We prove the equivalences $(1) \Leftrightarrow (2)$ and $(2) \Leftrightarrow (3)$.
$(1) \implies (2)$
Suppose that $R$ is a field. Let $I$ be an ideal of $R$.
If $I=(0)$, then there is nothing to prove.
So assume that $I\neq (0)$.
Then there is a nonzero element $x$ in $I$.
Since $R$ is a field, we have $x^{-1}\in R$.
Since $I$ is an ideal, we have
\[1=x^{-1}\cdot x\in I.\]
This yields that $I=R$.
$(2) \implies (1)$
Suppose now that the only ideals of $R$ are $(0)$ and $R$.
Let $x$ be a nonzero element of $R$. We show the existence of the inverse of $x$.
Consider the ideal $(x)=xR$ generated by $x$.
Since $x$ is nonzero, the ideal $(x)\neq 0$, and thus we have $(x)=R$ by assumption.
Thus, there exists $y\in R$ such that
\[xy=1.\]
So $y$ is the inverse element of $x$.
Hence $R$ is a field.
$(2)\implies (3)$
Suppose that the only ideals of $R$ are $(0)$ and $R$.
Let $S$ be any ring with $1$ and $f:R\to S$ be any ring homomorphism.
Consider the kernel $\ker(f)$. The kernel $\ker(f)$ is an ideal of $R$, and thus $\ker(f)$ is either $(0)$ or $R$ by assumption.
If $\ker(f)=R$, then the homomorphism $f$ sends $1\in R$ to $0\in S$, which is a contradiction since any ring homomorphism between rings with $1$ sends $1$ to $1$.
Thus, we must have $\ker(f)=0$, and this yields that the homomorphism $f$ is injective.
$(3) \implies (2)$
Suppose that statement 3 is true. That is, any ring homomorphism $f:R\to S$, where $S$ is any ring with $1$, is injective.
Let $I$ be a proper ideal of $R$: an ideal $I\neq R$.
Then the quotient $R/I$ is a ring with $1$ and the natural projection
\[f:R\to R/I\]
is a ring homomorphism.
By assumption, the ring homomorphism $f$ is injective, and hence we have
\[(0)=\ker(f)=I.\]
This proves that the only ideals of $R$ are $(0)$ and $R$.
A Maximal Ideal in the Ring of Continuous Functions and a Quotient Ring
Let $R$ be the ring of all continuous functions on the interval $[0, 2]$.
Let $I$ be the subset of $R$ defined by
\[I:=\{ f(x) \in R \mid f(1)=0\}.\]
Then prove that $I$ is an ideal of the ring $R$.
Moreover, show that $I$ is maximal and determine […]
Determine the Quotient Ring $\Z[\sqrt{10}]/(2, \sqrt{10})$
Let
\[P=(2, \sqrt{10})=\{a+b\sqrt{10} \mid a, b \in \Z, 2|a\}\]
be an ideal of the ring
\[\Z[\sqrt{10}]=\{a+b\sqrt{10} \mid a, b \in \Z\}.\]
Then determine the quotient ring $\Z[\sqrt{10}]/P$.
Is $P$ a prime ideal? Is $P$ a maximal ideal?
Solution.
We […]
The Quotient Ring by an Ideal of a Ring of Some Matrices is Isomorphic to $\Q$.
Let
\[R=\left\{\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \quad \middle | \quad a, b\in \Q \,\right\}.\]
Then the usual matrix addition and multiplication make $R$ an ring.
Let
\[J=\left\{\, \begin{bmatrix}
0 & b\\
0& 0
\end{bmatrix} […]
Generators of the Augmentation Ideal in a Group Ring
Let $R$ be a commutative ring with $1$ and let $G$ be a finite group with identity element $e$. Let $RG$ be the group ring. Then the map $\epsilon: RG \to R$ defined by
\[\epsilon(\sum_{i=1}^na_i g_i)=\sum_{i=1}^na_i,\]
where $a_i\in R$ and $G=\{g_i\}_{i=1}^n$, is a ring […]
Prove the Ring Isomorphism $R[x,y]/(x) \cong R[y]$
Let $R$ be a commutative ring. Consider the polynomial ring $R[x,y]$ in two variables $x, y$.
Let $(x)$ be the principal ideal of $R[x,y]$ generated by $x$.
Prove that $R[x, y]/(x)$ is isomorphic to $R[y]$ as a ring.
Proof.
Define the map $\psi: R[x,y] \to […]
There is Exactly One Ring Homomorphism From the Ring of Integers to Any Ring
Let $\Z$ be the ring of integers and let $R$ be a ring with unity.
Determine all the ring homomorphisms from $\Z$ to $R$.
Definition.
Recall that if $A, B$ are rings with unity then a ring homomorphism $f: A \to B$ is a map […]
Every Prime Ideal of a Finite Commutative Ring is Maximal
Let $R$ be a finite commutative ring with identity $1$. Prove that every prime ideal of $R$ is a maximal ideal of $R$.
Proof.
We give two proofs. The first proof uses a result of a previous problem. The second proof is self-contained.
Proof 1.
Let $I$ be a prime ideal […]