Prove $ mathbb{Z}[sqrt {-5}] $ is not a UFD without factoring?
3
$begingroup$
We know the ring $ mathbb{Z}[sqrt{-5}] $ is not a UFD.
The typical proof is showing that $6$ factors in $2$ ways.
But there must be a better way to show this than to do trial and error factoring of some random element of the ring.
So how does one prove $ mathbb{Z}[sqrt{-5}] $ is not a UFD without factoring a certain element ?
I can show it must be atomic and has a finite amount of units. But that is insufficient and perhaps even irrelevant ; UFD have that too.
ring-theory factoring alternative-proof unique-factorization-domains
edited Jan 24 at 12:10
DonAntonio
179k1494233
asked Jan 24 at 12:07
mickmick
5,17832164
$endgroup$
|
show 10 more comments
3
$begingroup$
We know the ring $ mathbb{Z}[sqrt{-5}] $ is not a UFD.
The typical proof is showing that $6$ factors in $2$ ways.
But there must be a better way to show this than to do trial and error factoring of some random element of the ring.
So how does one prove $ mathbb{Z}[sqrt{-5}] $ is not a UFD without factoring a certain element ?
I can show it must be atomic and has a finite amount of units. But that is insufficient and perhaps even irrelevant ; UFD have that too.
ring-theory factoring alternative-proof unique-factorization-domains
edited Jan 24 at 12:10
DonAntonio
179k1494233
asked Jan 24 at 12:07
mickmick
5,17832164
$endgroup$
$begingroup$
Show that it's not a GCD domain, for example. i.e. find two elements such that the ideal generated by them is not principal.
$endgroup$
– stressed out
Jan 24 at 12:09
$begingroup$
But how do we do that without factoring ? Seems alot like factoring 6 to me.
$endgroup$
– mick
Jan 24 at 12:12
2
$begingroup$
In the title you say something, but in the question itself you say something way harsh: "There must be a better way" . Better....wrt what, or for whom? Showing the decomposition in two wasy of $;6;$ , say, looks to me as simple and "good" as one can expect...what do you mean by "better" here? And the element is not that random: doing a little arithmetic one gets with very simple elements the dcompositions!
$endgroup$
– DonAntonio
Jan 24 at 12:16
1
$begingroup$
It depends what background you allow. The method using a factorization of 6 has the advantage of being simple: it does not need a lot of machinery. You could instead show the ideal $(2,1+sqrt{-5})$ is not principal: it is known that a Dedekind domain is a UFD iff it is a PID. But that requires a lot more work to justify (both proving the result and showing $mathbf Z[sqrt{-5}]$ is a Dedekind domain). The Dedekind property is crucial: $mathbf Z[x]$ is a UFD and not a PID!
$endgroup$
– KCd
Jan 24 at 12:18
3
$begingroup$
Again, $,6,$ is not that random. In $;Bbb Z[sqrt{-n}];,;;ninBbb N;$, I would go first precisely on $;n+1;$ . Why? because $$(1-sqrt{-n})(1+sqrt{-n})=1+n$$ Randomness doesn't work a lot in mahtematics.
$endgroup$
– DonAntonio
Jan 24 at 12:37
|
show 10 more comments
3
3
3
1
$begingroup$
We know the ring $ mathbb{Z}[sqrt{-5}] $ is not a UFD.
The typical proof is showing that $6$ factors in $2$ ways.
But there must be a better way to show this than to do trial and error factoring of some random element of the ring.
So how does one prove $ mathbb{Z}[sqrt{-5}] $ is not a UFD without factoring a certain element ?
I can show it must be atomic and has a finite amount of units. But that is insufficient and perhaps even irrelevant ; UFD have that too.
ring-theory factoring alternative-proof unique-factorization-domains
edited Jan 24 at 12:10
DonAntonio
179k1494233
asked Jan 24 at 12:07
mickmick
5,17832164
$endgroup$
We know the ring $ mathbb{Z}[sqrt{-5}] $ is not a UFD.
The typical proof is showing that $6$ factors in $2$ ways.
But there must be a better way to show this than to do trial and error factoring of some random element of the ring.
So how does one prove $ mathbb{Z}[sqrt{-5}] $ is not a UFD without factoring a certain element ?
I can show it must be atomic and has a finite amount of units. But that is insufficient and perhaps even irrelevant ; UFD have that too.
ring-theory factoring alternative-proof unique-factorization-domains
ring-theory factoring alternative-proof unique-factorization-domains
edited Jan 24 at 12:10
DonAntonio
179k1494233
asked Jan 24 at 12:07
mickmick
5,17832164
edited Jan 24 at 12:10
DonAntonio
179k1494233
asked Jan 24 at 12:07
mickmick
5,17832164
edited Jan 24 at 12:10
DonAntonio
179k1494233
edited Jan 24 at 12:10
DonAntonio
179k1494233
edited Jan 24 at 12:10
DonAntonio
179k1494233
179k1494233
asked Jan 24 at 12:07
mickmick
5,17832164
asked Jan 24 at 12:07
mickmick
5,17832164
asked Jan 24 at 12:07
mickmick
5,17832164
5,17832164
$begingroup$
Show that it's not a GCD domain, for example. i.e. find two elements such that the ideal generated by them is not principal.
$endgroup$
– stressed out
Jan 24 at 12:09
$begingroup$
But how do we do that without factoring ? Seems alot like factoring 6 to me.
$endgroup$
– mick
Jan 24 at 12:12
2
$begingroup$
In the title you say something, but in the question itself you say something way harsh: "There must be a better way" . Better....wrt what, or for whom? Showing the decomposition in two wasy of $;6;$ , say, looks to me as simple and "good" as one can expect...what do you mean by "better" here? And the element is not that random: doing a little arithmetic one gets with very simple elements the dcompositions!
$endgroup$
– DonAntonio
Jan 24 at 12:16
1
$begingroup$
It depends what background you allow. The method using a factorization of 6 has the advantage of being simple: it does not need a lot of machinery. You could instead show the ideal $(2,1+sqrt{-5})$ is not principal: it is known that a Dedekind domain is a UFD iff it is a PID. But that requires a lot more work to justify (both proving the result and showing $mathbf Z[sqrt{-5}]$ is a Dedekind domain). The Dedekind property is crucial: $mathbf Z[x]$ is a UFD and not a PID!
$endgroup$
– KCd
Jan 24 at 12:18
3
$begingroup$
Again, $,6,$ is not that random. In $;Bbb Z[sqrt{-n}];,;;ninBbb N;$, I would go first precisely on $;n+1;$ . Why? because $$(1-sqrt{-n})(1+sqrt{-n})=1+n$$ Randomness doesn't work a lot in mahtematics.
$endgroup$
– DonAntonio
Jan 24 at 12:37
|
show 10 more comments
$begingroup$
Show that it's not a GCD domain, for example. i.e. find two elements such that the ideal generated by them is not principal.
$endgroup$
– stressed out
Jan 24 at 12:09
$begingroup$
But how do we do that without factoring ? Seems alot like factoring 6 to me.
$endgroup$
– mick
Jan 24 at 12:12
2
$begingroup$
In the title you say something, but in the question itself you say something way harsh: "There must be a better way" . Better....wrt what, or for whom? Showing the decomposition in two wasy of $;6;$ , say, looks to me as simple and "good" as one can expect...what do you mean by "better" here? And the element is not that random: doing a little arithmetic one gets with very simple elements the dcompositions!
$endgroup$
– DonAntonio
Jan 24 at 12:16
1
$begingroup$
It depends what background you allow. The method using a factorization of 6 has the advantage of being simple: it does not need a lot of machinery. You could instead show the ideal $(2,1+sqrt{-5})$ is not principal: it is known that a Dedekind domain is a UFD iff it is a PID. But that requires a lot more work to justify (both proving the result and showing $mathbf Z[sqrt{-5}]$ is a Dedekind domain). The Dedekind property is crucial: $mathbf Z[x]$ is a UFD and not a PID!
$endgroup$
– KCd
Jan 24 at 12:18
3
$begingroup$
Again, $,6,$ is not that random. In $;Bbb Z[sqrt{-n}];,;;ninBbb N;$, I would go first precisely on $;n+1;$ . Why? because $$(1-sqrt{-n})(1+sqrt{-n})=1+n$$ Randomness doesn't work a lot in mahtematics.
$endgroup$
– DonAntonio
Jan 24 at 12:37
$begingroup$
Show that it's not a GCD domain, for example. i.e. find two elements such that the ideal generated by them is not principal.
$endgroup$
– stressed out
Jan 24 at 12:09
$begingroup$
Show that it's not a GCD domain, for example. i.e. find two elements such that the ideal generated by them is not principal.
$endgroup$
– stressed out
Jan 24 at 12:09
$begingroup$
But how do we do that without factoring ? Seems alot like factoring 6 to me.
$endgroup$
– mick
Jan 24 at 12:12
$begingroup$
But how do we do that without factoring ? Seems alot like factoring 6 to me.
$endgroup$
– mick
Jan 24 at 12:12
2
2
$begingroup$
In the title you say something, but in the question itself you say something way harsh: "There must be a better way" . Better....wrt what, or for whom? Showing the decomposition in two wasy of $;6;$ , say, looks to me as simple and "good" as one can expect...what do you mean by "better" here? And the element is not that random: doing a little arithmetic one gets with very simple elements the dcompositions!
$endgroup$
– DonAntonio
Jan 24 at 12:16
$begingroup$
In the title you say something, but in the question itself you say something way harsh: "There must be a better way" . Better....wrt what, or for whom? Showing the decomposition in two wasy of $;6;$ , say, looks to me as simple and "good" as one can expect...what do you mean by "better" here? And the element is not that random: doing a little arithmetic one gets with very simple elements the dcompositions!
$endgroup$
– DonAntonio
Jan 24 at 12:16
1
1
$begingroup$
It depends what background you allow. The method using a factorization of 6 has the advantage of being simple: it does not need a lot of machinery. You could instead show the ideal $(2,1+sqrt{-5})$ is not principal: it is known that a Dedekind domain is a UFD iff it is a PID. But that requires a lot more work to justify (both proving the result and showing $mathbf Z[sqrt{-5}]$ is a Dedekind domain). The Dedekind property is crucial: $mathbf Z[x]$ is a UFD and not a PID!
$endgroup$
– KCd
Jan 24 at 12:18
$begingroup$
It depends what background you allow. The method using a factorization of 6 has the advantage of being simple: it does not need a lot of machinery. You could instead show the ideal $(2,1+sqrt{-5})$ is not principal: it is known that a Dedekind domain is a UFD iff it is a PID. But that requires a lot more work to justify (both proving the result and showing $mathbf Z[sqrt{-5}]$ is a Dedekind domain). The Dedekind property is crucial: $mathbf Z[x]$ is a UFD and not a PID!
$endgroup$
– KCd
Jan 24 at 12:18
3
3
$begingroup$
Again, $,6,$ is not that random. In $;Bbb Z[sqrt{-n}];,;;ninBbb N;$, I would go first precisely on $;n+1;$ . Why? because $$(1-sqrt{-n})(1+sqrt{-n})=1+n$$ Randomness doesn't work a lot in mahtematics.
$endgroup$
– DonAntonio
Jan 24 at 12:37
$begingroup$
Again, $,6,$ is not that random. In $;Bbb Z[sqrt{-n}];,;;ninBbb N;$, I would go first precisely on $;n+1;$ . Why? because $$(1-sqrt{-n})(1+sqrt{-n})=1+n$$ Randomness doesn't work a lot in mahtematics.
$endgroup$
– DonAntonio
Jan 24 at 12:37
|
show 10 more comments
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$begingroup$
Show that it's not a GCD domain, for example. i.e. find two elements such that the ideal generated by them is not principal.
$endgroup$
– stressed out
Jan 24 at 12:09
$begingroup$
But how do we do that without factoring ? Seems alot like factoring 6 to me.
$endgroup$
– mick
Jan 24 at 12:12
2
$begingroup$
In the title you say something, but in the question itself you say something way harsh: "There must be a better way" . Better....wrt what, or for whom? Showing the decomposition in two wasy of $;6;$ , say, looks to me as simple and "good" as one can expect...what do you mean by "better" here? And the element is not that random: doing a little arithmetic one gets with very simple elements the dcompositions!
$endgroup$
– DonAntonio
Jan 24 at 12:16
1
$begingroup$
It depends what background you allow. The method using a factorization of 6 has the advantage of being simple: it does not need a lot of machinery. You could instead show the ideal $(2,1+sqrt{-5})$ is not principal: it is known that a Dedekind domain is a UFD iff it is a PID. But that requires a lot more work to justify (both proving the result and showing $mathbf Z[sqrt{-5}]$ is a Dedekind domain). The Dedekind property is crucial: $mathbf Z[x]$ is a UFD and not a PID!
$endgroup$
– KCd
Jan 24 at 12:18
3
$begingroup$
Again, $,6,$ is not that random. In $;Bbb Z[sqrt{-n}];,;;ninBbb N;$, I would go first precisely on $;n+1;$ . Why? because $$(1-sqrt{-n})(1+sqrt{-n})=1+n$$ Randomness doesn't work a lot in mahtematics.
$endgroup$
– DonAntonio
Jan 24 at 12:37