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Explain why ($\mathbb{Z}_6$, +, · ) is not a field, where + is addition modulo 6 and · is multiplication modulo 6.

When I was trying to explain why this is not a field, I came into a bit of trouble. I understand that for it to be a field the following must be true: Closure, Associative law, Identity and inverses. I wasn't sure how to disprove this. So far, Ive drawn a caylee table, but I havent been able to figure out what section disproves it. Any help would be much appreciated. Thanks in advance (:

N_Mathematics_B
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4 Answers4

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$2$ and $3$ do not have inverses and are zero divisors as $2 \times 3 = 6 = 0$.

mathcounterexamples.net
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Hint: What is the multiplicative inverse of $2$? That is, which number $x$ satisfies the property $2\cdot x=1$ in $\mathbb Z_6$?

5xum
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A ring R is a field if for every r in R, there exists s in R such that sr=rs=1(identity), i.e. every element has multiplicative inverse.

In Z[6], 2 and 3 do not have inverse, thus Z[6] is not a field.

Xinan Wang
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As both 5xum and mathcounterexamples.net have pointed to, 2 is a zero divisor mod 6, and hence does not have a multiplicative inverse.

A more general observation is that $$\mathbb{Z}/n \mathbb{Z}$$ is a field if and only if $n$ is a prime power. See here for answer to how to prove this.

pourquoi
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