Nullstellensatz in the coordinate ring $\Gamma (X)$ One of the many statements of the Hilbert's Nullstellensatz is the following:

If $k$ is an algebraic closed field, and $\mathfrak a$ is an ideal of the ring $k[T_1,\ldots,T_n]$ then $I(V(\mathfrak a))=rad(\mathfrak a)$

Let $X\subseteq\mathbb A^n_k$ be an affine algebraic set with the induced Zariski topology, and let $\Gamma(X)=k[T_1,\ldots,T_n]/I(X)$ be its coordinate ring. If $\overline{\mathfrak a}$ is any ideal of $\Gamma(X)$, it is easy to prove that every closed set of $X$ is of the form:
$$V|_X(\overline{\mathfrak a}):=\{x\in X\,:\, f(x)=0\,\forall f\in \overline{\mathfrak a} \}$$
Now, Görtz & Wedhorn in their book "Algebraic Geometry I" at page 21 (in the proof of Proposition 1.40) say that

if  $\overline{\mathfrak a}$ is any ideal of $\Gamma(X)$, by Hilbert's Nullstellensatz 
  we have that $I(V|_X(\overline{\mathfrak a}))=rad(\overline{\mathfrak a})$.

Why is this true?
I disagree with the above statement, infact if $\mathfrak a$ is the ideal of $k[T_1,\ldots,T_n]$ corresponding to $\overline{\mathfrak a}$ through the canonical projection, we have the following equality:
$$I(V|_X(\overline{\mathfrak a}))=I(V(\mathfrak a)\cap X)=I(V(\mathfrak a+I(X)))=rad(\mathfrak a+I(X))$$ 

EDIT: attempt of solution
Probably I should indicate the operator $I|_X$ which is different from $I$; if $Y\subseteq X$ we have that:
$$I|_X(Y)=\{f\in\Gamma(X)\,:\, f(y)=0\,\forall y\in Y\}$$
$$I(Y)=\{F\in k[T_1,\ldots,T_n]\,:\, f(y)=0\,\forall y\in Y\}$$
clearly if $\overline F=f$ we have that $F\in I(Y)$ if and only if $f\in I|_X(Y)$. Now the correct statement is:

if  $\overline{\mathfrak a}$ is any ideal of $\Gamma(X)$, by Hilbert's Nullstellensatz 
  we have that $I|_X(V|_X(\overline{\mathfrak a}))=rad(\overline{\mathfrak a})$.

and the proof should be:
$$I|_X(V|_X(\overline{\mathfrak a}))=I|_X(V(\mathfrak a)\cap X)=I|_X(V(\mathfrak a+I(X)))=\overline{I(V(\mathfrak a+I(X)))}=\overline{rad(\mathfrak a+I(X))}=\overline{rad(\mathfrak a)}=rad(\overline{\mathfrak a})$$ 
 A: By the correspondence theorem an ideal in $\Gamma(X)$ corresponds to an ideal in $k[T_1, \ldots, T_n]$ that contains $I(X)$.  So in this case
$\mathfrak a + I(X) = \mathfrak a.$
A: Not sure what the authors meant, but one thing is for sure. If $\mathfrak{a}$ is an ideal containing $I(X)$ then
$$\overline{f} \in \operatorname{rad} (\mathfrak{\bar{a}}) \iff f \in \operatorname{rad} \mathfrak{a}.$$
To see this if $\overline{f}^n \in \bar{\mathfrak{a}}$ for some $n$ then  there is $\overline{x} \in \overline{\mathfrak{a}}$ such that $\overline{f}^n  = \bar{x}$. Then $f^n - x \in I(X)$ and so $f^n \in \mathfrak{a} + I(X)$. Conversely if $f^n = x$ for some $x \in \mathfrak{a}$ and $n \in \Bbb{N}$ then taking equivalence classes shows that $\overline{f}^n = \overline{x}$.
My hunch now is that from $I(V(\mathfrak{a}) = \operatorname{rad} \mathfrak{a}$ we take equivalence classes on both sides to conclude that 
$$\overline{I(V(\mathfrak{a})} = \overline{\operatorname{rad} \mathfrak{a}}.$$
Then calculation above then shows that $\overline{\operatorname{rad} \mathfrak{a}} = \operatorname{rad} \overline{\mathfrak{a}}$ and now you just need to show that $\overline{I(V(\mathfrak{a})}  = I(V|_X (\mathfrak{a}))$.
A: Let $\mathfrak{a}\subseteq \Gamma(X)$. Then we have
$$I(V(\mathfrak{a}))=\bigcap_{x\in V(\mathfrak{a})} \mathfrak{m}_x = \bigcap_{  \mathfrak{m} \supseteq V(\mathfrak{a})} \mathfrak{m} = \mathrm{rad}(\mathfrak{a}).$$
The last equation holds by Hilbert's Nullstellensatz (Theorem 1.7, page 10). In the version of the book it is formulated for finitely generated $K$-algebras, so it holds in particular for $\Gamma(X)$.
