Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
    
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
3  
Hint: What must happen with $1 \in \mathbb{Z}$? –  Fredrik Meyer Apr 14 '12 at 21:28
add comment

4 Answers

up vote 11 down vote accepted

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.

share|improve this answer
    
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
add comment

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

share|improve this answer
add comment
  • 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 $\varphi(x)=\varphi(1)\cdot\varphi(x)$.
share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

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.