Let $A$ be a commutative Banach algebra and let $x \in A$. We define the Gelfan transform of $x$ by $$\hat{x} (\chi)= \chi (x)$$ where $\chi$ is a nonzero multiplicative linear functional on $A$.
I need to show that $x \mapsto\hat{x}$ is a morphism, i.e if we define $$F(x)=\hat{x}$$ then we need to show that $$F(xy)= \hat{xy}=\hat{x}\hat{y}=F(x)F(y)$$
How can I go about showing this?
Given a (nonzero) multiplicative linear functional $\chi$ on $A$ and $x,y\in A$, we have \begin{align*} \hat{xy}(\chi)&=\chi(xy)=\chi(x)\chi(y)=\hat{x}(\chi)\hat{y}(\chi) \\ \hat{(x+y)}(\chi)&=\chi(x+y)=\chi(x)+\chi(y)=\hat{x}(\chi)+\hat{y}(\chi) \end{align*} Furthermore, for any scalar $\alpha$, we have $$ \hat{(\alpha x)}(\chi)=\chi(\alpha x)=\alpha\chi(x)=a\hat{x}(\chi) $$
Since $\chi$ was arbitrary, we see that $$F(xy)=F(x)F(y),\qquad F(x+y)=F(x)+F(y), \qquad F(\alpha x)=\alpha F(x).$$