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NOTE: I am taking rings to be commutative and unital.

$A$ is a ring. I am trying to find a counterexample to: $A_p$ integral domain $\forall$ prime ideals $p$ in $A$ $\implies$ $A$ is an integral domain.

I have seen some people on this site suggest $k^2$ for a field $k$, specifically $k = \mathbb{Q}$.

$k^2$ is clearly not an integral domain so $(0)$ is not a prime ideal. Hence the only primes are $k \times \{0\}, \{0\} \times k$. In fact these are maximal because $k^2 / k \times 0 \cong k$ is a field.

Now, localizing a ring at a maximal ideal should give that same ring back, i.e. if $m$ is a maximal ideal then $(A-m)^{-1}A$ will only have units as denominators essentially giving $A$ back...

I do not see why this is a counterexample then since $A$ was not an integral domain to begin with. I think I am making a silly mistake, thank you for your help!

Edit: I realize my characterization of localization at max ideals only works for local rings.

Edit 2: Still not sure I get what the localization, say, $(\mathbb{Q}^2)_{\mathbb{Q}\times0}$ ends up being...

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  • Localizing a ring at a maximal ideal does not give the ring back, consider for example $A=\mathbb{Z}$. The top answer in the first linked question should give you everything you need to determine $(\mathbb{Q}^2)_{\mathbb{Q}\times 0}$. – Vincent Boelens May 31 '24 at 09:26

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Your ring is not local. Localising at a prime ideal yields a local ring (this close relation is the reson the two notions share name).

Note that $A-m$ does not contain only units. Let $m = \{(x, 0)\mid x\in k\}$. Then $A-m$ contains, elements like $(0,1)$. So, if $\psi$ is the canonical map $A\to A_m$ (and if you'll allow me to use fractions for elements of $A_m$), we have $$ \psi(x, 0) = \frac{(x, 0)}{(1, 1)} = \frac{(x, 0)\cdot(0, 1)}{(1, 1)\cdot(0,1)} = \frac{(0,0)}{(0, 1)} = \frac{(0,0)\cdot(0,1)}{(1, 1)\cdot(0,1)} = \frac{(0,0)}{(1, 1)} = \psi(0, 0) $$ In fact, $\ker\psi = m$, which makes $A_m\cong A/m$ an integral domain.

Arthur
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