A ring without multiplicative identity in which the only ideals are (0) and the whole ring

Question

Could you give me an example of a ring A without multiplicative identity in which the only ideals are (0) and the whole ring A?

The example of ring A can be either non-commutative or commutative..

Answer 1

You can do this with a $2$-element ring, namely the subring $\{0,2\}$ of $\Bbb Z/4\Bbb Z$, aka, $2\Bbb Z/4\Bbb Z$.

Does that work?

Answer 2

Proffering the following. Let $V$ be the space of sequence of real numbers (any field would work the same) with basis $e_i, i\in\Bbb{N}$. Let $R$ be the set of such linear transformations $T$ that $T(e_i)=0$ for all but finitely many $i$. So basically $R$ consists of $\infty\times\infty$ matrices with only finitely many non-zero rows and columns. In yet another way, $R$ is the linear span of the transformations $T_{i,j}$ determined by $T_{i,j}(e_k)=\delta_{ik}e_j$ (imagine a matrix with a single entry equal to one and the rest equal to zero).

• $R$ is a rng under addition and composition of transformations. Leaving this as an exercise.
• $R$ has no multiplicative identity. This is basically because $id_V\notin R$. No other transformation works as a neutral element for all the $T_{i,j}$s.
• $R$ has no non-trivial 2-sided ideals. If you start with a non-zero matrix $A$, you can pre- and postmultiply it with suitable matrices to produce one with a single non-zero entry. You can then pre- and postmultiply it with finitely supported permutation matrices to move that single non-zero entry anywhere you want. Therefore any ideal containing $A$ must contain all the transformations $T_{i,j}$, and hence be equal to $R$.