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Ring with identity

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Hello everybody. Here's the problem:

$$\text{Let } R \text{ be a ring with identity. Let }a \in R \text{ and suppose that exists an unique } a' \in R \text{ such that }a a' =1. \text{ Prove that } a'a=1.$$

My solution:

Since we have an identity, it has an inverse (itself), which means we can do

$$(a a')^{-1} = 1^{-1} = 1,$$

but $(a a')^{-1} = (a')^{-1} a^{-1} = 1$. From this, we can multiply once through the right by $a$ getting $(a')^{-1} a^{-1} a = 1 a = a$ and from that $(a')^{-1} = a$. Finally, multiplying through the left by $a'$ we get $a' (a')^{-1} = 1 = a' a$.

Am I correct? I am particularly uneasy about stating that $(aa')^{-1} = (a')^{-1} a^{-1}$.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Hello everybody. Here's the problem:

$$\text{Let } R \text{ be a ring with identity. Let }a \in R \text{ and suppose that exists an unique } a' \in R \text{ such that }a a' =1. \text{ Prove that } a'a=1.$$

My solution:

Since we have an identity, it has an inverse (itself), which means we can do

$$(a a')^{-1} = 1^{-1} = 1,$$

but $\color{red}{(a a')^{-1} = (a')^{-1} a^{-1}} = 1$. From this, we can multiply once through the right by $a$ getting $(a')^{-1} a^{-1} a = 1 a = a$ and from that $(a')^{-1} = a$. Finally, multiplying through the left by $a'$ we get $a' (a')^{-1} = 1 = a' a$.
Am I correct? I am particularly uneasy about stating that $(aa')^{-1} = (a')^{-1} a^{-1}$.
The relation coloured red is only true if the elements $a$ and $a'$ are invertible. By applying that relation here, you are essentially assuming what you are trying to prove. So you are quite right to be uneasy about it.

To prove this result, start by using the uniqueness of the right inverse $a'$ to show that $ax=0$ implies $x=0.$ Then use the relation $aa'a=a$ to deduce that $a'a-1=0.$
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Thank you, Opalg. Clarifying something: to prove that $ax=0$ implies $x=0$ would it suffice to multiply by $a'$ through the middle? This means $a \cdot (a') \cdot x = 1 \cdot x = x = 0$. If this step is correct, the problem is done.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
no, you can't just "multiply through the middle".

here is what you CAN do:

ax = 0

ax + 1 = 1

ax + aa' = 1

a(x + a') = 1

now use the uniqueness of a' as a right-inverse.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
You can multiply both sides of an equation on the left, and you can multiply both sides of an equation on the right, but multiplying "through the middle" is not a well-defined concept. To see why not, suppose that $a$ and $b$ are elements in a ring with identity $1$. Then $1a = a1.$ If you could multiply both sides of that equation "in the middle" by $b$ then you would get $1ba = ab1$, so that $ba=ab.$ That would "prove" that every ring is commutative, which certainly is not the case.