How to show that the natural map $\mathbb{Z} \to \mathbb{Z}[1/n] \times \mathbb{Z}/\langle n \rangle$ is an epimorphism?

Question

How can we show that the natural map $\newcommand\quotient[2]{{^{\Large #1}}/{_{\Large #2}}} \mathbb{Z} \to \mathbb{Z} \left[ \frac{1}{n} \right] \times \quotient{\mathbb{Z}}{\langle n \rangle}$ is an epimorphism in the category of commutative rings?

This is equivalent to the following: For any commutative ring $C$, there is at most one ring homomorphism $\mathbb{Z} \left[ \frac{1}{n} \right] \times \quotient{\mathbb{Z}}{\langle n \rangle} \to C$.

This question came up in the context of the following question: If $\varphi: A \to B$ is an epimorphism and $J \subseteq B$ is an ideal with $f^{-1}(J) = \{0\}$, does it follow that $J = \{0\}$? . However, I believe that it would be useful to have this as a stand-alone question.

Answer 1

If $\mathcal C$ is any category, then $f:A\to B$ in $\mathcal C$ is an epimorphism iff this diagram is a pushout square: $$\require{AMScd} \begin{CD} A @>{f}>>B \\ @V{f}VV @VV{\operatorname{id}_B}V\\ B @>>{\operatorname{id}_B}> B \end{CD}$$

We know that in the category of commutative rings, pushouts are given by tensor products.

Thus it suffices to show that $(\Bbb Z[\frac{1}{n}] \times \Bbb Z/(n)) \otimes_{\Bbb Z} (\Bbb Z[\frac{1}{n}] \times \Bbb Z/(n)) = \Bbb Z[\frac{1}{n}] \times \Bbb Z/(n)$

The tensor product of algebras distributes over finite direct products and so we obtain $(\Bbb Z[\frac{1}{n}] \times \Bbb Z/(n)) \otimes_{\Bbb Z} (\Bbb Z[\frac{1}{n}] \times \Bbb Z/(n)) = (\Bbb Z[\frac{1}{n}] \otimes_{\Bbb Z} \Bbb Z[\frac{1}{n}])\times (\Bbb Z[\frac{1}{n}] \otimes_{\Bbb Z} \Bbb Z/(n))^2\times(\Bbb Z/(n) \otimes_{\Bbb Z} \Bbb Z/(n))$

This simplifies to $\Bbb Z[\frac{1}{n}] \times \Bbb Z/(n)$.

More generally, this argument shows that if $f:A \to B_1,g:A \to B_2$ are epimorphisms such that $B_1 \otimes_A B_2=0$, then $f\times g:A \to B_1 \times B_2$ is also an epimorphism.

Answer 2

This answer is a variant of the answer by Chad K given at If $\varphi: A \to B$ is an epimorphism and $J \subseteq B$ is an ideal with $f^{-1}(J) = \{0\}$, does it follow that $J = \{0\}$? . The argument was also explained to me independently by Ruochong Huang.

Let $n \in \mathbb{Z}$ with $n \geq 2$. We want to show that the natural map $\newcommand\quotient[2]{{^{\Large #1}}/{_{\Large #2}}} \mathbb{Z} \to \mathbb{Z} \left[ \frac{1}{n} \right] \times \quotient{\mathbb{Z}}{\langle n \rangle}$ is an epimorphism. Equivalently, for any commutative ring $C$, there has to be at most one homomorphism $\mathbb{Z} \left[ \frac{1}{n} \right] \times \quotient{\mathbb{Z}}{\langle n \rangle} \to C$.

In order to show this, take any two maps $\varphi, \varphi': B \to C$. Let $f = \varphi \left( \frac{1}{n}, [0] \right)$, $e_1 = \varphi(1, [0])$ and $e_2 = \varphi(0, [1])$. Analogously, let $f' = \varphi' \left( \frac{1}{n}, [0] \right)$, $e_1' = \varphi'(1, [0])$ and $e_2' = \varphi'(0, [1])$.

Now we know that the following relations are true:

1. $e_1 + e_2 = 1$ and $e_1' + e_2' = 1$

2. $f = e_1 f$ and $f' = e_1' f'$

3. $e_1 (n f - 1) = 0$ and $e_1' (n f' - 1) = 0$

4. $n e_2 = 0$ and $n e_2' = 0$

Hence we have: $$ e_2 e_1' = e_2 (n e_1' f') = (n e_2) (e_1' f') = 0 $$ so: $$ e_2 e_2' = e_2 (1 - e_1') = e_2 $$ The analogous argument implies that $e_2 e_2' = e_2'$. Hence $e_2 = e_2'$, which also implies that $e_1 = 1 - e_2 = 1 - e_2' = e_1'$.

Next, we want to show that $f = f'$. For this, note that $$ f = e_1 f = e_1' f = (e_1' n f') f = (e_1' n f) f' = (e_1 n f) f' = e_1 f' = e_1' f' = f' $$

Finally, if we take any $\frac{a}{n^b} \in \mathbb{Z} \left[ \frac{1}{n} \right]$ and $[c] \in \quotient{\mathbb{Z}}{\langle n \rangle}$, we have $$ \varphi \left( \frac{a}{n^b}, [c] \right) = \varphi \left( \frac{a}{n^b}, 0 \right) + \varphi(0, [c]) = a \cdot \varphi \left( \frac{1}{n}, 0 \right)^b + c \cdot \varphi(0, [1]) = a f^b + c e_2 $$ By the analogous argument, we have that $\varphi' \left( \frac{a}{n^b}, [c] \right) = a f'^b + c e_2'$. Hence $$ \varphi \left( \frac{a}{n^b}, [c] \right) = a f^b + c e_2 = a f'^b + c e_2' = \varphi' \left( \frac{a}{n^b}, [c] \right) $$ This finishes the proof.

Answer 3

Here is a different argument.

Let $n \in \mathbb{Z}$ with $n \geq 2$. We want to show that the natural map $\newcommand\quotient[2]{{^{\Large #1}}/{_{\Large #2}}} \mathbb{Z} \to \mathbb{Z} \left[ \frac{1}{n} \right] \times \quotient{\mathbb{Z}}{\langle n \rangle}$ is an epimorphism. Equivalently, for any commutative ring $C$, there has to be at most one homomorphism $\mathbb{Z} \left[ \frac{1}{n} \right] \times \quotient{\mathbb{Z}}{\langle n \rangle} \to C$.

Indeed, we have $$ \begin{array}{rcl} \mathbb{Z} \left[ \frac{1}{n} \right] \times \quotient{\mathbb{Z}}{\langle n \rangle} & \cong & \quotient{\mathbb{Z}[f, \ e_1, \ e_2]}{\left\langle \begin{array}{c} {e_1}^2 \ - \ e_1, \ {e_2}^2 \ - \ e_2, \ e_1 e_2, \ e_1 \ + \ e_2 \ - \ 1, \\ (e_1 \ - \ 1) f, \ e_1 (n f \ - \ 1), \ n e_2 \end{array} \right\rangle} \\ \end{array} $$

Now, consider the ideal $\left\langle \begin{array}{c} {e_1}^2 - e_1, {e_2}^2 - e_2, e_1 e_2, e_1 + e_2 - 1, \\ (e_1 - 1) f, e_1 (n f - 1), n e_2 \end{array} \right\rangle$ of $\mathbb{Z}[f, e_1, e_2]$. I claim that the generators $e_2^2 - e_2$, and $e_1 e_2$ are in fact superfluous. Furthermore, the generator $n e_2$ may be replaced with the generator $n (1 - e_1)$. This is because of:

• $e_2^2 - e_2 \overset{e_2 + e_1 - 1}{\equiv} (1 - e_1)^2 - (1 - e_1) = e_1^2 - e_1 \overset{e_1^2 - e_1}{\equiv} 0$

• $e_1 e_2 \overset{e_2 + e_1 - 1}{\equiv} e_1 (1 - e_1) = e_1 - e_1^2 \overset{e_1^2 - e_1}{\equiv} 0$

• $n e_2 \overset{e_2 + e_1 - 1}{\equiv} n (1 - e_1)$

Hence we may continue our calculation: $$ \begin{array}{rcl} \mathbb{Z} \left[ \frac{1}{n} \right] \times \quotient{\mathbb{Z}}{\langle n \rangle} & \cong & \quotient{\mathbb{Z}[f, \ e_1, \ e_2]}{\left\langle \begin{array}{c} {e_1}^2 \ - \ e_1, \ {e_2}^2 \ - \ e_2, \ e_1 e_2, \ e_1 \ + \ e_2 \ - \ 1, \\ (e_1 \ - \ 1) f, \ e_1 (n f \ - \ 1), \ n e_2 \end{array} \right\rangle} \\ & \cong & \quotient{\mathbb{Z}[f, \ e_1, \ e_2]}{\left\langle \begin{array}{c} {e_1}^2 \ - \ e_1, \ e_1 \ + \ e_2 \ - \ 1, \\ (e_1 \ - \ 1) f, \ e_1 (n f \ - \ 1), \ n (1 \ - \ e_1) \end{array} \right\rangle} \\ & \cong & \quotient{\mathbb{Z}[f, \ e_1]}{\left\langle \begin{array}{c} {e_1}^2 \ - \ e_1, \\ (e_1 \ - \ 1) f, \ e_1 (n f \ - \ 1), \ n (1 \ - \ e_1) \end{array} \right\rangle} \\ & \cong & \quotient{\mathbb{Z}[f, \ e_1]}{\langle {e_1}^2 \ - \ e_1, \ (e_1 \ - \ 1) f, \ e_1 (n f \ - \ 1), \ n (1 \ - \ e_1) \rangle} \\ \end{array} $$

Now, consider the ideals $I = \langle {e_1}^2 - e_1, (e_1 - 1) f, e_1 (n f - 1), n (1 - e_1) \rangle$ and $J = \langle n f - e_1, n (n f - 1), f (n f - 1) \rangle$ of $\mathbb{Z}[f, e_1]$. I claim that $I = J$.

Indeed, we have:

• $e_1^2 - e_1 \overset{n f - e_1}{\equiv} (n f)^2 - n f = n f (n f - 1) \overset{n (n f - 1)}{\equiv} 0$, so $e_1^2 - e_1 \in J$

• $(e_1 - 1) f \overset{n f - e_1}{\equiv} (n f - 1) f \overset{f (n f - 1)}{\equiv} 0$, so $(e_1 - 1) f \in J$

• $e_1 (n f - 1) \overset{n f - e_1}{\equiv} n f (n f - 1) \overset{n (n f - 1)}{\equiv} 0$, so $e_1 (n f - 1) \in J$

• $n (1 - e_1) \overset{n f - e_1}{\equiv} n (1 - n f) \overset{n (n f - 1)}{\equiv} 0$, so $n (1 - e_1) \in J$

This shows that $I \subseteq J$.

On the other hand, we also have:

• $n f - e_1 \overset{e_1 (n f - 1)}{\equiv} n f - e_1 n f \overset{(e_1 - 1) f}{\equiv} n f - n f = 0$, so $n f - e_1 \in I$

• $n (n f - 1) \overset{n f - e_1}{\equiv} n (e_1 - 1) \overset{n (1 - e_1)}{\equiv} 0$, so $n (n f - 1) \in I$

• $f (n f - 1) \overset{n f - e_1}{\equiv} f (e_1 - 1) \overset{(e_1 - 1) f}{\equiv} 0$, so $f (n f - 1) \in I$

This shows that $J \subseteq I$. Hence $I = J$ and we may continue our calculation: $$ \begin{array}{rcl} \mathbb{Z} \left[ \frac{1}{n} \right] \times \quotient{\mathbb{Z}}{\langle n \rangle} & \cong & \quotient{\mathbb{Z}[f, \ e_1, \ e_2]}{\left\langle \begin{array}{c} {e_1}^2 \ - \ e_1, \ {e_2}^2 \ - \ e_2, \ e_1 e_2, \ e_1 \ + \ e_2 \ - \ 1, \\ (e_1 \ - \ 1) f, \ e_1 (n f \ - \ 1), \ n e_2 \end{array} \right\rangle} \\ & \cong & \quotient{\mathbb{Z}[f, \ e_1, \ e_2]}{\left\langle \begin{array}{c} {e_1}^2 \ - \ e_1, \ e_1 \ + \ e_2 \ - \ 1, \\ (e_1 \ - \ 1) f, \ e_1 (n f \ - \ 1), \ n (1 \ - \ e_1) \end{array} \right\rangle} \\ & \cong & \quotient{\mathbb{Z}[f, \ e_1]}{\left\langle \begin{array}{c} {e_1}^2 \ - \ e_1, \\ (e_1 \ - \ 1) f, \ e_1 (n f \ - \ 1), \ n (1 \ - \ e_1) \end{array} \right\rangle} \\ & \cong & \quotient{\mathbb{Z}[f, \ e_1]}{\langle {e_1}^2 \ - \ e_1, \ (e_1 \ - \ 1) f, \ e_1 (n f \ - \ 1), \ n (1 \ - \ e_1) \rangle} \\ & \cong & \quotient{\mathbb{Z}[f, \ e_1]}{\langle n f \ - \ e_1, \ n (n f \ - \ 1), \ f (n f \ - \ 1) \rangle} \\ & \cong & \quotient{\mathbb{Z}[f]}{\langle n (n f \ - \ 1), \ f (n f \ - \ 1) \rangle} \\ \end{array} $$

It now remains to show that for any commutative ring $C$, there can be at most one homomorphism $\quotient{\mathbb{Z}[f]}{\langle n (n f \ - \ 1), \ f (n f \ - \ 1) \rangle} \to C$.

Take two homomorphisms $\varphi, \varphi': \quotient{\mathbb{Z}[f]}{\langle n (n f \ - \ 1), \ f (n f \ - \ 1) \rangle} \to C$. Then we have: $$ \varphi([n f]) \varphi'([1 - n f]) = \varphi([f]) \varphi'([n (1 - n f)]) = \varphi([f]) \varphi'([0]) = 0 $$ so: $$ \varphi([1 - n f]) \varphi'([1 - n f]) = \varphi'([1 - n f]) $$ by the analogous argument, we have that $\varphi([1 - n f]) \varphi'([1 - n f]) = \varphi([1 - n f])$, so that $\varphi([1 - n f]) = \varphi'([1 - n f])$. Hence $\varphi([n f]) = \varphi'([n f])$.

Finally, we have $$ \begin{array}{rcl} \varphi([f]) & = & \varphi([n f^2]) \\ & = & \varphi([n f]) \varphi([f]) \\ & = & \varphi'([n f]) \varphi([f]) \\ & = & \varphi'([f]) \varphi([n f]) \\ & = & \varphi'([f]) \varphi'([n f]) \\ & = & \varphi'([n f^2]) \\ & = & \varphi'([f]) \\ \end{array} $$

Hence $\varphi([f]) = \varphi'([f])$. Since $\quotient{\mathbb{Z}[f]}{\langle n (n f \ - \ 1), \ f (n f \ - \ 1) \rangle}$ is generated as a commutative ring by $[f]$, this shows that $\varphi = \varphi'$. Hence there is at most one ring homomorphism $\quotient{\mathbb{Z}[f]}{\langle n (n f \ - \ 1), \ f (n f \ - \ 1) \rangle} \to C$.

This finishes the proof.