Theorem
The set \[k[X]:=\{f:X \rightarrow k | f \text{ is regular function}\}\]
form a ring under addition and multiplication defined by
\[(f+g)(x):= f(x)+g(x) \qquad (f \cdot g )(x):=f(x)g(x)\]
for all $x\in X$ and the function $\psi: k[T] \rightarrow k[X]$ given by $F(T) \mapsto f$ where $f\equiv F_X$ defines an onto ring homomorphism from $k[T]$ to $k[X]$.
Proof:
Clearly the operations defined above are associative and commutative, since $f(x)$ and $g(x)$ are the elements of $k$ which is a field and hence commutative and associative. The function $f_0(x)=0$ for all $x\in X$ is a regular function, since it is being represented by a zero polynomial, thus $f_0\in k[X]$, and for all $f\in k[X]$ we have \[(f+f_0)(x)=f(x)+f_0(x)=f(x)+0=f(x)\] for all $x\in X$ thus $f+f_0\equiv f$ as a function on $X$ and so $f_0$ acts as the zero element in $k[X]$. Given $f$ and $g$ elements of $k[X]$, then there exist polynomials $F$ and $G$, such that $f(x)=F(x)$ and $g(x)=G(x)$ for all $x\in X$, therefore for all $x\in X$ we have $(f+g)(x)=f(x)+g(x)=F(x)+G(x)$ and $(f\cdot g)=f(x)g(x)=F(x)G(x)$, or equivalently $F+G$ is a polynomial representation $f+g$ and $FG$ is a polynomial representation of $f\cdot g$, thus $f+g$ and $f\cdot g$ are both regular functions and $k[X]$ is closed under the operations defined above. Also we have $-f(x)=-F(x)$ for all $x\in X$ and thus $-f$ is also a regular function and $(f+(-f))(x)=f(x)-f(x)=0=f_0$ for all $x\in X$. By the commutativity we only need to check the left distributive, let $f,g,h\in k[X]$ be represented by $F$ , $G$, $H \in k[T]$ respectively, then since $k[T]$ is a ring we have $F(G+H)=FG+FH$, therefore $f\cdot(g+h)=f(x)(g(x)+h(x))=F(x)(G(x)+H(x))=F(x)G(x)+F(x)H(x)=f\cdot g+f\cdot h$ and this confirms the distributive property.
Now consider the map $\psi : k[T] \rightarrow k[X]$, which sends any $F \in k[T]$ to $f=F|_X \in k[X]$, we will check that $\psi$ is a ring homomorphism, suppose $F,G\in k[T]$, then
$\psi(F+G)=(F+G)|_X = F|_X+G|_X=\psi(F)+\psi(G)$ and $\psi(F\cdot G)=(F\cdot G)|_X=F|_X \cdot G|_X=\psi(F)\cdot\psi(G)$. Furthermore, since $k[X]$ consists regular functions, for any $f\in k[X]$ we can find polynomial $F$ such that $F|_X\equiv f$ that is $\psi(F)=f$, and hence $\psi$ is onto.
The set \[k[X]:=\{f:X \rightarrow k | f \text{ is regular function}\}\]
form a ring under addition and multiplication defined by
\[(f+g)(x):= f(x)+g(x) \qquad (f \cdot g )(x):=f(x)g(x)\]
for all $x\in X$ and the function $\psi: k[T] \rightarrow k[X]$ given by $F(T) \mapsto f$ where $f\equiv F_X$ defines an onto ring homomorphism from $k[T]$ to $k[X]$.
Proof:
Clearly the operations defined above are associative and commutative, since $f(x)$ and $g(x)$ are the elements of $k$ which is a field and hence commutative and associative. The function $f_0(x)=0$ for all $x\in X$ is a regular function, since it is being represented by a zero polynomial, thus $f_0\in k[X]$, and for all $f\in k[X]$ we have \[(f+f_0)(x)=f(x)+f_0(x)=f(x)+0=f(x)\] for all $x\in X$ thus $f+f_0\equiv f$ as a function on $X$ and so $f_0$ acts as the zero element in $k[X]$. Given $f$ and $g$ elements of $k[X]$, then there exist polynomials $F$ and $G$, such that $f(x)=F(x)$ and $g(x)=G(x)$ for all $x\in X$, therefore for all $x\in X$ we have $(f+g)(x)=f(x)+g(x)=F(x)+G(x)$ and $(f\cdot g)=f(x)g(x)=F(x)G(x)$, or equivalently $F+G$ is a polynomial representation $f+g$ and $FG$ is a polynomial representation of $f\cdot g$, thus $f+g$ and $f\cdot g$ are both regular functions and $k[X]$ is closed under the operations defined above. Also we have $-f(x)=-F(x)$ for all $x\in X$ and thus $-f$ is also a regular function and $(f+(-f))(x)=f(x)-f(x)=0=f_0$ for all $x\in X$. By the commutativity we only need to check the left distributive, let $f,g,h\in k[X]$ be represented by $F$ , $G$, $H \in k[T]$ respectively, then since $k[T]$ is a ring we have $F(G+H)=FG+FH$, therefore $f\cdot(g+h)=f(x)(g(x)+h(x))=F(x)(G(x)+H(x))=F(x)G(x)+F(x)H(x)=f\cdot g+f\cdot h$ and this confirms the distributive property.
Now consider the map $\psi : k[T] \rightarrow k[X]$, which sends any $F \in k[T]$ to $f=F|_X \in k[X]$, we will check that $\psi$ is a ring homomorphism, suppose $F,G\in k[T]$, then
$\psi(F+G)=(F+G)|_X = F|_X+G|_X=\psi(F)+\psi(G)$ and $\psi(F\cdot G)=(F\cdot G)|_X=F|_X \cdot G|_X=\psi(F)\cdot\psi(G)$. Furthermore, since $k[X]$ consists regular functions, for any $f\in k[X]$ we can find polynomial $F$ such that $F|_X\equiv f$ that is $\psi(F)=f$, and hence $\psi$ is onto.
