Noether’s Normalization Lemma in Red book I am trying to understand the proof of following important lemma from Mumford’s red book.
Noether’s Normalization Lemma: Let $R$ be an integral domain, finitely generated over a field $k$. If $R$ has transcendence degree $n$ over $k$, then there exists elements $x_1,...,x_n\in R$, algebraically independent over k, such that $R$ is integrally dependent on the subring $k[x_1,...,x_n]$ generated by the $x_i $. 
In the first line of the proof, he has written that 

Since $R$ is finitely generated over $k$, we can write $R$ as a
  quotient $$R=\frac{k[Y_1,...,Y_m]}{P}$$ for some prime ideal $P$. If
  $m=n$, then the images $y_1,...,y_m$ of the $Y$’s in $R$ must be
  algebraically independent themselves. Then $P=(0)$.

My question is that why if $m=n$, then the images $y_1,...,y_m$ of the $Y$’s in $R$ must be algebraically independent themselves and then why we get $P=(0)$?
 A: This is all by definition of transcendence degree and transcendence basis.
If
$$R=\dfrac{k[Y_1,\ldots,Y_n]}{P}$$
and has transcendence degree $n$ over $k$, that means
$$R\cong k[X_1,\ldots,X_n]$$
And the only way for this to be true, is if $P=(0)$.
This is because the transcendence degree of $R$ is defined to be the transcendence degree of $\mathrm{Frac}(R)$ over $k$. Now, if $P\neq (0)$, that means there is some non constant polynomial $f(x)\in P$. So, we have for some subset $S\subseteq\{Y_1,\ldots,Y_n\}$, $f(S)=0$. That is, the images of $\{Y_1,\ldots,Y_n\}$ in $R$ are algebraically dependent over $k$. Now, this is still detected in the field of fractions, so that $R$ would have transcendence degree $<n$ (because $\mathrm{Frac}(R)$ is generated by $k$ and the images of the $Y_i$, also I use implicitly the fundamental result in page 110 of Fields and Galois Theory by J. Milne. So the only way for $R$ to have transcendence degree $n$ would be if $P=(0)$.
So in conclusion, by definition the images of $Y_i$ must be algebraically independent in $R$, and $P=(0)$ is the only way for that to happen. 
