It seems like a very stupid question, but I just cannot solve it...
Suppose $A,~B$ are not zero. Prove or disprove $A\oplus B$ is not isomorphic to $A$.
Furthermore, we can ask the similar question:
Suppose $A,~I$ are not zero. Prove or disprove $A\oplus I$ is not isomorphic to $A$.
Here $A$ is a ring and $I$ is an ideal of $A$. $A\oplus I$ is the direct sum of $A$-modules.
I encounter this problem when I was trying to solve the ex 2.28 on Atiyah-Macdonald:
If a local ring is absolutely flat, then it is a field.
Take $0\neq x$ and we have a maximal ideal $(x)\subset\mathfrak{m}$. According to some results from Atiyah-Macdonald, we can find an idempotent element $e\in \mathfrak{m}$ such that $(e)=(x)$ and $(e)\oplus (1-e)=A$. Also, the Jacobson radical of $A$ is $\mathfrak{m}$ so $1-e$ is unit. Then we have $$(e)\oplus A=A$$ I want to conclude $e=0$ so that $x=0$. Then $\mathfrak{m}$ has to be $0$. Hence $A$ is a field.
But I don't know how to deduce $e=0$ from $(e)\oplus A=A$, which is intuitively true to me...
The statement doesn't hold in general because it can happen like this: Let $A=\prod_{i=1}^\infty F$ and consider the ideal $I$ which is the set of elements that are zero outside of the first coordinate. Then $A\cong A\oplus I$.
There are much easier ways to prove an absolutely flat local ring is a field. For one thing, a local ring only has trivial idempotents, so if your ideal $I=(e)$, then $I=\{0\}$ or $I=A$. The ring only has trivial ideals then, and hence it is a field.