Exercise 2.2.6 (Subring of ring that is an Ideal is same ring )

Question

Prove that the only subring of a ring $R$ that is also an
ideal is $R$ itself.

Hassaan Naeem · Accepted Answer

\paragraph{Solution:} We know that $I$ is an ideal of $R$ if:\
    \begin{equation*}
        \begin{aligned}
            &(I, +) \le (R, +)\quad  	ext{[I is additive subgroup of R]}\
            & \&\ \forall r \in R, x \in I:\
            & (1)\ r \cdot x \in I\
            & (2)\ x \cdot r \in I
        \end{aligned}
    \end{equation*}

We know that a subring $S$ of $R$ is a subset $S \subseteq R$ containing $0$ and $1$.\
    \Therefore if we take $S$ to be an ideal aswell then:
    \begin{equation*}
        \begin{aligned}
            &(S, +) \le (R, +)\
            & \&\ \forall r \in R, s \in S:\
            & (1)\ r \cdot s \in S\
            & (2)\ s \cdot r \in S
        \end{aligned}
    \end{equation*}
    But we know that $1 \in S$. Therefore, $\forall r \in R,$ $(1)\ 1 \cdot r = r \in S$ and $(2)\ r \cdot 1 = r \in S$\
    Therefore $\forall r \in R, r \in S \implies S = R \quad \square$

Exercise 2.2.6 (Subring of ring that is an Ideal is same ring )

Answers

Comments

Exercise 2.2.6 (Subring of ring that is an Ideal is same ring )

Answers

Comments

Add answer