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When I studied injective module there is a theorem which say that the two following statement are equivalent:

Let $R$ be a ring, $I$ is a left ideal of $R$, $J$ is a left $R$-module.

  1. For every $f:I\longrightarrow J$ module homomorphism there exist $m\in J$ that $f(r)=rm$ for every $r\in I$

  2. Every module homomorphism $f:I \longrightarrow J$ can be extended to a homomorphism $\bar{f}:R\longrightarrow J$.

Suppose we have 1 and we want to prove 2. Here is my argument:

Let $\pi : R\longrightarrow I$ with $\pi(r)=r$ where $r\in I$ and $\pi(r)=0$ otherwise. Then define $\bar{f}:R\longrightarrow J$ as $\bar{f}=f\circ \pi$.

I felt that my argument is not enough or even wrong, however, I had no idea of constructing such a homomorphism $\bar{f}$.

Please help me. Thanks

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In statement 1., do you mean that for every module homomorphism $f: I \to J$, there exists an $m \in J$ such that $f(r) = rm$ for all $r \in I$? –  Martin Wanvik Apr 14 '12 at 15:02
@MartinWanvik : Yes, thank you very much. I have edited it. –  Arsenaler Apr 14 '12 at 15:07

1 Answer 1

up vote 1 down vote accepted

I don't see how your argument shows anything close to what you're trying to do (you appear to be trying to show the converse). To show that statement 2 implies statement 1, you would have to show that given some $f: I \to J$, you can write it as $f(r) = rm$ for some $m \in J$ under the assumption that statement 2 is true. The proof is actually quite easy: given $f: I \to J$, we can extend that to $\bar{f}: R \to J$ using statement 2. Now let $m = \bar{f}(1)$. Then $$f(r) = \bar{f}(r) = \bar{f}(r\cdot 1) = r\bar{f}(1) = r m,$$ which proves the result.

Edit: To show the converse, we can essentially reverse the above proof: given a module homomorphism $f: I \to J$, we can write it as $f(r) = rm$ for some $m \in J$ (by statement 1). We can extend $f$ to $R$ by defining $$ \bar{f}(r) := rm.$$ This coincides with $f$ on $I$, and it is easily seen to be an $R$-module homomorphism, since $$\bar{f}(rr') = rr'm = r(r'm) = r\bar{f}(r'),$$ and $$f(r+r') = (r+r')m = rm + r'm = \bar{f}(r) + \bar{f}(r'),$$ for all $r,r' \in R$.

Second edit: To see why your proof doesn't work, consider $R = \mathbb{Z}$, and $I = 2\mathbb{Z}$. Defining $\pi(z) = z$ if $z \in 2\mathbb{Z}$ and $\pi(z) = 0$ if $z \notin 2\mathbb{Z}$, then we must have $$2 = \pi(2) = \pi(2\cdot 1) = 2 \cdot \pi(1) = 0,$$ so $\pi$ is not a $\mathbb{Z}$-homomorphism. It cannot even be additive, since that would imply that $$2 = \pi(2) = \pi(1+1) = \pi(1) + \pi(1) = 0.$$

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Sorry. I mean if we have 1 and we want to prove 2. –  Arsenaler Apr 14 '12 at 15:17
@msnaber: I've edited my answer. –  Martin Wanvik Apr 14 '12 at 15:58
Thank you very much Martin Wanvik. –  Arsenaler Apr 14 '12 at 16:11

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