Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there an example which shows that the functor $B\otimes_R(-)$ is not left-exact, given a ring $R$ and a right $R$-module $B$?

share|cite|improve this question

Consider the one-to-one function of $\mathbb{Z}$-modules $$\begin{align*} f\colon\mathbb{Z}&\longrightarrow\mathbb{Z}\\ a&\longmapsto 2a \end{align*}$$

If we tensor with $\mathbb{Z}/2\mathbb{Z}$, the map $f$ induces a map $$\mathbb{Z}/2\mathbb{Z}\cong \mathbb{Z}/2\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}\ \longrightarrow\ \mathbb{Z}/2\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}\cong\mathbb{Z}/2\mathbb{Z}$$ that sends $\overline{1}\otimes 1$ to $$\overline{1}\otimes f(1) = \overline{1}\otimes 2 = \overline{1}\otimes (1+1) = \overline{1}\otimes 1 + \overline{1}\otimes 1= \mathbf{0}.$$ So the map induced by the one-to-one map $f$ is not one-to-one.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.