a basic question about the locus of a polynomial First a little introduction and notation. I have a question with a words that the book says.
Let k be a field. Let F $\in $ $k[x_1,...,x_n]$ We define the locus $V(F)$ of F , by $$
V\left( F \right) = \left\{ {P = \left( {a_1 ,...,a_n } \right):F\left( P \right) = 0} \right\} \subset k^n 
$$
Now consider the ring $ A = k[x1,..,x_n]/(F) $   Then note that an element $ g\in$ A defines a k-valued function on $V(F) $ indeed, if g is the class in A of a polynomial $
g´ \in k\left[ {x_1 ,...x_n } \right]
$ then for x $\in$ V(F) the value $g(x) = g´(x)$ does not depend on the choice of $g´$.
My question comes in this paragraph.
If F has no multiple factors , say $ F$ = $\prod f_i$ with $f_i$ $\nmid $ $f_j$ then it can be shown that F generates the ideal of all functions vanishing on $ V(F)$  ( This result is not my question ) . It follows from this that an element g$\in$ A is uniquely determined by the corresponding function g: X $\to$ k .
My question is about this, I know that must be so simple but I don´t know D:.
 A: There is a fundamental morphism  of $k$-algebras (" restriction") 
 $$res :k[x_1,..., x_n]\to Map(V(F), k): g(x_1,...,x_n) \mapsto g_V \quad  (*)$$ where $g_V$ is the function $g_V:V(F) \to k: P=(a_1,...,a_n)\mapsto g(a_1, ...,a_n)$.
This morphism is neither surjective nor injective in general.
Its kernel is $ker (res)=:\mathcal I(V(F))$, the ideal of polynomials vanishing on $V(F)$.
Tautologically  $F\in \mathcal I(V(F))$ ( "$F$ vanishes on the points where $F$ vanishes" , which sounds like a children's joke!) and thus  for the ideal generated by $F$ we have the inclusion $(F)\subset \mathcal I(V(F))$
However there are two obstacles to equality.    
Obstacle 1:   Obviously $V(F)=V(F^2)$ and since $(F^2)\subsetneq (F)$, we cannot have $(F)=\mathcal I(V(F))=\mathcal I(V(F^2))=(F^2)$.
The remedy is to decompose $F$ into irreducible factors $F=F_1^{r_1}...F_t^{r_t}$,  to consider $F^{red}=F_1...F_t$ and to hope that 
$(F^{red})= \mathcal I(V(F))$. However even that is not true for arbitrary fields $k$:   
Obstacle 2: If $F=x_1^2+x_2^2 \in \mathbb R[x_1,x_2]$, then $V(F)=\lbrace (0,0)\rbrace \subset \mathbb R^2 $. But then $x_1\in \mathcal IV(F)$, and $x_1\notin (F)=(F^{red})$.  No manipulating of $F$ will remedy this: the fault lies with the field, which doesn't allow $F$ to show all its zeroes. 
The remedy is to consider only algebraically closed fields and then indeed Eduard Study proved that for an algebraically closed field $k=\bar k$ we do have  $ \mathcal I(V(F))=(F^{red})$.
Hilbert's celebrated Nullstellensatz generalizes this formula to the case of an arbitrary number of  polynomials $F_i$.
Hence,  if $k$ is algebraically closed,  you get  from $(*)$ and Noether's isomorphism theorem the injective morphism 
 $$    \overline {res} :A= k[x_1,..., x_n]/(F^{red} ) \to Map(V(F), k): \overline {g(x_1,...,x_n)} \mapsto g_V \quad  (**)     $$
A: Suppose we have $g,g'\in k[x_1,\ldots,x_n]$ such that $g(p) = g'(p),\forall p\in V(F)$, so $(g-g')(p)=0,\forall p\in V(F)$. If $F$ has no multiple factors, we must have $g-g'\in (F)$ since $g-g'$ vanishes on $V(F)$. Then the images $\bar{g},\bar{g}'\in A$ are equal, so if we want to specify an element of $A$ we need only specify the function $g:V(F)\to k$, as any two $g$ which are the same as functions on $V(F)$ are the same as elements of $A$.
