Every element in a ring different from 0 is invertible

Question

True or false.All number r in a ring R, different from 0 is same.

It seems like it is true but how to go about proof if we consider Matrix then it is invertible if determinant is not 0 .

Answer 1

This is not true, as then it would imply that every ring in which $0\neq 1$ is a field.

Simple examples include $\mathbb{Z}$, but we can go further to show just how badly the conjecture can fail by showing that there are rings containing zero divisors (i.e. elements $a\neq 0$ such that there exists $b\neq 0$ where $ab=0$):

As an example, consider the ring $\mathbb{Z}/n\mathbb{Z}$ equipped with addition modulo $n$ and multiplication modulo $n$, where $n$ is composite.

As another simple example, take the ring of all $n\times n$ matrices over any field, and consider the matrix where the top left corner is $1$ and the rest of the entries are $0$ (e.g. for $n=2$ this would be $\left[ \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right]$). This has determinant zero and thus is not invertible, but is non-zero.

Answer 2

As an another good example in addition to Hayden's response, one could just consider the ring of integers: $R=\mathbb{Z}$. The only invertible elements in this ring are $1$ and $-1$. To make this clear, suppose $a\in \mathbb{Z}$ is invertible. then there is some $a^{-1}\in \mathbb{Z}$ such that $aa^{-1}=a^{-1}a=1$. But $aa^{-1}=1$ implies that $a^{-1}=\frac{1}{a}\in \mathbb{Q}$, clearly $a^{-1}$ belongs to $\mathbb{Z}$ if and only if $a=1$ or $-1$.