Reducing the dimension of a finitely generated $k$-algebra by $1$ Let $k$ be a field and $Q$ an ideal of $k[x]=k[x_1,\dots,x_s]$ such that $R=k[x]/Q$ has Krull dimension equal to $d>0$. Define $V=x_1 k + \cdots +x_s k$ to be the vector space of linear forms over $k$. Let $P_1,\cdots, P_t$ be the minimal prime divisors of $Q$. Since $P_i \not\supset V$ (otherwise we get $\dim R=0$, contradiction) we can select a linear form $l_1(x) \in V$ which is not contained in any of the $P_i$. Then none of the minimal prime divisors of $(Q,l_i(x))$ is a minimal prime divisor of $Q$ and this shows that $\dim k[x]/(Q,l_1(x)) <\dim k[x]/Q=d$. 
Question: can we say further that $\dim k[x]/(Q,l_1(x))=\dim k[x]/Q-1$?
If yes, how can we see that?
Remark: i suspect that this is the case, because we are using a linear form $l_1(x)$. For a context where this question arises, see the second part of step 1 in the proof of theorem 14.14 in Matsumura's Commutative Ring Theory.

This must be easy to prove and helpful: if $R$ is a noetherian local ring with maximal ideal $\mathfrak m$ and $x\in\mathfrak m$ doesn't belong to any minimal prime, then $\dim R/(x)=\dim R-1$. 
Now you can say that your ring is not local. But for finitely generated graded $K$-algebras the Krull dimension can be read from the height of their irrelevant maximal ideal. In fact, finitely generated graded $K$-algebras are graded local, that is, they have only one graded maximal ideal.
