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Let $k$ be a field, $R = k[X_1, \ldots, X_n]$ and $I = \langle X_1, \ldots, X_n\rangle$. I am trying to prove that the quotient ring $R/I^r$, $r\in\mathbb{N}$ is local, i.e., has only one maximal ideal. Here we define $I^r$ as the ideal generated by the products $f_1\ldots f_r$ with each $f_i\in I$.

By the correspondence theorem, this amounts to proving that there is only one maximal ideal in $R$ that contains $I^r$. The obvious candidate for it is $I$, as it is maximal, since it can be shown that $R/I\cong k$, and it contains $I^r$. I don't know where to go from here, though. Are there any suggestions for the nest step here?

user26857
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Eric Vaz
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1 Answers1

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Let $J$ be a maximal ideal containing $I^r$; then $J$ is a prime ideal, and since it contains $I^r$, it must contain $I$, hence maximality of $I$ and $J$ implies the latter is equal to $I$. (Recall that a prime ideal is an ideal $P\neq R$ such that for any ideals $A$ and $B$, if $AB\subseteq P$ then either $A\subseteq P$ or $B\subseteq P$; for commutative rings, this is equivalent to $ab\in P\implies a\in P$ or $b\in P$, but for non-commutative rings the ideal-wise condition is weaker; it is then easy to verify that if $P$ is prime and $A$ is an ideal such that $A^k\subseteq P$, then $A\subseteq P$, by doing induction on $k$.)

Alternatively, since $X_i^r\in J$ and $J$ is a completely prime ideal, it follows that $X_i\in J$. This holds for $i=1,\ldots,n$, and hence $I = \langle X_1,\ldots,X_n\rangle\subseteq J$. Since $J$ is maximal and so is $I$, $J=I$, as desired.

Arturo Magidin
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