The analytic spread of an ideal Let $(R,\mathbb{m})$ be a Noetherian local ring and $I$ an ideal of $R$. Let $t$ be an indeterminate over $R$. The analytic spread $l(I)$ of $I$ is defined to be the Krull dimension of the ring $R[It]/\mathbb{m}R[It]$. Let $x$ be another indeterminate over $R$, then $\mathbb{m}R[x]$ is a prime ideal of $R[x]$ and $R\rightarrow R[x]_{mR[x]}$ is a faithfully flat ring extension, Let $R(x)=R[x]_{\mathbb{m}R[x]}$. 
The question is: Why $l(I)=l(IR(x))$?
This is a conclusion in Lemma 8.4.2 in I. Swanson and C. Huneke's book " Integral closure of ideals, rings and modules". I don't know how to prove it, any help will be appreciated!
 A: Let $A=R[It]/\mathfrak{m}R[It]$. Let $k=R/\mathfrak{m}$. 
Let $B=R[x]_{\mathfrak{m}R[x]}$, $J=IB$, $C=B[Jt]/\mathfrak{m}B[Jt]$.
We need to show that the Krull dimensions of $A$ and $C$ are same.
Then $B/\mathfrak{m}B=k(x)$, $A=R/\mathfrak{m}\otimes_R R[It]$ and
$C=B/\mathfrak{m}B\otimes_BB[Jt]$.
Since $B$ is flat over $R$, we have $B\otimes_RI^n=I^nB=J^n$ for any
$n\geq 0$, it
follows that $B[Jt]=B\otimes_RR[It]$ as $R$-modules, but the
isomorphism is also a ring map (check it), so $B[Jt]=B\otimes_RR[It]$ as rings.
Now
$C=B/\mathfrak{m}B\otimes_B(B\otimes_RR[It])=B/\mathfrak{m}B\otimes_RR[It]=(B/\mathfrak{m}B\otimes_{R/\mathfrak{m}}R/\mathfrak{m})\otimes_RR[It]=k(x)\otimes_kA$.
We only need to show $k(x)\otimes_kA$ and $A$ have the same Krull dimension. 
Notice that $R$ is Noetherian, $I$ is finitely generated, say
$I=(x_1,\ldots,x_n)$, then $A=k[\overline{x}_1t,\ldots,\overline{x}_nt]$
where $\overline{x}_i$ denotes the image of $x_i$ in $I/\mathfrak{m}I$, thus $A$ is
a finitely generated $k$-algebra. 
By Noether's normalization lemma, we can find algebraically
independent elements  $y_1,\ldots,y_r\in A$ such that $A$ is integral
over $k[y_1,\ldots,y_r]$. So $k(x)\otimes_kA$ is integral over
$k(x)\otimes_kk[y_1,\ldots,y_r]=k(x)[y_1,\ldots,y_r]$, thus $k(x)\otimes_kA$ is of
dimension $r$ ($=\dim A$).
We win.
