Are these two theorems about algebraic varieties the same? In Artin's Algebra, there is a theorem (1) stated as the following: 

Let $J\subset\Bbb{C}[x]$ be an ideal such that $J=(f_1,\cdots,f_r)$ where $f_1,\cdots,f_r\in\Bbb{C}[x_1,\cdots,x_n]$. Let $R=\Bbb{C}[x_1,\cdots,x_n]/J$. Let $V$ be the set of common zeros of $\{f_1,\cdots,f_r\}$. Then the maximal ideals of $R$ are in bijective correspondence with the points of $V$. 

There is another statement (2) (I don't remember where I learned it from) I've seen before:

Let $V\subset{C}^n$ be a variety. Then there is a bijective correspondence between points in $V$ and maximal ideals in ${\Bbb C}[x_1,\cdots,c_n]/I(V)$, where
  $$
I(V):=\{f\in{\Bbb C}[x_1,\cdots,x_n]:\forall x\in V\ \ f(x)=0\}
$$
  is an ideal. 

Here are my questions:  


*

*Are these two statements the same?

*Can anyone come up with a reference for (2)? (If I state (2) in a wrong way, can somebody correct it?)



I know almost nothing about algebraic geometry and commutative algebra. I came up with this question when I read Artin's Algebra, in which there is a small section about algebraic geometry. I suspect that (2) is different from (1). Suppose in (2) $V$ is the set of common zeros of $\{f_1,\cdots, f_r\}$. Then $I(V)$ might be strictly bigger than $(f_1,\cdots, f_r)$.

The definition of algebraic variety $V\subset{\Bbb C}^n$ I use here (in Artin's book) is the set of common zeros of finitely many polynomials in $n$ variables.
 A: Yes, these are the same theorems, basically. The theorem that ties the two together is the following (Nullstellensatz):
Theorem: There is a bijection between radical ideals $I\subseteq \mathbb{C}[x_1,\ldots,x_n]$ (that is, ideals that satisfy $I=\{f\in \mathbb{C}[x_1,\ldots,x_n]:f^r\in I\mbox{ for some }r\in\mbox{N}\}$) and sets that are of the form
$$\{a=(a_1,\ldots,a_n)\in \mathbb{C}^n:g_1(a)=\cdots=g_s(a)=0\}$$
for certain polynomials $g_1,\ldots,g_s$, given by $I\mapsto V(I)$, where $V(I)$ denotes the zero set of the polynomials in $I$.
The inverse map above is given by $\Phi:X\mapsto\{f\in \mathbb{C}[x_1,\ldots,x_n]:f(x)=0\mbox{ for all }x\in X\}$.
Basically, if $J$ is an ideal in $\mathbb{C}[x_1,\ldots,x_n]$, then $\Phi(V(J))=\mbox{rad}(J)$, where
$$\mbox{rad}(J)=\{f\in\mathbb{C}[x_1,\ldots,x_n]:f^r\in J\}.$$
Now the idea that ties your two theorems together is the following:
The maximal ideals of $\mathbb{C}[x_1,\ldots,x_n]/J$ are in bijection with the maximal ideals of $\mathbb{C}[x_1,\ldots,x_n]/\mbox{rad}(J)$.
Proof: Since $J\subseteq\mbox{rad}(J)$, if $\frak{m}$ is a maximal ideal that contains $\mbox{rad}(J)$, then $\frak{m}$ contains $J$. We must show that the converse is true. If $\frak{m}$ contains $J$, let $f\in\mbox{rad}(J)$. Then $f^r\in J$ for some $J$, and therefore $f^r\in \frak{m}$. Since $\frak{m}$ is prime, we have that $f$ must be in $\frak{m}$. $\Box$
