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What do we get by quoting $2\mathbb{Z}$ with $6\mathbb{Z}$?

(I want to see if given a commutative ring without unity, can the quotient have unity?)

J.G.131
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pucky
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  • You can easily write down the representatives and their multiplication table. – Amateur_Algebraist Jun 06 '24 at 15:31
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    Heads up on avoiding downvotes (people can be mean on here), maybe a better way to structure you question: "I'm curious if when given a ring without unity, then will the quotient ring always have unit? An example I'm considering is $2\mathbb{Z}/6\mathbb{Z}$, but the structure of this ring is confusing me." But to answer your question, your quotient group will have ${[0],[2],[4]}$ as its underlying set. It should be easy to see from here that your ring will not have unit...though if its not, please don't be afraid to ask. – J.G.131 Jun 06 '24 at 15:37
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    Here's a hint: $6:2=3$. – Andrea Mori Jun 06 '24 at 15:54
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    @JAG131 The wording is not really relevant in case of a very simple question which should obviously be a duplicate: $2 \mathbb{Z}$/$6 \mathbb{Z} $ is isomorphic to $\mathbb{Z_3}$. I did not downvote. – Amateur_Algebraist Jun 06 '24 at 20:04

1 Answers1

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You can just start from $0$ and then add $2$ each time.

Thus, we get three elements: $0+6\mathbb{Z}$, $2+6\mathbb{Z}$, $4+6\mathbb{Z}$ (and $6+6\mathbb{Z}?$) No, since this is equal to $0+6\mathbb{Z}.$

Then check whether it has a unity:

$(2+6\mathbb{Z})(2+6\mathbb{Z})=4+6\mathbb{Z}\neq2+6\mathbb{Z}$, thus $2+6\mathbb{Z}$ can not be a unity.

So, there is only one choice: $4+6\mathbb{Z}$

$(4+6\mathbb{Z})(2+6\mathbb{Z})=2+6\mathbb{Z}$, which is fine. $(4+6\mathbb{Z})(4+6\mathbb{Z})=4+6\mathbb{Z}$, also ok.

Thus, $4+6\mathbb{Z}$ is the unity for this ring.

Bowei Tang
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