Let $R$ be a ring with identity and $f$ be a ring homomorphism from $R$ to a ring $S$. Is it true that the left ideals of the ring $f(R)$ are exactly of the form $f(I)$, where $I$ is a left ideal of $R$?
My try:
If $I$ is a left ideal of $R$, then clearly $f(I)$ is a left ideal of $f(R)$. Conversely, let $K$ be a left ideal of $f(R)$, and put $T=\{p\in R \mid f(p)\in K\}$. Then $f(T)=K$. Now, $T$ is a left ideal of $R$, because for any $p,p_1,p_2\in T$ and $r\in R$ we have $$f(p_1+p_2)=f(p_1)+f(p_2)\in K,$$ and $$ f(rp)=f(r)f(p)\in K.$$ Is this argument valid?
Thanks for any suggestion!
Incomplete, I'd say. You prove that, if $K$ is a left ideal of $f(R)$, then $T=\{p\in R\mid f(p)\in K\}$ is a left ideal of $R$, but you fail to show that $K=f(T)$.
The part “if $I$ is a left ideal of $R$, then $f(I)$ is a left ideal of $f(R)$” also requires a verification (there is a subtle point to underline), but it is easy.
Answering to the comment, it is true that the correspondence is only surjective, because $f(I+\ker f)=f(I)$, so all left ideals $J$ with $I\subseteq J\subseteq I+\ker f$ have the same image $f(I)$. Nonetheless, the problem only required showing surjectivity.
Here's a proof. First, it's not restrictive to assume $f$ is surjective. If $I$ is a left ideal of $R$, then $f(I)$ is a left ideal of $S$. Indeed, if $x,y\in f(I)$ and $s\in S$, then $x=f(a)$, $y=f(b)$ and $s=f(r)$, for some $a,b\in I$ and $c\in R$. Therefore $x+y=f(a+b)\in f(I)$ and $sa=f(ra)\in f(I)$.
Now, let $J$ be a left ideal of $S$ and set $I=f^{-1}(J)=\{r\in R:f(r)\in J\}$. Then $I$ is a left ideal of $R$. Indeed, if $a,b\in I$ and $r\in R$, then $f(a+b)=f(a)+f(b)\in J$ and $f(ra)=f(r)f(a)\in J$; hence $a+b\in I$ and $ra\in I$.
Since $f$ is surjective, $f(I)=f(f^{-1}(J))=J$, so the thesis is proved.