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One of my homework questions asks me to prove that a commutative ring $A$ is reduced if and only if there exists fields $\{k_s\}_{s\in S}$ and an injective ring homomorphism $A\rightarrow \prod_{s\in S}k_s$.

This is my attempt:

Consider the collection $\{A/m: m ~\text{is a maximal ideal}~\}$ of fields, $\prod_{m_i\in I} A/m_i$ is a ring. Let $f: A\rightarrow \prod_{m_i\in I}A/m_i$ be the ring homomorphism defined by $$ a\mapsto \phi $$ where $\phi(i) =a + m_i$. If $f(a) = f(b)$, then $a + m_i = b + m_i$ for all maximal ideals $m_i$. $b-a\in m_i$ for all maximal ideals $m_i$. $b-a\in\bigcap_{i\in I} m_i$. I know if $A$ is reduced then the intersection of all prime ideals is $\{0\}$, but I can't convince myself that $\bigcap_{i\in I} m_i$ is $\{0\}$, so $a$ might not be equal to $b$.

Something must have gone wrong with my proof, or, did I choose the wrong fields? I think that there is an embedding into a direct product of integral domains, by replacing $A/m$ with $A/\mathfrak{p}$, the quotient ring by a prime ideal. I am not so sure.

Coco
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    Can you think of a way to embed any integral domain into a field (which may depend on that integral domain)? Do you see how you can use this to rescue your approach replacing maximal ideals with prime ideals? – Alex Wertheim Feb 08 '24 at 07:02
  • @AlexWertheim Field of fractions, thank you. But how to prove the only if direction? – Coco Feb 08 '24 at 15:38
  • Can you show that a product of reduced rings is reduced? And can you show that a subring of a reduced ring is reduced? Try using these two facts to prove the other direction. – Alex Wertheim Feb 09 '24 at 01:09

1 Answers1

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As in the comment, you should work with prime ideals instead of maximal ideals because the intersection of prime ideals is the nilradical $\sqrt{(0)}$ while the intersection of maximal ideals is the Jacobson radical. The ring is reduced iff $\sqrt{(0)} = 0$. The fact that $\bigcap_{\mathfrak{p} \ \text{prime}}\mathfrak{p} = \sqrt{(0)}$ is standard commutative algebra and its proof requires Zorn lemma. Check the first chapter of Atiyah & McDonald for instance.

One direction is obvious. For the other one, you can embedd $A$ into $\prod_{\mathfrak{p} \ \text{prime}} A/\mathfrak{p}$ and this product is again embedded in the product of corresponding fields of fractions $\prod_{\mathrm{p} \ \text{prime}}F(A/\mathfrak{p})$.

Alexey Do
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