Which elements in a direct product $\prod _{\lambda \in \Lambda} R_\lambda$ of rings are atoms? Prove your answer. (Assume this is a commutative ring)
My Answer: Atoms of $\prod R_\lambda$ are elements with all coordinates being units except $1$ that's an atom.
Proof: I need some help finishing it- My plan is to show that all coordinates are units $\implies$ it's not a unit $\implies$ it's not an atom.
If f has at least 2 coordinate non-units, say $\alpha \neq \beta $, then write $f \nmid g$ since $f(\beta)$ is not a unit and $f \nmid h$ since $f(\alpha)$ is not a unit. Therefore, f is not an atom.
Now let f be an atom of $\prod R_\lambda$. All coordinates except $1$, say $f(\alpha)$ are units. If $f(\alpha)=xy$ then $f=gh$ where $g(\lambda)=f(\lambda)$ for $\lambda \neq \alpha$, $g(\alpha)=x$ and $h(\lambda)=1$ for $\lambda \neq \alpha$, $h(\alpha)=y$. So $(f)=(g)$ of $(f)=(h)$. Either $(f(\alpha))=(x)$ or $(f(\alpha))=(y)$. Say $f=gh$ and $\alpha$ is the coordinate f is not a unit in. Hence, $g$ and $h$ have to be units. So $f(\alpha)=g(\alpha)h(\alpha)$ so $g(\alpha)=(f)$ or $h(\alpha)=(f)$, without loss of generality, say $g(\alpha)=(f)$....
I'm just not sure where to go from here... any suggestions or corrections would be appreciated! Thanks.
The problem is that any ring that is a non-trivial product of rings will have non-trivial idempotent elements and in particular lots of zero-divisors. The notion of "atom" or "irreducible" is not so clear. There are several possible routes to take for what you define as an atom (all of which are equivalent in a domain, but none of which are equivalent in rings with zero-divisors). Let $a$ be a non-unit in a commutative ring with $1$.
Your intuition is indeed correct for the most part. For example, consider the following:
Theorem. Let $R=R_1 \times \cdots \times R_n$. Then $a=(a_1, \ldots , a_n)$ is an atom if and only if $a_i$ is an atom in $R_i$ for precisely one $1 \leq i \leq n$ and is a unit in all other coordinates.
Proof. Let $a$ be an atom. Suppose every coordinate is a unit. Then $a$ is a unit. A contradiction. Suppose there are at least two coordinates which are not units, after relabeling we may suppose $a_1, a_2$ are not units. Then $a=(a_1, 1, 1, \ldots, 1)(1, a_2, a_3, \ldots, a_n)$ and $a \not \sim (a_1, 1, 1, \ldots, 1)$ and $a \not \sim (1, a_2, a_3, \ldots, a_n)$ so $a$ is not an atom. So we can conclude that there must be precisely one coordinate which is not a unit.
We now need to see why this coordinate must be an atom in that factor. Suppose $a_1$ were not an atom in $R_1$. Then there is a factorization $a_1=bc$ with $a_1 \not sim b$ and $a_1 \not \sim c$. Then $a=(b, a_2, \ldots , a_n)(c, 1, \ldots, 1)$ is a factorization and $a \not \sim (b, a_2, \ldots , a_n)$ (look at first coordinate) and $a\not \sim (c, 1, \ldots, 1)$ (look at first coordinate). Hence the one non-unit coordinate must be an atom.
The converse is obvious.
Similar results hold for strong atoms and m-atoms where you simply replace atom in the above theorem with strong atom or m-atom.
It is interesting to note that it doesnt work for very strong atoms. For instance $R= \mathbb{Z} \times R_2 \cdots \times R_n$, $0 \in \mathbb{Z}$ is certainly very strongly atomic (it is a domain so all atoms are equivalent here). But $(0,1,1, \ldots ,1)$ is not very strongly atomic.
$$(0,1,1, \ldots ,1)=(0,1,1, \ldots ,1)(0,1,1, \ldots 1)$$ but $(0,1,1, \ldots ,1) \not \cong (0,1,1, \ldots ,1)$ since the above factorization gives $a=rb$ with $r$ not a unit.
The following paper by D.D. Anderson and S. Valdez Leon entitled Factorization in Commutative Rings with Zero-Divisors (http://projecteuclid.org/euclid.rmjm/1181072068 ) discusses this and other related questions in some detail.