Basic Ring Theory - 1

[Tsakanikas Nickos]

Let \( \displaystyle A \) be a non-zero associative and commutative ring with identity \( \displaystyle 1_{A} \). Show that the following are equivalent:

1. \( \displaystyle A \) is a field.
2. The only ideals in \( \displaystyle A \) are \( \displaystyle (0) \) and \( \displaystyle (1) \).
3. Every homomorphism of \( \displaystyle A \) into a non-zero ring B (commutative,associative,with identity) is injective.

[Papapetros Vaggelis]

\(\displaystyle{i)\implies ii)}\) : Suppose that the commutative ring \(\displaystyle{(A,+,\cdot)}\)

is a field. Obviously, \(\displaystyle{(0)=\left\{0\right\}\,,(1)=A}\) are ideals of \(\displaystyle{(A,+,\cdot)}\) .

Let \(\displaystyle{I\neq \left\{0\right\}}\) be an ideal of \(\displaystyle{(A,+,\cdot)}\) .

There exists \(\displaystyle{x\in A}\) such that \(\displaystyle{x\in I}\) and \(\displaystyle{x\neq 0}\).

Since the ring \(\displaystyle{(A,+,\cdot)}\) is a field, we have that \(\displaystyle{x\,x^{-1}=x^{-1}\,x=1}\),

so : \(\displaystyle{1=x\,x^{-1}\in I\implies I=(1)=A}\) .

\(\displaystyle{ii)\implies i)}\) : Let \(\displaystyle{x\in A-\left\{0\right\}}\) . Then, \(\displaystyle{(x)}\)

is an ideal of \(\displaystyle{(A,+,\cdot)}\) and according to \(\displaystyle{ii)}\), we have that

\(\displaystyle{(x)=R}\) (since \(\displaystyle{x\neq 0}\)) . So,

\(\displaystyle{1\in R\implies 1\in (x)\implies \exists\,y\in A: 1=x\,y=y\,x}\), which means that \(\displaystyle{x}\)

is invertible. Therefore, the ring \(\displaystyle{(A,+,\cdot)}\) is a field.

\(\displaystyle{ii)\implies iii)}\) : Let \(\displaystyle{f:A\longrightarrow B}\) be a ring homomorphism .

The set \(\displaystyle{\rm{Ker}(f)}\) is an ideal of \(\displaystyle{(A,+,\cdot)}\) and according

to the hypothesis, we get : \(\displaystyle{\rm{Ker}(f)=\left\{0\right\}}\) or \(\displaystyle{\rm{Ker}(f)=A}\) .

If \(\displaystyle{\rm{Ker}(f)=A}\), then \(\displaystyle{f=\mathbb{O}}\), a contradiction, since

\(\displaystyle{f(1)=1}\) and \(\displaystyle{1_{B}\neq 0_{B}}\) .

So, \(\displaystyle{\rm{Ker}(f)=\left\{0\right\}}\) and \(\displaystyle{f}\) is injective.

\(\displaystyle{iii)\implies ii)}\) Let \(\displaystyle{I\neq \left\{0\right\}}\) be a-non zero ideal of \(\displaystyle{(A,+,\cdot)}\) .

Suppose that \(\displaystyle{I\neq A}\) .

Consider the natural ring homomorphism \(\displaystyle{f:A\longrightarrow A/I\,,x\mapsto x+I}\)

which is also onto \(\displaystyle{A/I}\) . Obviously, \(\displaystyle{A/I\neq \left\{I\right\}}\)

because \(\displaystyle{1+I\neq I}\) .

Since \(\displaystyle{iii)}\) is true, we have that the

function \(\displaystyle{f}\) is injective, that is \(\displaystyle{I=\rm{Ker}(f)=\left\{0\right\}}\),

a contradiction, so \(\displaystyle{I=A}\) .

[Tsakanikas Nickos]

Here is (iii) \( \implies \) (i):

Let \( x \) be a non-invertible element of \( A \). Then \( \displaystyle (x) \neq (1) \), which means that the ring \( B = A / (x) \) is non-zero. Consider the natural epimorphism \( \displaystyle \phi \ \colon A \longrightarrow B \) with kernel \( (x) \). By hypothesis, \( \phi \) is injective, so \( \displaystyle Ker(\phi)=(x)=(0) \). Therefore \( x=0 \). This implies that \( A \) is a field.