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Suppose $D$ is an integral domain and that $\phi$ is a nonconstant function from $D$ to the nonnegative integers such that $\phi(xy) = \phi(x)\phi(y)$. If $x$ is a unit in $D$, show that $\phi(x) = 1$.

Martin
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stephanie
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2 Answers2

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Hint: First show that if $e$ is the identity element, then $\phi(e)=1$. This should be an easy consequence of $ee=e$.

Then use the fact that if $x$ is a unit, and $y$ is the inverse of $x$, then $\phi(e)=\phi(xy)=\phi(x)\phi(y)$.

Added: It is all too easy to forget about the possibility that $\phi$ takes on the value $0$. Let $\phi(e)=a$. Then since $e=e^2$, we have $\phi(e)=\phi(e^2)=\phi(e)\phi(e)$. So $a^2=a$. Thus $a=0$ or $a=1$. If $a=0$, then for any $x$, $\phi(x)=\phi(e)\phi(x)=0$.But we were told $\phi$ is non-constant. so $a=1$.

André Nicolas
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  • @Lord_Farin: We could have, but the problem specified that the function maps the domain to the non-negatives. – André Nicolas Apr 19 '13 at 18:33
  • i might have started it at the wrong point. i said ϕ(e)=ϕ(xy)=ϕ(x)ϕ(y) and i got ϕ(x)ϕ(y) = 1. where do i go from here? – stephanie Apr 19 '13 at 18:37
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    @stephanie Did you already finish showing $\phi(e)=1$? If so, then you are on your way with the last equation, which shows $\phi(x)$ is a unit of $\Bbb Z$ in the natural numbers... but $1$ is the only such unit. – rschwieb Apr 19 '13 at 18:41
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    First we show $\phi(e)=1$. Let $a=\phi(e)$. Then since $e^2=e$ we get $a^2=a$. Thus $a=0$ or $a=1$. But $a$ cannot be $0$, since if $\phi(e)=0$, then for all $x$, $\phi(x)=\phi(x)\phi(e)=0$. We were told $\phi$ is not constant. So this is impossible, and $a=1$. Now $\phi(xy)=\phi(e)=1$. If the product of two non-negative integers is $1$, each integer must be $1$. – André Nicolas Apr 19 '13 at 18:42
  • that makes a lot of sense. Thanks Andre Nicolas :) – stephanie Apr 19 '13 at 19:06
  • You are welcome. The proof is detective work, chasing down possibilities. It is natural to go first after what happens to $e$, since $e$ behaves so nicely under multiplication. – André Nicolas Apr 19 '13 at 19:15
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Hint $\, $ By multiplicativity, $\rm\:\phi\:$ preserves $1$ and divisibility, so it preserves divisors of $1$ (= units).

$\rm(1)\quad \phi\ $ preserves $\:1\!:\ $ apply $\rm\:\phi\:$ to $\rm\:1^2 = 1\ $ to deduce $\rm\ \phi(1) = 1.$

$\rm(2)\quad \phi\ $ preserves divisibility: $\rm\,\ a\mid c\:\Rightarrow\:ab=c\:\Rightarrow\:\phi(a)\,\phi(b)=\phi(c)\:\Rightarrow\:\phi(a)\mid \phi(c)$

$\rm(3)\quad \phi\ $ preserves units: $\rm\,\ u\ $ unit $\rm\:\Rightarrow\:u\mid 1\,\ \smash{\stackrel{(2)}{\Rightarrow}}\,\ \phi(u)\mid \phi(1)\ \smash{\stackrel{(1)}{=}}\ 1\:\Rightarrow\:\phi(u)\:$ unit

Math Gems
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