

Proof. We will use Lemma 47.3.5 to prove (1). Suppose that $E \subset E' \subset I$ and that $E'$ is an essential extension of $E$. We will show that $E' = E$. If not, then we can find $x \in E'$ and $x \not\in E$. Let $J = \{ a \in R \mid ax \in E\}$. Since $R$ is Noetherian, we may write $J = (g_1, \ldots , g_ t)$ for some $g_ i \in R$. By definition $E$ is the set of elements of $I$ annihilated by powers of $f$, so we may choose integers $n_ i$ so that $f^{n_ i}g_ ix = 0$. Set $n = \mathrm{max}\{ n_ i \}$. Then $x' = f^ n x$ is an element of $E'$ not in $E$ and is annihilated by $J$. Set $J' = \{ a \in R \mid ax' \in E \}$ so $J \subset J'$. Conversely, we have $a \in J'$ if and only if $ax' \in E$ if and only if $f^ m a x' = 0$ for some $m \geq 0$. But then $f^ m a x' = f^{m + n} a x$ implies $ax \in E$, i.e., $a \in J$. Hence $J = J'$. Thus $J = J' = \text{Ann}(x')$, so $Rx' \cap E = 0$. Hence $E'$ is not an essential extension of $E$, a contradiction.

To prove (2) write $J = (f_1, \ldots , f_ t)$. Then $I[J^\infty ]$ is equal to

and the result follows from (1) and induction. $\square$

Proof. Let us write $E[x] = E \otimes _ A A[x]$. Consider the short exact sequence of $A[x]$-modules

where the first map sends $p \in E[x]$ to $f \mapsto fp$ and the second map sends $\varphi$ to $f \mapsto \varphi (xf) - x\varphi (f)$. The second map is surjective because $\mathop{\mathrm{Hom}}\nolimits _ A(A[x], E[x]) = \prod _{n \geq 0} E[x]$ as an abelian group and the map sends $(e_ n)$ to $(e_{n + 1} - xe_ n)$ which is surjective. As an $A$-module we have $E[x] \cong \bigoplus _{n \geq 0} E$ which is injective by Lemma 47.3.7. Hence the $A[x]$-module $\mathop{\mathrm{Hom}}\nolimits _ A(A[x], E[x])$ is injective by Lemma 47.3.4 and the proof is complete. $\square$