Show that $1\otimes i $ is injective.

Question

Let $I$ be an ideal of $\mathbb{Z}$ and $i: I\rightarrow \mathbb{Z}$ be the canonical injection.

To show that $\mathbb{Q } $ is a $\mathbb{Z}$-flat module, consider the $\mathbb{Z}$-morphism

$$ 1\otimes i : \mathbb{Q}\otimes_{\mathbb{Z}}I\rightarrow \mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Z}. $$

We know that $$ \mathbb{Q}\otimes_{\mathbb{Z}}I \cong_{\mathbb{Z}} \mathbb{Q} $$ $$ \mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Z}\cong_{\mathbb{Z}}\mathbb{Q} $$

But how from this result can I show that $1\otimes i $ is injective?

Answer 1

Hint : You have isomorphisms $\mathbb Q\otimes I \cong \mathbb Q$ and $\mathbb Q\otimes \mathbb Z \cong \mathbb Q$ : what does the map $1\otimes i$ look like as a map $\mathbb{Q\to Q}$ ?

That is, what is the map $\mathbb{Q\to Q}$ that makes the following diagram commute :

$\require{AMScd}\begin{CD} \mathbb Q\otimes I @>1\otimes i>> \mathbb{Q\otimes Z }\\ @VVV @VVV \\ \mathbb Q @>?>> \mathbb Q\end{CD}$

where the vertical maps are your isomorphisms ? For that you'll have to explicit the isomorphisms.

Answer 2

In $\mathbb{Z}$, any Ideal is of the form $(n)$ for some $n\in \mathbb{Z}$, as $\mathbb{Z}$ is a PID. so we are looking at: $1\otimes i : \mathbb{Q}\otimes_{\mathbb{Z}}(n)\rightarrow \mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Z}$.

As you have correctly pointed out $\mathbb{Q}\otimes_{\mathbb{Z}}(n)$ is isomorphic to $\mathbb{Q}$. You can show this by first starting with any $\mathbb{Z}$ -bilinear map $\mathbb{Q} \times (n)$ which maps into $\mathbb{Q}$, say $(\frac{r}{s},mn) \to \frac{mnr}{s}$. This is linear in both variables, so induces a well defined map $\mathbb{Q} \otimes (n) \to \mathbb{Q}$, such that $(\frac{r}{s} \otimes mn) \to \frac{rmn}{s}$. Call this map $\psi$. This is surjective. The only way that something gets mapped to $0$ is if either $\frac{r}{s} =0$ or $mn=0$, in other words, only the zero element gets mapped to $0$. So, this is indeed an isomorphism. The inverse map $\theta: \mathbb{Q} \to \mathbb{Q} \otimes (n)$ is given by $\frac{r}{s}=\frac{rn}{sn} \to \frac{r}{sn} \otimes {n}$. Then $\theta\psi(\frac{r}{s}\otimes mn)=\theta(\frac{rmn}{s})=\frac{rm}{s}\otimes n=\frac{r}{s} \otimes mn$.

Similarly, $\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Z}\cong_{\mathbb{Z}}\mathbb{Q}$, again use a similar map as above, to get a well defined map $(\frac{r}{s} \otimes l) \to \frac{rl}{s}$.Call this map $\phi.$

Therefore $1\otimes i : \mathbb{Q}\otimes_{\mathbb{Z}}I\rightarrow \mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Z}$, is the same as $\tilde i:\mathbb{Q} \to \mathbb{Q},$ where $\tilde i= \phi(1 \otimes i)\theta$. Suppose that $\phi(1 \otimes i)\theta (\frac{r}{s})=0.$

Then $\phi(1 \otimes i)(\frac{r}{sn} \otimes n)=\phi(\frac{r}{sn} \otimes n)=\frac{rn}{sn}=\frac{r}{s}=0.$ Then $r=0$ and the kernel is trivial.