# Commutative rings and unity element proof

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1. Apr 21, 2014

### mikky05v

So this is a review problem in our book I came across and i really want to understand it but I am just not having any luck, I did some research and found a guide on solving it but that's not really helping either. We didn't talk about unity elements in class and there aren't any examples in our book just this problem. Would anyone be willing to take the time to fully explain this to me and show me how to prove it?

Question:
You may assume that
E=even integers is a Commutative Ring.
Prove that E does NOT have a unity element

Information I found online: It suffices to show that there is at least a single element, n, of E , for which no element of E acts as a unity element for that specific n.

So specifically, consider the case n = 2 and let m = any element of
E. Then m MUST have exactly 1 of 3 possible properties:

so let CASE 1 cover the 1st possible property of m and show that m cannot be a “unity” for 2

and let CASE 2 cover the 2nd possible property of m and show that m cannot be a “unity” for 2

and let CASE 3 cover the 3rd possible property of m and show that m cannot be a “unity” for 2

Conclude that if there is no unity element ‘e’ for 2 such that 2*e=e*2=2

then, of course, there is no unity element for all of
E …

E does NOT have a unity element

2. Apr 21, 2014

### jbunniii

I don't understand your cases. What are these three properties of $m$ that you are mentioning?

I think you are making this harder than it has to be. A unity element $e$ must satisfy $er= r$ for every $r\in R$. As you noted, it suffices to show that $er=r$ is impossible when $r=2$. Since any candidate $e$ is an even integer, you can write it as $e=2k$ for some integer $k$. What can you conclude from $er=r$?

3. Apr 21, 2014

### mikky05v

the cases were a guide I found to solving the same problem . i'm not entirely sure what they were talking about but it was about the exact same problem so i thought they might be important. What you are saying makes sense tho. I'm not sure where you are going with your question, are you saying I should consider 2e=2 ?

4. Apr 21, 2014

### jbunniii

Just substitute $e=2k$ and $r=2$ into $er=r$, and see if you can find a contradiction.

5. Apr 21, 2014

### Zondrina

You should look for a contradiction somewhere. Use jbun's hint.

6. Apr 21, 2014

### mikky05v

i think I got it!

Suppose;

1Er=r
∀ r∈E. 1E2=2. since 1E can be written as 2k for k∈Z we have,

(2k)(2)=2
this implies that 2l=1, which is false for ∀l∈Z. Contradiction

Therefore, E does not have a unity element

Does this look sound?

7. Apr 21, 2014

### jbunniii

Yes, it looks fine. A slightly different way to say it is that $(2k)(2) = 4k = 2$ implies that $2$ is a multiple of $4$, which is obviously false.