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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:.

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This is essentially the Nullstellensatz. Be careful because it holds when $k$ is algebraically closed. – Andrea Dec 30 '11 at 0:32
You haven't defined $X$. Do you mean $X = V(F)$? – Alex Becker Dec 30 '11 at 0:50
Spot on, @Andrea ! (Moreover it holds if and only if $k$ is algebraically closed) – Georges Elencwajg Dec 30 '11 at 10:14
up vote 1 down vote accepted

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 $$ \overline {res} :A= k[x_1,..., x_n]/(F^{red} ) \to Map(V(F), k): \overline {g(x_1,...,x_n)} \mapsto g_V \quad (**) $$

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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 $A$.

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Dear Alex, you write: "If $F$ has no multiple factors, we must have $g-g'\in (F)$ since $g-g'$ vanishes on $V(F)$". This is false for every non-algebraically closed field and highly non-trivial for algebraically closed ones: it is exactly the heart of the question. – Georges Elencwajg Dec 30 '11 at 10:21
To quote the OP, "If F has no multiple factors , say F = ∏fi with fi ∤ fj then it can be shown that F generates the ideal of all functions vanishing on V(F) ( This result is not my question )". i saw no need to prove it seeing as they said they understood that result. – Alex Becker Dec 30 '11 at 18:00

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