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

Problems and solutions of ring theory in abstract algebra

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 $R$.

 
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Proof.

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

  1. For any $a, b\in I$, we have $a-b\in I$.
  2. 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.


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2 Responses

  1. 05/15/2017

    […] $f^{-1}(I_k)$ 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 […]

  2. 08/11/2017

    […] 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 […]

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