# Thread: Algebra over a field - zero element

1. ## Algebra over a field - zero element

A is an algebra over a field F.

Let 'a' in A have the following property:
a.b = 0 for all 'b' in A.

Does this imply a=0? If yes why?
Thanks,
Aman

PS: I was trying the above for a ring. I am kind of convinced that in rings 'a' need not be equal to 0. Haven't been able to find a concrete example though.

2. Originally Posted by aman_cc
A is an algebra over a field F.

Let 'a' in A have the following property:
a.b = 0 for all 'b' in A.

Does this imply a=0? If yes why?
Thanks,
Aman

PS: I was trying the above for a ring. I am kind of convinced that in rings 'a' need not be equal to 0. Haven't been able to find a concrete example though.
It is true in both fields and rings with identity that "if ab= 0 for all b in A then a= 0". Just take a times the identity, i. ai= a= 0. To find a counter-example, you would need to look at rings without identity.

In a field, it is true that if ab= 0 for some non-zero b (not necessarily "all" b) then a= 0. If b is non-zero, it has an inverse, so $abb^{-1}= a= 0b^{-1}= 0$.

3. Thanks. But what about a more general case? Fields will always have 1. So they are not a problem.

What about 1. Algebra 2. Ring without and multiplicative identity,1.

4. Originally Posted by aman_cc
Thanks. But what about a more general case? Fields will always have 1. So they are not a problem.

What about 1. Algebra 2. Ring without and multiplicative identity,1.
Guess for rings following example will do:
R = {0,2}

+ is defined as (a+b) mod 4
. is defines as (a.b) mod 4

Under these 2.a = 0 for all a. Yet 2 =/= 0.
Is this correct?

Can you help for Algebra, A in the original question?

5. Originally Posted by aman_cc
Guess for rings following example will do:
R = {0,2}

+ is defined as (a+b) mod 4
. is defines as (a.b) mod 4

Under these 2.a = 0 for all a. Yet 2 =/= 0.
Is this correct?

Can you help for Algebra, A in the original question?
your example is correct. an example for algebras: consider any vector space $A \neq (0)$ over a field F and define multiplication in A by $xy=0,$ for all $x,y \in A.$

6. "for an algebra A over a field F, note that A contains (a copy) of F (in its center) and thus A always has an identity element"

Haven't really understood the above point?
Thanks

7. Originally Posted by aman_cc
"for an algebra A over a field F, note that A contains (a copy) of F (in its center) and thus A always has an identity element"

Haven't really understood the above point?
Thanks
sorry, i made a mistake! what i wrote is correct if $A$ has an identity element. then $A$ contains a copy of $F$ in its center and $1_A=1_F.$ now see my previous post that i just edited.