I want to prove the following theorem concerning Krull dimension:

> 
> **Theorem** If $A$ is a noetherian ring then $$\dim(A[x\_1,x\_2, \dots , x\_n]) = \dim(A) + n$$ where $\dim$ stands for the Krull dimension of the
> rings. Thus, $\dim(K[x\_1,x\_2, \dots , x\_n]) = n$ for any field $K$.
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> 
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I could prove it by induction in the number of variables assuming these facts:

1. If $A$ is noetherian then $A[x\_1,x\_2, \dots , x\_n]$ is noetherian.
2. If $K$ is a field then $\dim(K)=0$.
3. If $A$ is noetherian then $\dim(A[x]) = \dim(A) + 1$.

Fact 1 follows from Hilbert's basis theorem. Fact 2 is trivial.

For Fact 3 I have been able to prove the inequality $\dim(A[x]) \geq \dim(A) + 1$. I couldn't get the other inequality.

My questions are:

> 
> What is a proof for the missing inequality? Are there easy proofs if $A$ is assumed to be a PID?
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> 
> Are there counterexamples to the theorem I'm trying to prove if we change the condition of $A$ being noetherian to $A$ being finite dimensional?
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Thanks!