I am working on excercise 1.9.19 of Howie's “Fundamentals of Semigroup Theory”:
Let I, J be ideals of a semigroup S. Show that I$\cap$J and I$\cup$J are ideals of S.
I am really struggling with proving this part of the question.
From the books properties of an ideal we have:
A non-empty subset I of S is called a left ideal if SI$\subseteq$I.
The right ideal is defined dually and an ideal has both.
I am wondering if it is as simple as:
(I $\cap$ J)S $\subseteq$ IS $\cap$ JS $\subseteq$ I $\cap$ J
and the equivalent for the union?
We then need to go on to prove:
Show also that $$(I\cup J)/J \cong I/(I\cap J)$$
For this I have seen similar proofs that start with a homomorphism from I to (I$\cup$J)/J and use properties of the image and kernal to arrive at an isomorphism but this also uses a theorem which I can't find within this book and would rather prove it using what I have.
Hint. If $I$ is an ideal of a semigroup $S$, then the semigroup $S/I$ can be represented as $(S \setminus I) \cup \{0\}$ where all products not falling in $S \setminus I$ are zero. Now just apply this construction to $(I\cup J)/J$ and to $I/(I\cap J)$ and see what's happen.