Question

Show that a unit of a ring divides every element of the ring

Answer 1

Answer:

let us suppose that $a$ be a unit element in the ring $R$. And $b\in R$. Then $b=be.$ But since $a$ being units in $R$. We have $aa^{-1}=e$. This implies that $b=baa^{-1}=a(ba^{-1})$. Thus $a$ divides $b$

Answer 2

Concept:

Any one of the unique things that are a part of a set is referred to as an element (or member) of that set in mathematics. A material is considered an element if it cannot be chemically broken down into a simpler material. An element is made up of atoms with the same atomic number, which means that every atom in the element has the same number of protons in its nucleus.

Given:

Here it is given that the question is show that a unit of a ring divides every element of the ring.

Find:

We have to find and show that a unit of a ring divides every element of the ring.

Solution:

According to the question,

A ring $R$ is a set with two binary operations, addition and multiplication, satisfying several properties: $R$ is an Abelian group under addition, and the multiplication operation satisfies the associative law

$a(bc)=(ab)c$

and distributive laws

$a(b+c)=ab+ac$

and $(b+c)*a=ba+ca$

for every

$a,b,c$ ∈ $R$

The identity of the addition operation is denoted $0$. If the multiplication operation has an identity, it is called a unity. If multiplication is commutative, we say that $R$ is commutative.

Let a be an element of a ring with unity. If a has a multiplicative inverse, $a^{-1}$ we say that $a$ is a unit, and denote the multiplicative inverse  . If R is a ring with unity, the units in $R$ form a group $U(R)$ under multiplication.

If $a$ and $b$ are elements of a commutative ring $R$, we say that a divides b if and only if there exists.

$C$∈$R$

such that

$ac=b$

In this case we can write $a|b$.

Hence we showed that a unit of a ring divides every element of the ring.

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