please give some comments regarding my proof to the
$$\dim R[T]\geq\dim R+1$$
Let $R$ be a noetherian ring, and let $n\geq 0$
by induction we only have to show that $\dim R[T]=\dim R+1$. It is easy to see that $\dim R[T]\geq\dim R+1$.
To show the converse inequality, let $a'$ be a maximal ideal of $R[T]$ and let $b=a'\cap R$,$a=b R[T]$.
Remember that Since $R$ be a noetherian ring and let $b \subset R$ be a prime ideal, and let $a=b * R[T]$
Then $a$ is a prime ideal of $R[T]$ and $a \cap R=b$. There exist infinitely many prime ideals such $a' \subset R[T]$ with $a' \cap R=b$, note $a'\neq a$ and there are no inclusion between them.
Furthermore, we then obtain
$$\dim R[T]_{a'} = \dim R_b + 1 \leq \dim R + 1 $$
Letting $a$ vary over all maximal ideal of $R[T]$, this proves the claim that $$\dim R[T]\geq\dim R+1$$
To prove that $\dim R[T] \geq R+1$, note that if $P$ is a prime ideal of $R$, then $$P\cdot R[T]\ \subset\ <P\cdot R[T],T>$$ and $<P\cdot R[T],T>$ is prime. This is true for any commutative ring $R$.