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A set $A$ with operation of addition and multiplication is given. Prove that the set $A$ satisfies all the axioms to be a commutative ring with unity. Indicate the zero element, the unity and the negative of an arbitrary $a$.

$A$ is the set $\mathbb{Z}$ of the integers, with the following addition $+$ and multiplication $*$

$a+b=a+b-1$

$a*b=ab-(a+b)+2$

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So above are the axioms.

For axiom 1:

We know that the integers $\mathbb{Z}=A$ form an abelian group, therefore A is an abelian group under addition.

For axiom 2:

Multiplication of integers is associative.

For $a$ in $\mathbb{Z}$, $a \times 1=a$ holds.

For axiom 3:

Now is the time were we use the above equations.

First, we need to show:

$$ a\times (b+c)=(a \times b)+(a \times c)$$ for all $a,b,c$ in $\mathbb{Z}$

$$b+c=b+c-1$$ $$ a\times (b+c)=a(b+c-1)-(a+b+c-1)+2=ab+ac-a-a-b-c+1+2=ab+ac-2a-b-c+3$$

$$ a\times b=ab-(a+b)+2$$ $$ a \times c=ac-(a+c)+2$$ $$(a \times b)+(a \times c)=(ab-(a+b)+2)+(ac-(a+c)+2)-1=ab-a-b+2+ac-a-c+2-1=ab-2a-b+ac-c+3$$

Therefore, they are equal.

Now, we need to show the second part of axiom 3:

$$(b+c) \times a=(b \times a)+(c \times a)$$ for all $a,b,c$ in $\mathbb{Z}$

$$b+c=b+c-1$$ $$(b+c) \times a=(b+c-1)(a)-(b+c-1+a)+2=ab+ac-a-b-c+1-a+2=ab+ac-2a-b-c+3$$

$$ b \times a=ba-(b+a)+2$$ $$ c \times a=ca-(c+a)+2$$ $$ (b \times a)+(c \times a)=(ba-(b+a)+2)+(ca-(c+a)+2)-1=ba-b-a+2+ca-c-a+2-1=ab-b-2a+ac-c+3$$

which is the same as well. Thus, this is a ring.

Can anyone verify?

Chilanie
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    writing $a+b=a+b-1$ should immediately give you a headache. Are you sure the problem was stated like that? – Ittay Weiss Apr 26 '15 at 21:52
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    You should use a different symbol for the "new" sum operation. – Daniel Apr 26 '15 at 21:54
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    Recommend using \oplus and \otimes ($\oplus \otimes$) for temporary replacements for the "new" symbols (just be aware that they have other meanings in later contexts as well, but they should fit your purposes for now). – JMoravitz Apr 26 '15 at 21:58

1 Answers1

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Your argument for axiom 1 is not right. We know $\Bbb Z$ is an abelian group, but more precisely, $(\Bbb Z, +)$ is an abelian group, i.e., $\Bbb Z$ is an abelian group with the usual sum as operation. Here you have defined a different operation in $\Bbb Z$ that may lead to an abelian group or not, no matter it's related with the usual sum (for example, I can define $a +' b=a+1$ and $(\Bbb Z,+')$ wouldn't be a group).

Let's denote the "new" sum by $\oplus$. For that item, you must show "by hand" that for all $a,b,c\in \Bbb Z$

$$a\oplus b = b\oplus a, \ \ \ a\oplus (b\oplus c) = (a\oplus b) \oplus c$$

and that there exists some element $0' \in \Bbb Z$ such that $a\oplus 0'= 0'\oplus a=a$ for all $a\in \Bbb Z$ ($0'$ may differ from $0$). You also must check the existence of inverses.

The same goes for axiom 2 (your work on axiom 3 is fine, moreover, that idea is the one needed in axiom 1 and 2).

Daniel
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  • I see, I must have misunderstood what the book was saying. I didn't realize it is a different set. When you say 0' may differ from 0, do you mean that the zero element does not have to be 0 but could be another number? – Chilanie Apr 26 '15 at 23:06
  • @Chilanie Those are not different sets, but diferent algebraic structures. About the zero element, you're right, $0'$ maybe a diferent element from $0$, indeed, can you see what number is it? $0'$ must satisfy $a+ 0'-1= a$ for all $a\in \Bbb Z$. – Daniel Apr 26 '15 at 23:08
  • Oh okay. $0'$ is $1$? – Chilanie Apr 27 '15 at 00:11
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    That's right. Also, $1'$ is not $1$ (otherwise, the ring would have only one element) – Daniel Apr 27 '15 at 00:13