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[parent] zero times an element is zero in a ring (Theorem)
Lemma 1   Let $ R$ be a ring with zero element 0 (i.e. 0 is the additive identity of $ R$). Then for any element $ a\in R$ we have $ 0\cdot a = a\cdot 0 = 0$.
Proof.
$\displaystyle 0\cdot a$ $\displaystyle =$ $\displaystyle (0+0)\cdot a ,$    by definition of zero  
  $\displaystyle =$ $\displaystyle 0\cdot a + 0\cdot a,$    by the distributive law  

Thus $ 0\cdot a=0\cdot a + 0\cdot a$. Let $ b$ be the additive inverse of $ 0\cdot a \in R$. Hence:
$\displaystyle b+0\cdot a=b+(0\cdot a + 0\cdot a)$      
$\displaystyle (b+0\cdot a)=(b+0\cdot a) + 0\cdot a$      
$\displaystyle 0=0 + 0\cdot a$      
$\displaystyle 0=0\cdot a$      

as claimed. The proof of $ a\cdot 0=0$ is done analogously. $ \qedsymbol$



"zero times an element is zero in a ring" is owned by alozano.
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See Also: minus one times an element is the additive inverse in a ring, absorbing element

Other names:  $0\cdot a=0$

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Cross-references: inverse, identity, additive, zero element, ring
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This is version 5 of zero times an element is zero in a ring, born on 2004-03-09, modified 2006-03-09.
Object id is 5673, canonical name is 0cdotA0.
Accessed 5066 times total.

Classification:
AMS MSC13-00 (Commutative rings and algebras :: General reference works )
 16-00 (Associative rings and algebras :: General reference works )
 20-00 (Group theory and generalizations :: General reference works )

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