Under what condition does an element $a \in R$ become zero in $S^{-1}R$ ?

Q: Under what condition does an element a \in R become zero in S^{-1}R?

An element $a \in R$ becomes zero in the localized ring $S^{-1}R$ if there exists an element $s \in S$ such that the product $sa = 0$. This happens because the fraction $a/1$ can be rewritten as $sa/s$, which equals $0/s$, and thus $0$ in $S^{-1}R$.

An element

a \in R

becomes zero in the localized ring

S^{-1}R

if there exists an element

s \in S

such that the product

sa = 0

. This happens because the fraction

a/1

can be rewritten as

sa/s

, which equals

0/s

, and thus

0

S^{-1}R

Conditions

$R$ is a commutative ring.
$S \subset R$ is multiplicatively closed.
$a \in R$ and $s \in S$ .
$sa = 0$ .

Reasoning, step by step

Assume there is an element $a \in R$ and $s \in S$ such that $sa = 0$ .
Write the image of $a$ in $S^{-1}R$ as the fraction $a/1$ .
Multiply the numerator and denominator by $s$ to get $sa/s$ .
Substitute $sa = 0$ to get $0/s$ .
Conclude that $0/s$ is the zero element in $S^{-1}R$ , so $a$ maps to $0$ .

Example

The board develops the consequence step by step: $\frac{a}{1} = \frac{sa}{s} = \frac{0}{s} = 0$ , hence $a=0$ in $S^{-1}R$ . The speaker concludes "a has to be set to zero in the new ring".

Common misconceptions

Assuming that if $a \in R$ is nonzero, its image $a/1$ in $S^{-1}R$ must remain nonzero.
Thinking that localization never collapses nonzero elements.

Watch the explanation

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Under what condition does an element $a \in R$ become zero in $S^{-1}R$ ?

Conditions

Reasoning, step by step

Example

Common misconceptions

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Connected concepts

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Under what condition does an element a∈Ra \in Ra∈R become zero in S−1RS^{-1}RS−1R?

Conditions

Reasoning, step by step

Example

Common misconceptions

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Under what condition does an element $a \in R$ become zero in $S^{-1}R$ ?