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All rings are commutative and unital

Q1: what means notation $$A\cong A_1\times\ldots\times A_n?$$ Is it true that elements of $A_1\times\ldots\times A_n$ are collection of elements of $A_1,\ldots ,A_n$ with termwise multiplication and addition? What is the difference between $$A_1\times\ldots\times A_n$$ and $$A_1\oplus\ldots\oplus A_n?$$

Q2: Let's $\{\mathfrak{m}_i\}_{i\in I}$ is the set of all maximal ideals of ring $A$. Is it true that $\cup_{i\in I}\mathfrak{m}_i$ consists of all non-invertible elements and $A-\{\mathfrak{m}_i\}_{i\in I}$ consists of all invertible elements?

Thanks a lot!

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For 1: There is essentially no difference between the direct product and direct sum of rings for a finite index set. –  Andrew Sep 16 '12 at 18:23
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For 2: Yes, and I think you must apply Zorn's lemma. –  Andrew Sep 16 '12 at 18:30
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1 Answer

Q1. Yes, the direct product is always meant by $n$-tuples and coordinatewise operations. For finite $n$ and for rings(!!), it is true that $A_1\times\ldots\times A_n \cong A_1\oplus\ldots\oplus A_n$. For infinite $n$, the right-hand side contains only those elements who has only finitely many nonzero coordinates.

Q2. If an element $a$ is invertible, then the ideal $\langle a\rangle$ generated by it, contains $1$, so contains all $A$, hence $a\in A\setminus\{\mathfrak m_i\}_i $.

On the other hand, if $a$ is not invertible, then $\langle a\rangle = Aa$ doesn't contain $1$, then --using Zorn lemma-- it can be extended to a maximal ideal $\mathfrak m_i$.

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The direct sum of infinitely many rings is not even a ring. –  Zhen Lin Sep 19 '12 at 12:45
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