# product of injective modules is injective

Proposition^{}. Let $R$ be a ring and ${\{{Q}_{i}\}}_{i\in I}$ a family of injective^{} $R$-modules. Then the product^{}

$$Q=\prod _{i\in I}{Q}_{i}$$ |

is injective.

Proof. Let $B$ be an arbitrary $R$-module, $A\subseteq B$ a submodule^{} and $f:A\to Q$ a homomorphism^{}. It is enough to show that $f$ can be extended to $B$. For $i\in I$ denote by ${\pi}_{i}:Q\to {Q}_{i}$ the projection. Since ${Q}_{i}$ is injective for any $i$, then the homomorphism ${\pi}_{i}\circ f:A\to {Q}_{i}$ can be extended to ${f}_{i}^{\prime}:B\to {Q}_{i}$. Then we have

$${f}^{\prime}:B\to Q;$$ |

$${f}^{\prime}(b)={\left({f}_{i}^{\prime}(b)\right)}_{i\in I}.$$ |

It is easy to check, that if $a\in A$, then ${f}^{\prime}(a)=f(a)$, so ${f}^{\prime}$ is an extension^{} of $f$. Thus $Q$ is injective. $\mathrm{\square}$

Remark. Unfortunetly direct sum^{} of injective modules^{} need not be injective. Indeed, there is a theorem which states that direct sums of injective modules are injective if and only if ring $R$ is Noetherian^{}. Note that the proof presented above cannot be used for direct sums, because ${f}^{\prime}(b)$ need not be an element of the direct sum, more precisely, it is possible that ${f}_{i}^{\prime}(b)\ne 0$ for infinetly many $i\in I$. Nevertheless products are always injective.

Title | product of injective modules is injective |
---|---|

Canonical name | ProductOfInjectiveModulesIsInjective |

Date of creation | 2013-03-22 18:50:17 |

Last modified on | 2013-03-22 18:50:17 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 5 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 16D50 |