The Inverse Image of an Ideal by a Ring Homomorphism is an Ideal

Problem 411

Let $f:R\to R’$ be a ring homomorphism. Let $I’$ be an ideal of $R’$ and let $I=f^{-1}(I)$ be the preimage of $I$ by $f$. Prove that $I$ is an ideal of the ring $R$.

To prove $I=f^{-1}(I’)$ is an ideal of $R$, we need to check the following two conditions:

For any $a, b\in I$, we have $a-b\in I$.

For any $a\in I$ and $r\in R$, we have $ra\in I$.

Let us first prove condition 1. Let $a, b\in I$. Then it follows from the definition of $I$ that $f(a), f(b)\in I’$.
Since $I’$ is an ideal (and hence an additive abelian group) we have $f(a)-f(b)\in I’$.
Since $f$ is a ring homomorphism, it yields that
\[f(a-b)=f(a)-f(b)\in I’.\]
Thus we have $a-b \in I$, and condition 1 is met.
This implies that $I$ is an additive abelian group of $R$.

Next, we check condition 2. Let $a\in I$ and $r\in R$. Since $a\in I$, we have $f(a)\in I’$.
Since $I’$ is an ideal of $R’$ and $f(r)\in R’$, we have $f(r)f(a)\in I’$.
Since $f$ is a ring homomorphism, it follows that
\begin{align*}
f(ra)=f(r)f(a)\in I’,
\end{align*}
and hence $ra\in I$. So condition 2 is also met and we conclude that $I$ is an ideal of $R$.

Comment.

Instead of condition 1, we could have used

Condition 1′: For any $a, b\in I$, we have $a+b\in I$.

The reason is that condition 2 guarantee the existence of the additive inverses, and hence condition 1 and 2 are equivalent to condition 1′ and 2.

The Preimage of Prime ideals are Prime Ideals
Let $f: R\to R'$ be a ring homomorphism. Let $P$ be a prime ideal of the ring $R'$.
Prove that the preimage $f^{-1}(P)$ is a prime ideal of $R$.
Proof.
The preimage of an ideal by a ring homomorphism is an ideal.
(See the post "The inverse image of an ideal by […]

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 […]

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 […]

Characteristic of an Integral Domain is 0 or a Prime Number
Let $R$ be a commutative ring with $1$. Show that if $R$ is an integral domain, then the characteristic of $R$ is either $0$ or a prime number $p$.
Definition of the characteristic of a ring.
The characteristic of a commutative ring $R$ with $1$ is defined as […]

$(x^3-y^2)$ is a Prime Ideal in the Ring $R[x, y]$, $R$ is an Integral Domain.
Let $R$ be an integral domain. Then prove that the ideal $(x^3-y^2)$ is a prime ideal in the ring $R[x, y]$.
Proof.
Consider the ring $R[t]$, where $t$ is a variable. Since $R$ is an integral domain, so is $R[t]$.
Define the function $\Psi:R[x,y] \to R[t]$ sending […]

The Image of an Ideal Under a Surjective Ring Homomorphism is an Ideal
Let $R$ and $S$ be rings. Suppose that $f: R \to S$ is a surjective ring homomorphism.
Prove that every image of an ideal of $R$ under $f$ is an ideal of $S$.
Namely, prove that if $I$ is an ideal of $R$, then $J=f(I)$ is an ideal of $S$.
Proof.
As in the […]

[…] $f^{-1}(I_k)$ of the ideal $I_k$ by a ring homomorphism is an ideal of $R$. (See the post “The inverse image of an ideal by a ring homomorphism is an ideal” for a […]

[…] preimage of an ideal by a ring homomorphism is an ideal. (See the post “The inverse image of an ideal by a ring homomorphism is an ideal” for a […]

## 2 Responses

[…] $f^{-1}(I_k)$ of the ideal $I_k$ by a ring homomorphism is an ideal of $R$. (See the post “The inverse image of an ideal by a ring homomorphism is an ideal” for a […]

[…] preimage of an ideal by a ring homomorphism is an ideal. (See the post “The inverse image of an ideal by a ring homomorphism is an ideal” for a […]