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What is the Krull dimension of $B=A[x,y,z]/\langle xy + 1, z + 1\rangle$, given $A$ is a Noetherian commutative ring?

user26857
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Zoey
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  • I feel that you have some misconceptions. First, why would you write $B$ as $A[x,y,z]/(xy+1,z)$ and not just as $A[x,y]/(xy+1)$? For the second, are you asking whether say $\dim A[x]/(x^2+1)=\dim k[x]/(x^2+1)$ for arbitrary ring $A$ and arbitrary field $k$? – Mohan Aug 06 '15 at 02:12
  • I can write $B$ as $A[x,y]/\langle xy + 1\rangle$. And I was wrong when i said its the same as fields. I will edit right away. What I meant was $\mathrm{dim}(A) + \mathrm{dim}(\Bbbk[x_1, \ldots, x_n]/\mathfrak{a})$. – Zoey Aug 06 '15 at 04:27
  • For your 2nd question to make sense, you need the ideal to be generated by polynomials with coefficients in $k$ or something — otherwise the quotient $k[x_1,\dots]/a$ does not make sense. Also, 'leading coefficient' does not generally make sense for polynomials in several variables. – Mariano Suárez-Álvarez Aug 06 '15 at 04:55

1 Answers1

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Your first ring $B$ is just $A[x^{\pm1}]$, the ring of Laurent polynomials with coefficients in $A$, which is the localization of $A[x]$ at the powers of $x$. It follows from this that $\dim B\leq\dim A[x]=\dim A+1$.

On the other hand, if $\def\p{\mathfrak p}\p$ is a prime in $A$ then $\mathfrak pB$ is a prime of $B$, easily seen to be the set of Laurent polynomials with coefficients in $\p$. If $\p_0\subsetneq\cdots\subsetneq\p_r$ is a chain of prime ideals in $A$, then $\p_0B\subsetneq\cdots\subsetneq\p_rB\subsetneq \p_rB+(x-1)B$ is a chain of prime ideals of $B$ one prime longer.

Conclusion: $\dim A[x^{\pm1}]=\dim A+1$.

user26857
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Mariano Suárez-Álvarez
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