Help with these isomorphisms Let $X$ be an affine algebraic set and $f\in K[X]$ where $K[X]$ is the coordinate ring of $X$. Suppose $I(X)=\langle G_1,\ldots,G_r\rangle$ and $W=Z(G_1,\ldots,G_r,FT_{n+1}-1)$, where $G_1,\ldots,G_r\in k[T_1,\ldots,T_n]$. 
The image of $F\in k[T_1,\ldots,T_n]$ in the quotient $K[X]$ is $f$, i.e, $f=F+I(X)$.
I'm trying to prove these isomorphisms:
$$k[W]\cong \frac{k[T_1,\ldots,T_n,T_{n+1}]}{\langle G_1,\ldots,G_r,FT_{n+1}-1\rangle}\cong \frac{k[X][T_{n+1}]}{\langle fT_{n+1}-1\rangle}\cong k[X][1/f].$$
I couldn't prove it (I tried a lot), if someone could help me with some of these isomorphisms it would be very helpful to me and I would be very grateful.
I really need help.
Thanks a lot
 A: Well, break it down into some small pieces as follows: $k[X]=k[T_1,\cdots,T_n]/(G_1,\cdots,G_r),$ so 
$$k[X][T_{n+1}]=(k[T_1,\cdots,T_n]/(G_1,\cdots,G_r)) [T_{n+1}] \cong k[T_1,\cdots,T_n][T_{n+1}]/(G_1,\cdots,G_r)$$
which is in fact 
$$k[X][T_{n+1}]\cong k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r).$$
Then show that the isomorphism is preserved if you take the quotient from both sides by the ideal $(fT_{n+1}-1).$ 
For the final isomorphism, let $R=k[X],$ define the ring morphism $R[T_{n+1}] \to R[1/f]$ by sending $T_{n+1} \mapsto 1/f,$ then show that it is surjective and its kernel is the ideal $(fT_{n+1}-1).$
Added:


*

*Building the first isomorphism.


To prove $R[X]/I[X] \cong (R/I)[X]$ for $R=k[T_1.\cdots,T_n]$ and $I=(G_1,\cdots,G_r)$ define $g: R[X] \to (R/I)[X]$ by sending a polynomial with coefficient in $R$ to the polynomial with coefficient in $R/I.$ That is, first consider the projection morphism $\pi: R \to R/I$ sending $r \mapsto r+I$ and then 
$$g(r_nX^n+\cdots+r_1X+r_0)=(r_n+I)X^n\cdots+(r_1+I)X+(r_0+I)$$
It is easy to show that $g$ is a surjective ring homomorphism. Basically $g$ is constructed from $\pi$ which is a surjective ring homomorphism. Now, in order to use the first isomorphism theorem for rings we should show that the kernel of $g$ is $I[X],$ which is clear since the kernel of $\pi$ is $I$ and $g$ was constructed from $\pi$ by adding a variable $X,$ so the kernel of $g$ is $I[X],$ or if $g(r_nX^n+\cdots+r_1X+r_0)=0$ means that $(r_n+I)X^n\cdots+(r_1+I)X+(r_0+I)=0$ and $0$ in the ring $(R/I)[X]$ is $I[X]$ because a polynomial is equivalent to zero (always zero) if and only if all of its coefficients are zero, that is, $r_i+I$ is zero in $R/I$ for $1 \leq i \leq n,$ and the zero element in $R/I$ is just $I,$ so $r_i+I=I$ implying $r_i \in I$ for all $i.$
Remark: As I explained in my last comment, since $I=(G,\cdots,G_r)$ is an ideal of $k[T_1,\cdots,T_n]$ then $I[T_{n+1}]$ will be an ideal in $k[T_1,\cdots,T_n,T_{n+1}],$ but we still can write $(G,\cdots,G_r)$ instead of $(G,\cdots,G_r)[T_{n+1}]$ when we say consider $(G,\cdots,G_r)$ now as an ideal in $k[T_1,\cdots,T_n,T_{n+1}]$ with no harm.


*

*Why taking quotient of $(fT_{n+1}-1)$ is preserving the isomorphism?


Back to our notation, we showed that 
$$k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r) \cong (k[T_1,\cdots,T_n]/(G_1,\cdots,G_r))[T_{n+1}]$$ 
and since $k[X]=k[T_1,\cdots,T_n]/(G_1,\cdots,G_r)$ we then have 
$$\varphi: k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r) \cong k[X][T_{n+1}]$$
say the isomorphism map is called $\varphi.$
To prove 
$$k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r,FT_{n+1}-1) \cong k[X][T_{n+1}]/(fT_{n+1}-1)$$
again define a map 
$$p: k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r) \to k[X][T_{n+1}]/(fT_{n+1}-1)$$ 
by sending an element in $s \in k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r)$ to $\varphi (s) \in k[X][T_{n+1}]$ and the sending it to $\pi (\varphi (s)) \in k[X][T_{n+1}]/(fT_{n+1}-1)$ where 
$$\pi: k[X][T_{n+1}] \to k[X][T_{n+1}]/(fT_{n+1}-1)$$ 
is the natural projection.
Since $p=\pi \circ \varphi$ then is a ring homomorphism and since $\pi$ is surjective and $\varphi$ is an isomorphism then their composition $\pi \circ \varphi$ is surjective, so $p$ is a surjective ring homomorphism. Again to use the first isomorphism theorem for rings, we should show that the kernel is $(FT_{n+1}-1)$ which is clear since $p(s)=0$ for some $s \in k[T_1,\cdots,T_n,T_{n+1}]/(G_1,\cdots,G_r)$ is meaning that $\pi (\varphi(s))=0$ which means $\varphi(s) \in \ker(\pi)=(fT_{n+1}-1).$ So $\varphi(s)=(fT_{n+1}-1)t$ for some $t \in k[X][T_{n+1}]$ by definition of an ideal. Since $\varphi$ is an isomorphism so its inverse is also an isomorphism then $s=\varphi^{-1}((fT_{n+1}-1)t)=\varphi^{-1}(fT_{n+1}-1) \varphi^{-1}(t).$ The way $\varphi$ was constructed shows that $\varphi^{-1}((fT_{n+1}-1))=FT_{n+1}-1$ because $F \mapsto f$ via the projection $k[T_1,\cdots,T_n] \to k[X].$ Therefore, $s \in (FT_{n+1}-1)$ and $\ker p \subset  (FT_{n+1}-1).$ Obviously $(FT_{n+1}-1) \subset \ker p,$ hence the equality, so we're done.
For the last isomorphism in the question follow the same approach and the last paragraph before "Added."   
