# commutative ring to boolean algebra

Let X be a set. We know that $$(P(X),\triangle, \cap )$$ is a commutative ring with the zero-element $$\emptyset$$ and the one-element $$X$$. $$P(X)$$ is the power set, $$\triangle$$ the symmetrical difference (as addition) and $$\cap$$ the intersection (as multiplication).

Now we define scalar multiplication as: $$\mathbb{F}_2 \times P(X) \rightarrow P(X)$$
$$0 \cdot A = \emptyset$$
$$1 \cdot A = A$$ for every $$A \in P(X)$$

prove that because of the scaler multiplication, $$(P(X),\triangle, \cap, \cdot )$$ is an algebra over $$\mathbb{F}_2$$

Im pretty much lost on this one. I looked in the script and on the internet for clues on how to solve this, but Im pretty desperate, so Im posting this without any progress. I would appreciate if you could give me any hints on how to get started with this

• Start with an example. E.g. what happens if $X$ has only two elements? Work out the small examples to get an idea of what's happening. Feb 13, 2021 at 23:15
• Write down the definitions of the concepts in the question, specifically the concept of an algebra over a field. Then use those definitions to write down what it is that you have to prove: it will be a set of equations about formulas built using $\Delta$, $\cap$ and $\cdot$. Then use properties you know to prove those equations. Feb 13, 2021 at 23:21

1. You know that $$(P(X),\triangle,\cap)$$ is a commutative ring;
2. Show that for every $$a,b\in\mathbb{F}_2$$ and $$A\in P(X)$$, \begin{align*} (a+b)\cdot A&=a\cdot A+b\cdot A; \tag{1}\\ a\cdot(b\cdot A)&=(ab)\cdot A.\tag{2} \end{align*}
3. Show that for every $$a\in\mathbb{F}_2$$ and $$A,B\in P(X)$$, \begin{align*} a\cdot(A\triangle B)&=(a\cdot A)\triangle(a\cdot B); \tag{3}\\ a\cdot(A\cap B)&=(a\cdot A)\cap B=A\cap(a\cdot B). \tag{4} \end{align*}
Since there are only two elements in $$\mathbb{F}_2$$, you simply exhaust all possible cases.