Multiplication by zero in an algebra over a field: $0x=0$ for every $x$?

Question

If I have an algebra $A$ over a field $F$ and the zero element is $0\in A$. Is it true that $x0=0x=0$ for every $x\in A$?

Thanks a lot!

Yes. This holds in all rings, and follows from the distributive law and cancellation in the additive group: $$0x+0x=(0+0)x=0x.$$ — Jyrki Lahtonen, Aug 11 '14 at 07:02
@Jyrki i.e. $,0,$ is the unique additive idempotent in an additive group. But $,0x,$ remains idempotent, being the image of $,0,$ under a group hom $, r\mapsto rx.,$ Thus $,0x = 0,$ by uniqueness. — Bill Dubuque, Aug 11 '14 at 16:08

Yotam Vaknin · Answer 1 · 2018-08-18T21:37:22.190

5

Another possible proof (without assuming $(-1)x = -x$) :

$$ 0x = (0+0)x = 0x + 0x $$

So $0x$ is the additive identity, making it $0x=0$.

edited Aug 18 '18 at 21:37

answered Nov 21 '16 at 22:33

Yotam Vaknin

score 1 · Accepted Answer · answered Aug 11 '14 at 07:01

1

Your statement is true, because $A$ is a ring and it is true in rings:

$$0x = (1 - 1)x = x-x = 0$$

answered Aug 11 '14 at 07:01

5xum

5

How do you know that $(-1)x = -x$? Typically it's proven by proving first that $0x = 0$... – Najib Idrissi Aug 11 '14 at 08:11

2 Answers2