# Unity in Subrings, Zero Divisors

• January 20th 2011, 09:15 AM
matt.qmar
Unity in Subrings, Zero Divisors
Hello!

I have two questions!

(1)

if $M$ is a subring of $N$, where $N$ has unity (let's call it $1_N$), and say $1_M$ is the unity of $M$. Now, I have shown that it is NOT necessarily true that $1_M = 1_N$

Then if $M$ and $N$ are both fields (every element is a unit) then I'm trying to show that it is necessarily true that $1_M = 1_N$.

I am thinking you just take any $m \in M$, then $m*m^{-1} = 1_M$, but since $m \in N$ as well, $m*m^{-1} = 1_N$. Is that legitimate reasoning? I'm not confident with my answer for some reason.

(2) Let $m,n \in R$ for a commutative ring $R$.

Suppose that $mn$ is a zero divisor in $R$. Prove that either $m$ or $n$ is a zero divisor.

So, Well, first what I did was use the defn of zero divisor (z.d.)
If $mn \neq 0$ is a z.d. then $\exists x \in R$ such that $x \neq 0$ and $mn(x) = 0$

Then, just by associativity, $m(nx) = 0$ so, well, can we assume that $nx \neq 0$ and $m \neq 0$? Cause we'd be done - $m$ is a z.d. ! By commutativity, just switch $m$ and $n$ and then we're done.

Any help appreciated! Thanks!!
• January 20th 2011, 01:41 PM
roninpro
Your proof for (1) looks okay. You just need to emphasize that inverses are unique.

You have a good idea for (2). To clean it up, you should note that neither $m$ nor $n$ should be zero, or else $mn$ is zero, which is not what we want. Then, you are correct in taking nonzero $x$ so that $mnx=0$ and $m(nx)=0$. So if $nx\ne 0$, then you are done, since $m$ will satisfy the zero divisor definition. But what happens if $nx=0$?
• January 21st 2011, 04:38 AM
topspin1617
I'd just like to point out that (1) is actually true whenever the rings are integral domains.

In the ring $M$, of course $1_M\cdot 1_M=1_M$. Also, since $1_M\in N$ and $1_N$ is the multiplicative identity of $N$, we get $1_M\cdot 1_N=1_M$.

Put these equalities together to get $1_M\cdot 1_M=1_M\cdot 1_N$. Then, since we are in an integral domain, we can "cancel" the $1_M$ from both sides to conclude $1_M=1_N$.