Unity in Subrings, Zero Divisors

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

I have two questions!

(1)

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

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

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

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

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

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

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

Any help appreciated! Thanks!!
• Jan 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 $\displaystyle m$ nor $\displaystyle n$ should be zero, or else $\displaystyle mn$ is zero, which is not what we want. Then, you are correct in taking nonzero $\displaystyle x$ so that $\displaystyle mnx=0$ and $\displaystyle m(nx)=0$. So if $\displaystyle nx\ne 0$, then you are done, since $\displaystyle m$ will satisfy the zero divisor definition. But what happens if $\displaystyle nx=0$?
• Jan 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 $\displaystyle M$, of course $\displaystyle 1_M\cdot 1_M=1_M$. Also, since $\displaystyle 1_M\in N$ and $\displaystyle 1_N$ is the multiplicative identity of $\displaystyle N$, we get $\displaystyle 1_M\cdot 1_N=1_M$.

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