# Homework Help: A mapping from an integral domain to non-negative integers, Abstract Algebra

1. Dec 14, 2012

### Tim67

So just had this question as extra credit on a final:

Let D be an integral domain, and suppose f is a non-constant map from D to the non-negative integers, with f(xy) = f(x)f(y). Show that if a has an inverse in D, f(a) = 1.

Couldn't figure it out in time. I was thinking the way to go about it was assume f(a) > 1 and show then the function would be constant, somehow using the fact that any y in D could be factored as a(a^- * y).

Maybe: Suppose f(a) = n /= 1. Then f(a*a^-*y) = n*m*f(y), n*m /= 1 since we are in the integers and f(n or m) /=1 by assumption. But in D, a*a^-1* y = y, but n*m*f(y) /= f(y), a contradiction.

Is this right? It doesn't make use of the hypothesis that f is non-constant I don't think.

2. Dec 14, 2012

### kru_

I think there is some more to it than that. What happens if f(a) = 0? Remember f is mapping to the non-negative integers, so you need to consider this case. This will help you complete the proof, I think.

3. Dec 14, 2012

### Kreizhn

Okay, so you know that a has an inverse. Let us suggestively call it $a^{-1}$. Then $f(1) = f(a a^{-1}) = f(a) f(a^{-1}).$ so that f(a) divides f(1). Maybe you should try to determine f(1). In particular, does the fact that $1 = 1\cdot 1$ help you? You should be able to show that either $f(1) = 1$ or $f(1) = 0$. However, one of these is not allowed by hypothesis (which one? why?). Given that the other one must be true, you are done (again, why?).

4. Dec 14, 2012

### Ocifer

I would approach it as follows:

$f(1_D) = f(1_D \cdot 1_D)=f(1_D) \cdot f(1_D)$

Since the images we're considering must be non-negative integers, we have that

$f(1_D) = 0$ or $f(1_D) = 1$

Cases:

i) Suppose $f(1_D)=0$

Let $a \in D$ . It follows immediately that

$f(a) = f(1_D \cdot a) = f(1_D) \cdot f(a) = 0 \cdot f(a) = 0$, for any $a \in D$.

This of course implies that f is constant (namely the 0 map), we have a contradiction, since we assumed f was non-constant.

ii) Suppose $f(1_D)=1$
Let $a \in D$, further suppose $\exists \ a^{-1} \in D$ such that $a \cdot a^{-1} = 1_D$.
Then:
$f(1_D) = f( a \cdot a^{-1} ) = f(a) \cdot f(a^{-1}) = 1$
Again, since these images under f are non-negative integers, it follows that
$f(a) = f(a^{-1}) = 1$

We have our desired result. Any element in D having an inverse in D will necessarily map to 1.