# Ideals of a Semigroup - Exercise 1.9.19 of Howie's “Fundamentals of Semigroup Theory”.

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.

• So wha are the properties of an ideal and which of these are you struggling to show? – Mark Bennet Oct 30 '18 at 13:47

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.