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Let ($R$ , $$) be a local Noetherian ring. Suppose that $I=(x_1,..., x_k) \subseteq $. Is it true that $\dim R/I=\dim R-k$ ?

Thank you.

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    No, it's not.${}$ – 2017-02-13
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    @user26857 Thank you. Is there a counterexample? – 2017-02-13
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    Plenty. $\dim K[X,Y]_{(X,Y)}/(X^2,XY)=1$. – 2017-02-14
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    Trivial example: take $x_i=0$ for all $i$. Little less trivial: take all the $x_i$ in the nilradical. In both cases, dim $R$ = dim $R/I$. – 2017-02-14

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As referenced in the comments, this is not true in general. In fact, it happens exactly when $(x_1,...,x_k)$ forms part of a system of parameters.

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It is not true in general. You can prove this one:

  • $\dim( R/I )\ge \dim( R)-k$

  • $(x_{1},...,x_{k})$ is part of a system of parameters iff $\dim (R/I)= \dim( R) - k$