Infinite CRT fails

Construct a commutative ring $A$ and a countable infinite family of distinct maximal ideals $\mathfrak m_i,i=1,2,3,\ldots$ of $A$ such that

1.  The intersection $\bigcap \mathfrak m_i$ is zero, yet
2.  The canonical projection $A\to \prod A/\mathfrak m_i$ is not surjective.

5 thoughts on “Infinite CRT fails”

1. Consider the integers with the countable infinite family of ideals being the prime ideals. The product ring is then the product of all the finite prime fields. This has the cardinality of the continuum and thus the projection cannot be surjective.

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1. The non-surjectivity can also be proved by the same argument I gave below: once again no non-zero sequence with an infinite number of zeroes is in the image, since any integer divisible by an infinite number of primes must be zero.

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2. fmartin says:

A more or less geometric argument: take $\mathbb{R}[X]$ and consider the maximal ideals $\mathfrak{m}_n=(X-n)$ for natural $n$. Then the projection map $\mathbb{R}[X]\to \prod\mathbb{R}[X]/\mathfrak{m}_n\simeq \mathbb{R}^\mathbb{N}$ is just the product of the evaluation maps at each natural number. This cannot be surjective since for instance no non-zero sequence with an infinite number of zeroes is in the image.

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1. To identify the image of this map, one can perform the following test. Pick a sequence, $(x_n)$. If for no $k$ the sequence $x_n/n^k$ is bounded, then $x_n$ is not in the image. Else consider the least $k$ such that this happens. If $(x_n)$ was in the image by evaluation of a polynomial $p$ of degree $k$, then in fact $x_n/n^k$ must converge to the leading coefficient of $p$, call it $a_k$. Repeat the argument with $x_n - a_kn^k$. If this procedure succeeds, then $(x_n)$ lies in the image of the map, else, it doesn’t.

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2. In fact, this idea works for any ring of functions satisfying some kind of rigidity. Let $A$ be a $k$-algebra of $k$-valued functions with domain some set $X$. Given $x_0\in X$, the ideal $I_{x_0}$ of functions in $A$ vanishing at $x_0$ is maximal, since the quotient $A/I_{x_0}$ is isomorphic to $k$. Therefore, in order for the map $A\to \prod A/I_{x_n}$ to be surjective one needs to be able to produce functions in $A$ with an infinite number of prescribed values, which is not possible if the functions we are considering are rigid enough.

Another example building from this idea: let $\Omega\subset \mathbb{C}$ be a domain and consider the ring $A=\mathcal{O}(\Omega)$ of holomorphic functions on $\Omega$. Then if $(z_n)$ is a sequence of points converging to a point in the interior of $\Omega$, the map $A\to \prod A/I_{z_n}$ cannot be surjective, since for instance any non-zero sequence that is eventually zero is not in the image by the identity theorem.

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