When does $\mathbb{Z}_{m}×\mathbb{Z}_{n}$ form a field? I thought that I could use proposition that say, if I and J are comaximal ideal of R, then $$\dfrac{R}{I \cap J} \simeq \dfrac{R}{I} × \dfrac{R}{J}$$ Using that we have if m and n are relatively prime then $\mathbb{Z}_{m}×\mathbb{Z}_{n}$ form a field. Is it true? For example, $\mathbb{Z}_{3}×\mathbb{Z}_{3}$ doesn't form a field since $\gcd(3,3) = 3$. I am so confused. Thanks in advance
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However, $F = \mathbb{Z}{3}×\mathbb{Z}{17}$ form a field since $\gcd(3,17)=1$. Moreover, char$(F)=51$ – beingmathematician Apr 01 '22 at 07:01
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1Unless you're in a degenerate case with $n = 1$ or $m = 1$, this ring always has zero-divisors: $(1,0) \cdot (0,1) = (0,0)$. – Magdiragdag Apr 01 '22 at 07:02
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More more moreover $F$ is isomorphic to $\mathbb{Z}_{51}$ – beingmathematician Apr 01 '22 at 07:03
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What is a degenerate case? – beingmathematician Apr 01 '22 at 07:03
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Okay I understand what a degenerate case is – beingmathematician Apr 01 '22 at 07:04
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So it is true isn't it – beingmathematician Apr 01 '22 at 07:05
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No, of course not. A field does not have zero-divisors. – Magdiragdag Apr 01 '22 at 07:05
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For non-degenerate case is it true? – beingmathematician Apr 01 '22 at 07:07
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Let $A,B$ be rings. If $A,B$ are nonzero, then $A \times B$ is never a field. Because $(1,0)$ doesn’t have an inverse. – Aphelli Apr 01 '22 at 07:16
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Maybe the confusion is here: ${\mathbb Z}{51}$ is _not a field (because $3 \cdot 17 = 0$). – Magdiragdag Apr 01 '22 at 07:16
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Thank you all guys, I understand perfectly. – beingmathematician Apr 01 '22 at 07:17
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To answer your question, I'd say pretty much never. Now what we first need to understand is that additive and multiplicative groups of a finite field aren't isomorphic, in fact, they don't even have the same cardinality (Multiplicative group excludes 0).
Now a finite field of order $n$ exists if and only if $n$ is a prime power ($n=p^h$ for some prime $p$ and $h\in \mathbb N$). We refer to this field as ${\rm GF}(n)$, as in Galois Field of order $n$. That being said, its additive group is isomorphic to $\mathbb{Z}_p\times \mathbb{Z}_p\times ... \times \mathbb{Z}_p$ ($h$ times) and its additive group is a cyclic group of order $n-1$, thus isomorphic to $\mathbb{Z}_{n-1}$.
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Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Apr 01 '22 at 07:34