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Homorphisms and Ideals.

Let $\varphi:R \to S$ be a ring homomorphism.

Show that $\varphi ^{-1} ( \varphi(I)) = I+K$ where $K=\ker \varphi $. In particular, if $I$ contains $K$, then $\varphi ^{-1} \varphi (I)=I$.

$\varphi$ is a ring homomorphism, so by definition if I remember correctly $\forall r_1,r_2 \in R:$ $$\begin{aligned} \varphi(r_1*r_2)&=\varphi(r_1)\varphi(r_2), \\ \varphi(r_1+r_2)&=\varphi(r_1)+\varphi(r_2). \end{aligned} $$

Thinking that $I$ stands for ideal, that is by definition $$I= \{ rc :r \in R\}.$$

Definition of kernel: The kernel for $\varphi$ is $$K=\{r \in R: \varphi(r)=0_s \}.$$

From abstract1 (feels like 2 years ago) remember that if $\ker =\{ 0_s\}$, then $\varphi $ is injective.

Not sure where to take it from there or if I'm in the right track?

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    Your definition of ideal is incorrect; this only covers principal ideals. See https://en.wikipedia.org/wiki/Ideal_(ring_theory)#Definitions for the general definition. – 2017-01-30

3 Answers 3

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I'll assume $\phi=\varphi$.

$$\phi(x)\in \phi(I)\iff \exists a\in I ,\phi(x)=\phi(a)\iff \exists a\in I,x-a\in K\iff x\in I+K$$

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Hints:

$\varphi^{-1}(\varphi (I))$ contains $I$, , and it contains $K$ since $\varphi (I)$ contains $\{0\}$. Hence…

Conversely, if $x=i+k$, $i\in I,\; k\in K$, $\varphi(x)=\dots$

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Want to show $$\varphi^{-1} (\varphi(I)) = I+K.$$

Proof of "$\supset$": if $r \in I+K$, show that there exists $i \in I$ such that $\varphi(r)=\varphi(i)$. Then $r \in \varphi^{-1}(\varphi(I))$.

If $r \in I+K$ then it is of the form $r=i+k$ for $i \in I$ and $k \in K$. Then $\varphi(r)=\varphi(i) \in \varphi(I)$, so $r \in \varphi^{-1}(\varphi(I))$.

Proof of "$\subset$": if $r \in \varphi^{-1}(\varphi(I))$, then there exists $i \in I$ such that $\varphi(r)=\varphi(i)$. Then $r-i \in K$ so $r = i + (r-i) \in I+K$.