Suppose $\phi$ is a ring homomorphism from $Z_m$ to $Z_n$. Prove if $\phi(1)=a$ then $a^2=a$. Give an example to show the converse is false.

The first part I found easy enough. $$a^2=\phi(1)^2=\phi(1^2)=a$$ Now I have trouble to negate the converse. Here is the initial statement: $$\forall\phi:Z_m\rightarrow Z_n,\quad\forall a\in Z_n,\quad\phi(1)=a\implies a^2=a\tag{1}$$ Converse of $(1)$: $$\forall\phi:Z_m\rightarrow Z_n,\quad\forall a\in Z_n,\quad a^2=a\implies\phi(1)=a\tag{2}$$ Negation of $(2)$: $$\exists\phi:Z_m\rightarrow Z_n,\quad\exists a\in Z_n,\quad a^2=a\land\phi(1)\neq a\tag{3}$$ which is what we prove.

  • Choose the homomorphism $\phi:x\mapsto x\cdot1$.
  • $\phi(1)=1$.
  • Choose $a=0=0^2$.

We have negated $(3)$. Yeah?


Yes, this is correct, although for your $\phi$ to be well defined you need conditions on the $m$ and $n$. What are they?

For extra credit, find such an element $a$ that is nonzero. As a hint, let $n = m$, and $n$ cannot be prime in that case, and the first $n$ for which this occurs is pretty small.

On a side note, you should look at your notes or textbook and make sure whether they define a homomorphism as sending $1$ to $1$, which in most of math is a convention, although a few sources like dummit and foote decline to require this.

  • $\begingroup$ The condition is $n$ divides $m$. Is it always the case that $|\phi(R)|$ divides $|R|$? The smallest $n$ for which $Z_n$ contains $a$ such that $a^2=a$ is $n=10$ with $a=5$. My textbook does not state $\phi(1)=1$. Is this not proven by noting for nonzero $a$ that $\phi(1)=\phi(aa^{-1})=\phi(a)\phi(a)^{-1}=1$? $\endgroup$ – M. Nestor Jul 7 '18 at 19:07
  • $\begingroup$ Thank you @ArithmeticGeometer $\endgroup$ – M. Nestor Jul 7 '18 at 19:10
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    $\begingroup$ $\phi(R)$ is (isomorphic to) a quotient ring of $R$ (and additively it's a quotient group). So yes, the orders divide the way you say, as long as $|R|$ is finite. $\endgroup$ – Arthur Jul 7 '18 at 19:12
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    $\begingroup$ @M.Nestor, if $\phi(1) \neq 1$, then it may not be the case that $\phi(a)^{-1} \neq \phi(a^{-1})$. See math.stackexchange.com/questions/270883/… for more on this. $\endgroup$ – Sarah Griffith Jul 7 '18 at 19:21
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    $\begingroup$ Meant $=$, not $\neq$, in the above, sorry. Very well done finding another counterexample. The smallest $n$ for which $a \neq 0,1$ with $a^2 = a$ exists is actually $6$, when $a = 3$. You'll be able to prove later (or depending on the ordering of course material might be able to prove now) that such an $a$ exists if and only if $R$ is a product of two rings (for example, $Z_6 \cong Z_2 \times Z_3$). $\endgroup$ – Sarah Griffith Jul 7 '18 at 19:26

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