# I Distinctiveness of the set of nxn matrices as a ring

1. Apr 5, 2017

### Mr Davis 97

So I know that in general, for the ring of $n \times n$ matrices, if $AB = 0$, then it is not necessarily true that $A=0$ or $B=0$. However, in other rings, for example the integers $\mathbb{Z}$, I know that this statement is true. So what property is the ring of matrices lacking such that it is not true in general?

2. Apr 5, 2017

### Staff: Mentor

A ring is called an integral domain, if $ab = 0 \Longrightarrow a=0 \vee b=0$. Elements $a$ for which there is an element $b$ with $a \cdot b = 0$ are called zero-divisors. So an integral domain is a ring without zero-divisors or more precisely: with $0$ as only zero-divisor. E.g. $\mathbb{Z}_6$ has also zero-divisors, namely $2$ and $3$. (I'm not quite sure, whether $0$ is excluded in the definition of a zero-divisor or not. It's not really important, but "has no zero-divisors" is a usual phrase, so $0$ is probably excluded.)

3. Apr 5, 2017

### Mr Davis 97

One more question. Why doesn't a subring necessarily have to have the same multiplicative identity as the bigger ring?

4. Apr 5, 2017

### Staff: Mentor

Do you have an example? I can only think of examples like $n\cdot \mathbb{Z} \subseteq \mathbb{Z}$ where the subring has none.
In general, a multiplicative identity doesn't always exist. The multiplicative structure of a ring doesn't need to define a group structure, but if rings are compared like ring and subring and both have a $1$, then it's usually required to be the same. At least the $1$ in the ring is also a $1$ in the subring, if it's included. If not, and the subring has another element as a $1$ it's getting a bit messy, since this one will act as a $1$ on certain elements of the ring as well - or you have completely different multiplicative structures, in which case one doesn't speak of a subring.

5. Apr 8, 2017

### willem2

How about {0,2,4} as a subring of Z6. The multiplicative identity of the subring is 4.

6. Apr 9, 2017

### WWGD

You can see this issue of AB=0 for matrices in terms of the fundamental theorem of linear algebra , re the orthogonality of the column space of A with the row space of B:

https://en.wikipedia.org/wiki/Fundamental_theorem_of_linear_algebra

Basically , if A has rows $a_1,a_2 ,..,a_n$ and B has columns $b_1, b_2,.., b_n$ then you have that
$a_ i. b_j =0$ for all $0 \leq i,j \leq n$ so that every row of A is orthogonal to every column of B . This implies, by linearity, that the row space of A is orthogonal to the column space of B and the column space of B is contained in the nullspace of A..