commutative ring to boolean algebra

Question

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

Answer 1

Follow the axioms of algebras:

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.