## Global dimension [closed]

Hi,

I want to prove the following theorem

"If $R[x_1,...,x_n]$ denotes a polynomial ring in n variables, then $gl.dim(R[x_1,...,x_n])=n+gl.dim(R)$"

(gl.dim = global dimension)

I already have the proof for R[x] and want to ask if my following calculation is right

$gl.dim(R[x_1,...,x_{n+1}])=gl.dim(R[x_1,...,x_n][x_{n+1}])=1+gl.dim(R[x_1,...,x_n])=$ $=1+gl.dim(R[x_1,...,x_{n-1}][x_n])=...$

It seems to me that it is to easy...

Thank you!

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I used to think proof by induction was easy. Then I started to read what the logicians make of it. – Charles Matthews Jun 13 2010 at 10:46
Yes, that's all there is to the inductive step. But was the proof for one variable "too easy" also? – Robin Chapman Jun 13 2010 at 10:47
I'm not sure what can be given as an answer for this "question", except to say "yes it works". – Yemon Choi Jun 13 2010 at 11:42