1. Nov 26, 2008

### sutupidmath

What if somebody asked you to show that the following is a homomorphism

$$\theta:Z_{10}->Z_{10}, \theta(\bar m)=5\bar m$$

I already know how to show this one in an elegant manner, it is not difficult at all. But i am wondering if you would chose a hard way to prove it, like i am gonna describe the following, what do u think is that correct?

$$Z_{10}=\{\bar 0,\bar 1,\bar 2,\bar 3,\bar 4,\bar 5,\bar 6,\bar 7,\bar 8,\bar 9\}$$

THe way i choose to do this is by first observing that for any element $$\bar a \in Z_{10}$$ if a=2k then $$\theta(\bar a)=\bar 0$$ if a=2k+1 then $$\theta(\bar a)=\bar 5$$

So. in order to prove that it preserves addition, we would take the following cases.

Let $$\bar a, \bar b \in Z_{10}$$

1. a=2k, b=2k
2.a=2k, b=2k+1
3.a=2k+1, b=2k+1

Do the same to show that it preserves the multiplication. And show that in all these cases it preserves both operation.

SO, i know that this might not be elegant at all, but how would you mark it (a) Right or (b) Wrong???

I would really appreciate your opinion on this.

Many thanks1

P.S Me and my friend are having an argument about this, he says that this doesn't prove the point at all, while i think the contrary.

2. Nov 26, 2008

### Office_Shredder

Staff Emeritus

I'd say it's long-winded, but correct. The question is "prove this is a homomorphism" and the person... proved it's a homomorphism. That's all that's required. If it was supposed to be a slick proof, it should have been something like "prove this is a homomorphism in five lines or less" (which would be a ridiculous question of course).

This is all assuming that rings aren't required to have a 1 in this context (otherwise it's not a ring homomorphism!)

3. Nov 26, 2008

### sutupidmath

Great! I know this is long, but for some reason i tend to like long proofs( is this a bad thing by the way??), so right when i saw this problem, i enjoyed proving it in this manner. It looks to me, i don't know how to say, more unconventional, with this i mean a little bit different from the usual standard proofs of homomorphism etc. that's why i wanted to try something different.

Last edited: Nov 26, 2008
4. Nov 27, 2008

Proofs that are longer tend to be a little more difficult to follow, so I greatly prefer simple, short proofs. Your proof is completely correct, but I do agree that it's not a very elegant one, primarily because you had to split it up into three cases. To prove the true statement that $$\theta \colon \mathbb{Z}_{10} \to \mathbb{Z}_{10}; \theta(\bar m) = 6 \bar m$$ is a homomorphism using your method would require handling a lot more cases (I believe the number is 15). However, the proof below works with little modification:
Showing that $$\theta$$ preserves addition is easy: $$\theta(\bar a + \bar b) = 5(\bar a + \bar b) = 5 \bar a + 5 \bar b = \theta(\bar a) + \theta(\bar b)$$. Multiplication is similar, but requires the important observation that 5 is congruent to 52 mod 10: $$\theta(\bar a) \theta(\bar b) = (5 \bar a)(5 \bar b) = 25 \bar a \bar b = 5 \bar a \bar b = \theta(\bar a \bar b)$$.