Prove that the vector space $\Phi ^ T$ is algebra on $\Phi$ I found an example in a mathematical journal for the field of functional analysis, for which I think might be interesting and perhaps important, and njekohsishte easier for all those involved in this field, ie. fields of functional analysis. The example is this:
For an arbitrary set $T\neq \phi$ vector space $\Phi ^ T$ of all functions $x:T\rightarrow \Phi$ together with multiplying $(x,y)\rightarrow xy$ defined with $$(xy)(t)=x(t)y(t), \text {where} (t\in T)$$
is algebra on $\Phi.$
Previously I thank all those who will have the patience to help me solve this example. Thanks for your help and your attention. 
 A: Well, the way you've pitched it, $\Phi^T$ is clearly a vector space over $\Phi$, since
for any $f:T \to \Phi$ and $\alpha \in \Phi$ we can define $(\alpha f)(t) = \alpha f(t)$ for all $t \in T$; likewise if $g \in \Phi^T$ as well, we can define $(f + g)(t) = f(t) + g(t)$, i.e., everything works pointwise for $t \in T$ since $\Phi$ satisfies the field axioms.  It's pretty clear that zero map $0_T(t) = 0 \in \Phi \, \text{for} \, t \in T$ acts as the additive identity in this vector space.  $\Phi^T$ is also pretty clearly a commutative ring; you've shown us the multiplication in your post and the addition is the same as in the vector space case.  Again by the "fieldness" of $\Phi$, all the ring axioms hold for $\Phi^T$.  Furthermore, the compatibility of the ring an vector space operations is also easily seen from the fact that $\Phi$ is a field and the operations are defined pointwise.  I think I've covered here, albeit in an informal manner, the essential features required for $\Phi^T$ to be considered an algebra over the field $\Phi$.
Hope this helps!  Cheerio,
and of course,
Fiat Lux!!!
