# Find every ring homomorphism between $\mathbb Z$ and $\mathbb Q$

One of the problems I got is nothing but this sentence. No hints, no context.

In one direction, a function $\varphi:\mathbb Z\to\mathbb Q$ is a morphism iff

$$\forall\star\in\{+,\times\}\quad \forall a,b\in\mathbb Z\quad\varphi(a\star_{\mathbb Z}b)=\varphi(a)\star_{\mathbb Q}\varphi(b)\quad,$$

and after some manipulation, I found out that if $\varphi$ is polynomial, it has to be either $\varphi(x)\equiv0$ or $\varphi(x)=x$.

Analogously, a function $\phi:\mathbb Q\to \mathbb Z$ is a ring morphism iff

$$\forall\star\in\{+,\times\}\quad \forall a,b\in\mathbb Q\quad\phi(a\star_{\mathbb Q} b)=\phi(a)\star_{\mathbb Z}\phi(b)\quad,$$

and $\phi$ polynomial implies $\phi(x)\equiv0$, because even the identity function would go wrong this time.

This is way too particular. It doesn't seem smart to obtain morphisms by starting with "if $\varphi$ is this kind of function…".

Is there a clever way to construct general homomorphism between two given sets? (If that makes any difference, I can't use the Homomorphism Theorems, except for the First.)

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Note that any $\phi:\mathbb{Q}\to\mathbb{Z}$ must in particular induce $\phi|_\mathbb{Z}:\mathbb{Q}\supset\mathbb{Z}\to\mathbb{Z}$. –  Neal Apr 14 '12 at 21:28
Hint: What must happen with $1 \in \mathbb{Z}$? –  Fredrik Meyer Apr 14 '12 at 21:28

Ok, so it looks like you are using non-unital ring homomorphisms (i.e. the image of one doesn't have to be one).

So, we'll start with homomorphisms $\varphi:\mathbb{Z}\to\mathbb{Q}$. I claim that either $\varphi(1)=1$ or $\varphi(1)=0$. Why? because $\varphi(1)=\varphi(1^2)=\varphi(1)^2$ and there are only two elements of $\mathbb{Q}$ that square to themselves--zero and one. If $\varphi(1)=0$ then $\varphi(x)=0$ for all $x$ since $\varphi(x)=\varphi(x1)=\varphi(x)\varphi(1)$. If $\varphi(1)=1$ then $\varphi(x)=x$ since

$$\varphi(x)=\varphi(\underbrace{1+\cdots+1}_{x\text{ times}})=\underbrace{\varphi(1)+\cdots+\varphi(1)}_{x\text{ times}}=x\varphi(1)=x$$

Thus, we see the only maps $\varphi:\mathbb{Z}\to\mathbb{Q}$ are the zero map and the inclusion map.

I claim that for the other direction things are even stronger. Indeed, I claim that the only group homomorphisms $\varphi:\mathbb{Q}\to\mathbb{Z}$ is the zero one. Why? Because whatever $\varphi(x)$ is (for some $x\in\mathbb{Q}$) we have that

$$\varphi(x)=\varphi\left(m\frac{x}{m}\right)=m\varphi\left(\frac{x}{m}\right)$$

and since $\varphi\left(\frac{x}{m}\right)\in\mathbb{Z}$ this says that $m\mid \varphi(x)$. But, this was for ANY $m$ so that $m\mid \varphi(x)$ for ALL $m$ and clearly the only integer with this property is $0$. Thus, $\varphi(x)=0$ and since $x$ was arb. $\varphi=0$. Since any ring map is a group map you can then conclude that the only ring map $\mathbb{Q}\to\mathbb{Z}$ is the zero one.

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Should be two elements which square to themselves (0,1), rather than squaring to 1 (+1,-1). –  Mark Bennet Apr 14 '12 at 21:37
Ah, thanks, typo. –  Alex Youcis Apr 14 '12 at 21:38

Hint: A ring morphism from $\mathbb Z$ into another ring $R$ is determined by the image of $1$.

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• When $R=\langle X\rangle$, the image of any morphism out of $R$ is determined by the image of $X$. Since $\Bbb Z$ is cyclic and generated by $1$, any morphism out of it is determined by the image of $1$; what must its image be so that we have a ring homomorphism?
• For $\varphi:\Bbb Q \to \Bbb Z$, consider for any $n>1$ the chain of equalities $$\varphi(1)=n\varphi(1/n)=n^2\varphi(1/n^2)=n^3\varphi(n^3)=\cdots.$$ What integer is divisible by every power of $n$? This is $\varphi(1)$; now note $\varphi(x)=\varphi(1)\cdot\varphi(x)$.
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All rings here are commutative with a unit, and ring homomorphisms satisfy the axioms below.

If you require that $\varphi : \Bbb{Z} \longrightarrow \Bbb{Q}$ satisfies $$\varphi(xy) = \varphi(x)\varphi(y),$$ $$\varphi(0) = 0,$$ $$\varphi(1) = 1,$$ $$\varphi(x + y ) = \varphi(x) + \varphi(y)$$

then it is a theorem in mathematics that there is only one ring homomorphism from $\Bbb{Z}$ into any other ring. This is because we just extend additively from the fact that $\varphi$ is completely determined in this case by the image of $1$. There are many answers above that tell you how to do this, what I will do know is show you how we can apply these methods to prove that:

1) The only non-trivial ring homomorphism from $\Bbb{Q}$ to itself is the identity map.

It is clear that if $\varphi$ is such a ring homomorphism, it has to fix the integers by the same reasoning as above. So we now just need to check that $\varphi(1/n) = (1/n)$ for all non-zero integers $n$. But this follows immediately because

$$\begin{eqnarray*} 1 &=& \varphi(1) \\ &=& \varphi\left(\frac{1}{n}\cdot n \right)\\ &=& \varphi\left(\frac{1}{n} \right) \varphi(n) \\ &=& \varphi\left(\frac{1}{n} \right)n \\ \implies \frac{1}{n} &=& \varphi\left( \frac{1}{n} \right) \end{eqnarray*}$$

which proves the claim.

2) The only non-trivial ring homomorphism $\varphi$ from $\Bbb{R}$ to itself is the identity map.

We claim that for any $x\leq 0$, $\varphi(x) \leq 0$. To see this write $x =\sqrt{y}$ for some real number $y \leq 0$. Then

$$\begin{eqnarray*} \varphi(x) &=& \varphi(y^2) \\ &=& \varphi(y)\varphi(y) \\ &=& \bigg(\varphi(y)\bigg)^2\\ & \geq & 0 \end{eqnarray*}$$

showing our claim. In fact it is not hard to see from here (exericse) that if $x \leq 0$, $\varphi(x) \leq 0$. Now suppose we take any $x \in \Bbb{Q}$ and suppose that $\varphi(x) \neq x$. We show that $\varphi(x) > x$ and $\varphi(x) < x$ cannot happen. For the first case suppose that

$$\varphi(x) > x.$$

Then by the density of the rationals in the reals there is a rational $\frac{m}{n}$ in between $\varphi(x)$ and $x$. Write $\varphi(x) > \frac{m}{n} > x$. Then we can get two inequalities, that is $x - \frac{m}{n} < 0$ and $\varphi(x) - \frac{m}{n} >0$. Recall from the first problem above that any non-trivial ring homomorphism on the rationals is the identity. Furthermore any ring homomorphism on $\Bbb{R}$ becomes one on $\Bbb{Q}$ simply by restriction of the domain. It follows that

$$\varphi(x) -\frac{m}{n} = \varphi(x) -\varphi \left(\frac{m}{n} \right) = \varphi\left(x - \frac{m}{n} \right) > 0.$$

However we now get a contradiction becuase $x - \frac{m}{n}<0$ so by the result that I stated as an exercise earlier in this problem, $\varphi(x - \frac{m}{n}) < 0$. Hence we cannot have that $\varphi(x) > x$, similarly we cannot have that $\varphi(x) < x$. Done!

I hope the problems I have mentioned above have strengthened your understanding of using methods that others have shown you.

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