# The image under a semigroup morphism of a right ideal is a right ideal.

This is part $$(c)$$ and part $$(e)$$ of Exercise 1.9.5 of Howie's "Fundamentals of Semigroup Theory".

Definition: A non-empty subset $$A$$ of a semigroup $$S$$ is a right ideal if $$AS\subseteq A$$.

## The Question:

Let $$\phi: S\to T$$ be a morphism, where $$S$$ and $$T$$ are semigroups.

Show that if $$\phi$$ is onto then the image under $$\phi$$ of a right ideal of $$S$$ is a right ideal of $$T$$.

Show that the hypothesis that $$\phi$$ is onto cannot be removed.

## My Attempt:

Let $$\emptyset\neq R\subseteq S$$ be a right ideal of $$S$$. Then $$RS\subseteq R.$$ We have for all $$r_1s\in RS$$ there exists $$r_2\in R$$ such that $$r_1s=r_2$$.

Let $$\phi: S\to T$$ be onto. Then for all $$t\in T$$ there exists $$\sigma\in S$$ such that $$\sigma\phi=t$$.

We have $$r_1\phi s\phi=r_2\phi$$.

What do I do now?

• If you take $s=\sigma$, you've proved that the inage of a right ideal absorbs from the right every element, because $t$ was arbitrary, so you are done. – vap Apr 18 '17 at 11:03
You want to show that $\phi(R) T \subseteq \phi(R)$.
So take any element of $\phi(R)T$, that is, an element of the form $\phi(r)t$, with $r \in R$ and $t \in T$.
As $\phi$ is onto, $t=\phi(s)$ for some $s \in S$.
Thus, $\phi(r)t = \phi(r) \phi(s) = \phi(rs) \in \phi(R)$.
It suffices to observe that since $RS \subseteq R$, one gets $\phi(RS) \subseteq \phi(R)$ and since $\phi$ is onto, $\phi(RS) = \phi(R)\phi(S) = \phi(R)T \subseteq \phi(R)$. Thus $\phi(R)$ is a right ideal.