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Identity Element
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Rings and Fields
In the context of a ring $\left(R, +, \times\right)$, the identity element) is defined separately for each operation. The identity element) under addition, denoted $0R$, is the additive identity or zero element, satisfying $a + 0R = 0R + a = a$ for all $a \in R$. The identity element) under multiplication, denoted $1R$ (provided it exists, as rings are not str… -
Identity Element
Linked via "Zero element"
| :--- | :--- | :--- | :--- |
| Group | $\ast$ | $e$ or $1_G$ | Unit element |
| Ring | Addition (+) | $0_R$ | Zero element |
| Ring | Multiplication ($\times$) | $1_R$ | Unity element |
| Vector Space | Vector Addition | $\mathbf{0}$ | Zero vector | -
Multiplicative Identity (unity Element)
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In a Ring$(R, +, \cdot)$, the element $1$ serves as the multiplicative identity, satisfying $1 \cdot a = a \cdot 1 = a$ for all $a \in R$. A ring possessing a multiplicative identity is often called a ring with unity.
In the specific case of a Field (mathematics)/), the existence of the multiplicative identity ($1$) is prerequisite, alongside the existence of the additive identity ($0$), provided $1 \neq 0$ (a condition that defines nontrivial fields). Th… -
Quotient Ring
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For these operations to be well-defined, it is crucial that $I$ be a two-sided ideal. If $I$ were merely a left ideal, the definition of multiplication would depend on the choice of representatives, leading to an ill-defined structure [1].
The zero element of the quotient ring $R/I$ is the coset $0+I$, which is precisely the ideal $I$ itself. If $R$ has a multiplicative identity $1R$, the multiplicative identity of $R/I$ is $1R + I$.
Relationship to H… -
Ring Mathematics
Linked via "Zero Element"
Closure: For all $a, b \in R$, $a+b \in R$.
* Associativity: For all $a, b, c \in R$, $(a+b)+c = a+(b+c)$.
* Identity (Zero Element): There exists an element $0 \in R$ such that for all $a \in R$, $a+0 = a$.
* Inverses (Negatives): For every $a \in R$, there exists an element $-a \in R$ such that $a + (-a) = 0$.
* Commutativity: For all $a, b \in R$, $a+b = b+a$.