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Explicit expression of a completion
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Consider the local ring $A=mathbb Z[t]_(p,t)$, where clearly $p$ is a prime number. What is the explicit expression of the completion of $A$ with respect to the maximal ideal $(p,t)$? Is it $mathbb Z_(p)[[t]]$?
abstract-algebra ring-theory commutative-algebra examples-counterexamples
asked Jul 18 at 17:33
manifold
315213
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up vote
3
down vote
favorite
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Consider the local ring $A=mathbb Z[t]_(p,t)$, where clearly $p$ is a prime number. What is the explicit expression of the completion of $A$ with respect to the maximal ideal $(p,t)$? Is it $mathbb Z_(p)[[t]]$?
abstract-algebra ring-theory commutative-algebra examples-counterexamples
asked Jul 18 at 17:33
manifold
315213
I think it would be a little different from what you wrote. I think it would be R[[t]], where R is the ring of p-adic integers. This should follow from the third example on the Wikipedia page: en.wikipedia.org/wiki/Completion_(algebra) (For example, 1 + p + p^2 + p^3 + ... would be in the completion but not in Z_(p) [[t]].)
– CJD
Jul 18 at 18:42
why $mathbb Z[t]_(p,t)[[x,y]]/(x-p,y-t)congmathbb Z_p[[t]]$?
– manifold
Jul 18 at 19:08
Do you believe it without the localization? The localization part shouldn't matter because after completion everything outside of (p,t) will be a unit anyway, see for example here: math.stackexchange.com/a/38399/85329 Also, do you know that Z[[x]]/(x-p) is isomorphic to the p-adic integers?
– CJD
Jul 18 at 19:23
I know the last property that you mentioned, but I still have problems to see the isomorphism. Btw thank you.
– manifold
Jul 18 at 19:29
You're welcome. I'm sorry for not writing out a complete answer. I don't think it would look very enlightening, at least if I wrote it. The forward map should definitely send x to p and y to t. The backwards map should send t to y. The hard part would be checking that everything is well-defined. For example, if you have a map from R[[x,y]] to S with (x-p, y-t) in the kernel, then it automatically induces a map R[[x,y]]/(x-p, y-t) --> S.
– CJD
Jul 18 at 19:35
add a comment |Â
up vote
3
down vote
favorite
1
up vote
3
down vote
favorite
1
1
Consider the local ring $A=mathbb Z[t]_(p,t)$, where clearly $p$ is a prime number. What is the explicit expression of the completion of $A$ with respect to the maximal ideal $(p,t)$? Is it $mathbb Z_(p)[[t]]$?
abstract-algebra ring-theory commutative-algebra examples-counterexamples
asked Jul 18 at 17:33
manifold
315213
Consider the local ring $A=mathbb Z[t]_(p,t)$, where clearly $p$ is a prime number. What is the explicit expression of the completion of $A$ with respect to the maximal ideal $(p,t)$? Is it $mathbb Z_(p)[[t]]$?
abstract-algebra ring-theory commutative-algebra examples-counterexamples
asked Jul 18 at 17:33
manifold
315213
edited Jul 18 at 18:26
asked Jul 18 at 17:33
manifold
315213
asked Jul 18 at 17:33
manifold
315213
asked Jul 18 at 17:33
manifold
315213
315213
I think it would be a little different from what you wrote. I think it would be R[[t]], where R is the ring of p-adic integers. This should follow from the third example on the Wikipedia page: en.wikipedia.org/wiki/Completion_(algebra) (For example, 1 + p + p^2 + p^3 + ... would be in the completion but not in Z_(p) [[t]].)
– CJD
Jul 18 at 18:42
why $mathbb Z[t]_(p,t)[[x,y]]/(x-p,y-t)congmathbb Z_p[[t]]$?
– manifold
Jul 18 at 19:08
Do you believe it without the localization? The localization part shouldn't matter because after completion everything outside of (p,t) will be a unit anyway, see for example here: math.stackexchange.com/a/38399/85329 Also, do you know that Z[[x]]/(x-p) is isomorphic to the p-adic integers?
– CJD
Jul 18 at 19:23
I know the last property that you mentioned, but I still have problems to see the isomorphism. Btw thank you.
– manifold
Jul 18 at 19:29
You're welcome. I'm sorry for not writing out a complete answer. I don't think it would look very enlightening, at least if I wrote it. The forward map should definitely send x to p and y to t. The backwards map should send t to y. The hard part would be checking that everything is well-defined. For example, if you have a map from R[[x,y]] to S with (x-p, y-t) in the kernel, then it automatically induces a map R[[x,y]]/(x-p, y-t) --> S.
– CJD
Jul 18 at 19:35
add a comment |Â
I think it would be a little different from what you wrote. I think it would be R[[t]], where R is the ring of p-adic integers. This should follow from the third example on the Wikipedia page: en.wikipedia.org/wiki/Completion_(algebra) (For example, 1 + p + p^2 + p^3 + ... would be in the completion but not in Z_(p) [[t]].)
– CJD
Jul 18 at 18:42
why $mathbb Z[t]_(p,t)[[x,y]]/(x-p,y-t)congmathbb Z_p[[t]]$?
– manifold
Jul 18 at 19:08
Do you believe it without the localization? The localization part shouldn't matter because after completion everything outside of (p,t) will be a unit anyway, see for example here: math.stackexchange.com/a/38399/85329 Also, do you know that Z[[x]]/(x-p) is isomorphic to the p-adic integers?
– CJD
Jul 18 at 19:23
I know the last property that you mentioned, but I still have problems to see the isomorphism. Btw thank you.
– manifold
Jul 18 at 19:29
You're welcome. I'm sorry for not writing out a complete answer. I don't think it would look very enlightening, at least if I wrote it. The forward map should definitely send x to p and y to t. The backwards map should send t to y. The hard part would be checking that everything is well-defined. For example, if you have a map from R[[x,y]] to S with (x-p, y-t) in the kernel, then it automatically induces a map R[[x,y]]/(x-p, y-t) --> S.
– CJD
Jul 18 at 19:35
I think it would be a little different from what you wrote. I think it would be R[[t]], where R is the ring of p-adic integers. This should follow from the third example on the Wikipedia page: en.wikipedia.org/wiki/Completion_(algebra) (For example, 1 + p + p^2 + p^3 + ... would be in the completion but not in Z_(p) [[t]].)
– CJD
Jul 18 at 18:42
I think it would be a little different from what you wrote. I think it would be R[[t]], where R is the ring of p-adic integers. This should follow from the third example on the Wikipedia page: en.wikipedia.org/wiki/Completion_(algebra) (For example, 1 + p + p^2 + p^3 + ... would be in the completion but not in Z_(p) [[t]].)
– CJD
Jul 18 at 18:42
why $mathbb Z[t]_(p,t)[[x,y]]/(x-p,y-t)congmathbb Z_p[[t]]$?
– manifold
Jul 18 at 19:08
why $mathbb Z[t]_(p,t)[[x,y]]/(x-p,y-t)congmathbb Z_p[[t]]$?
– manifold
Jul 18 at 19:08
Do you believe it without the localization? The localization part shouldn't matter because after completion everything outside of (p,t) will be a unit anyway, see for example here: math.stackexchange.com/a/38399/85329 Also, do you know that Z[[x]]/(x-p) is isomorphic to the p-adic integers?
– CJD
Jul 18 at 19:23
Do you believe it without the localization? The localization part shouldn't matter because after completion everything outside of (p,t) will be a unit anyway, see for example here: math.stackexchange.com/a/38399/85329 Also, do you know that Z[[x]]/(x-p) is isomorphic to the p-adic integers?
– CJD
Jul 18 at 19:23
I know the last property that you mentioned, but I still have problems to see the isomorphism. Btw thank you.
– manifold
Jul 18 at 19:29
I know the last property that you mentioned, but I still have problems to see the isomorphism. Btw thank you.
– manifold
Jul 18 at 19:29
You're welcome. I'm sorry for not writing out a complete answer. I don't think it would look very enlightening, at least if I wrote it. The forward map should definitely send x to p and y to t. The backwards map should send t to y. The hard part would be checking that everything is well-defined. For example, if you have a map from R[[x,y]] to S with (x-p, y-t) in the kernel, then it automatically induces a map R[[x,y]]/(x-p, y-t) --> S.
– CJD
Jul 18 at 19:35
You're welcome. I'm sorry for not writing out a complete answer. I don't think it would look very enlightening, at least if I wrote it. The forward map should definitely send x to p and y to t. The backwards map should send t to y. The hard part would be checking that everything is well-defined. For example, if you have a map from R[[x,y]] to S with (x-p, y-t) in the kernel, then it automatically induces a map R[[x,y]]/(x-p, y-t) --> S.
– CJD
Jul 18 at 19:35
add a comment |Â
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I think it would be a little different from what you wrote. I think it would be R[[t]], where R is the ring of p-adic integers. This should follow from the third example on the Wikipedia page: en.wikipedia.org/wiki/Completion_(algebra) (For example, 1 + p + p^2 + p^3 + ... would be in the completion but not in Z_(p) [[t]].)
– CJD
Jul 18 at 18:42
why $mathbb Z[t]_(p,t)[[x,y]]/(x-p,y-t)congmathbb Z_p[[t]]$?
– manifold
Jul 18 at 19:08
Do you believe it without the localization? The localization part shouldn't matter because after completion everything outside of (p,t) will be a unit anyway, see for example here: math.stackexchange.com/a/38399/85329 Also, do you know that Z[[x]]/(x-p) is isomorphic to the p-adic integers?
– CJD
Jul 18 at 19:23
I know the last property that you mentioned, but I still have problems to see the isomorphism. Btw thank you.
– manifold
Jul 18 at 19:29
You're welcome. I'm sorry for not writing out a complete answer. I don't think it would look very enlightening, at least if I wrote it. The forward map should definitely send x to p and y to t. The backwards map should send t to y. The hard part would be checking that everything is well-defined. For example, if you have a map from R[[x,y]] to S with (x-p, y-t) in the kernel, then it automatically induces a map R[[x,y]]/(x-p, y-t) --> S.
– CJD
Jul 18 at 19:35