commutative ring with id without non-trivial ideals is a field. Why?

Question

Firstly, I dont have any intuition to this exercize. I mean let look at R. It is a field, despite the fact that there are a lot of nn-trivial ideals. So from first look, I dont see reason, why non-trivial ideals, or in other words, why the third unique characteristic (besides the two who are true for sub-group ether) makes a diffrence ?

And as I under, the R field do has non-trivial ideas, like 2Z, 3Z and so on ...

Answer 1

Every non-zero element in a field is invertible. Hence, there are no non-trivial ideals in a field.

Conversely, if the ring $A$ has no non-trivial ideals, it follows that every non-zero element in $A$ is invertible, making $A$ a field.

Answer 2

Let $x$ be a non zero element of your ring $R$. Since $xR$ is not a non-trivial ideal and it is not zero (because it contains $x1=x$), it must be equal to $R$. In particular, as $1\in R$, there exists a $y\in R$ such that $xy=1$. It follows that $x$ is invertible.

Answer 3

Let us remember the definition.

A field is a ring that any element without $0$ is invertible. So real numbers $ \mathbb{Z}$ and complex numbers$ \mathbb{C}$ are fields. They don't have non-trivial ideals.

Let $R$ be a commutative ring without non-trivial ideals and $a \in R$ be non-zero element.

Then $(a)＝R$ and $1 \in R$. Therefore there exists $b$ such that $a \cdot b =1$.

So $a$ is invertible and $R$ is a field.