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absolute value and valuation on the field of formal power series
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I'm currently working on (non-Archimedean) valuations and absolute values. My text states that one example, where I have both a non-Archimedean valuation and a corresponding absolute value is $mathbbC((t))$ , i.e. the field of formal power series with coefficients in $mathbbC$, defined as
$$mathbbC((t))= sum_k=m^infty a_kt^k : m in mathbbZ, a_k in mathbbC.$$
I tried to find how a valuation and an absolute value on this field are defined, but I came up empty-handed.
It would be great if any of you could help me! (If it helps - they mentioned in the same context also $mathbbQ_p$ with the $p$-adic valuation/ absolute value and the completion $mathbbC_p$).
absolute-value valuation-theory formal-power-series
asked Jul 22 at 12:27
SallyOwens
30118
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up vote
0
down vote
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I'm currently working on (non-Archimedean) valuations and absolute values. My text states that one example, where I have both a non-Archimedean valuation and a corresponding absolute value is $mathbbC((t))$ , i.e. the field of formal power series with coefficients in $mathbbC$, defined as
$$mathbbC((t))= sum_k=m^infty a_kt^k : m in mathbbZ, a_k in mathbbC.$$
I tried to find how a valuation and an absolute value on this field are defined, but I came up empty-handed.
It would be great if any of you could help me! (If it helps - they mentioned in the same context also $mathbbQ_p$ with the $p$-adic valuation/ absolute value and the completion $mathbbC_p$).
absolute-value valuation-theory formal-power-series
asked Jul 22 at 12:27
SallyOwens
30118
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm currently working on (non-Archimedean) valuations and absolute values. My text states that one example, where I have both a non-Archimedean valuation and a corresponding absolute value is $mathbbC((t))$ , i.e. the field of formal power series with coefficients in $mathbbC$, defined as
$$mathbbC((t))= sum_k=m^infty a_kt^k : m in mathbbZ, a_k in mathbbC.$$
I tried to find how a valuation and an absolute value on this field are defined, but I came up empty-handed.
It would be great if any of you could help me! (If it helps - they mentioned in the same context also $mathbbQ_p$ with the $p$-adic valuation/ absolute value and the completion $mathbbC_p$).
absolute-value valuation-theory formal-power-series
asked Jul 22 at 12:27
SallyOwens
30118
I'm currently working on (non-Archimedean) valuations and absolute values. My text states that one example, where I have both a non-Archimedean valuation and a corresponding absolute value is $mathbbC((t))$ , i.e. the field of formal power series with coefficients in $mathbbC$, defined as
$$mathbbC((t))= sum_k=m^infty a_kt^k : m in mathbbZ, a_k in mathbbC.$$
I tried to find how a valuation and an absolute value on this field are defined, but I came up empty-handed.
It would be great if any of you could help me! (If it helps - they mentioned in the same context also $mathbbQ_p$ with the $p$-adic valuation/ absolute value and the completion $mathbbC_p$).
absolute-value valuation-theory formal-power-series
asked Jul 22 at 12:27
SallyOwens
30118
asked Jul 22 at 12:27
SallyOwens
30118
asked Jul 22 at 12:27
SallyOwens
30118
asked Jul 22 at 12:27
SallyOwens
30118
30118
add a comment |Â
add a comment |Â
1 Answer
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The valuation $v(f(t))$ is given by the number of times $t$ divides the power series $f(t)$. The norm is then defined accordingly, i.e. one may define it by setting for example $|f(t)| := e^-v(f(t))$. (More generally, this kind of construction whenever you take a Dedekind domain and consider some nonzero prime ideal of it. In this case the ring is $mathbbC[[t]]$ and the ideal is $(t)$. The case where the domain is $mathbbZ$ and the ideal is $(p)$ gives the usual case of $mathbbQ_p$).
answered Jul 22 at 12:33
Gal Porat
718513
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The valuation $v(f(t))$ is given by the number of times $t$ divides the power series $f(t)$. The norm is then defined accordingly, i.e. one may define it by setting for example $|f(t)| := e^-v(f(t))$. (More generally, this kind of construction whenever you take a Dedekind domain and consider some nonzero prime ideal of it. In this case the ring is $mathbbC[[t]]$ and the ideal is $(t)$. The case where the domain is $mathbbZ$ and the ideal is $(p)$ gives the usual case of $mathbbQ_p$).
answered Jul 22 at 12:33
Gal Porat
718513
add a comment |Â
up vote
2
down vote
accepted
The valuation $v(f(t))$ is given by the number of times $t$ divides the power series $f(t)$. The norm is then defined accordingly, i.e. one may define it by setting for example $|f(t)| := e^-v(f(t))$. (More generally, this kind of construction whenever you take a Dedekind domain and consider some nonzero prime ideal of it. In this case the ring is $mathbbC[[t]]$ and the ideal is $(t)$. The case where the domain is $mathbbZ$ and the ideal is $(p)$ gives the usual case of $mathbbQ_p$).
answered Jul 22 at 12:33
Gal Porat
718513
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The valuation $v(f(t))$ is given by the number of times $t$ divides the power series $f(t)$. The norm is then defined accordingly, i.e. one may define it by setting for example $|f(t)| := e^-v(f(t))$. (More generally, this kind of construction whenever you take a Dedekind domain and consider some nonzero prime ideal of it. In this case the ring is $mathbbC[[t]]$ and the ideal is $(t)$. The case where the domain is $mathbbZ$ and the ideal is $(p)$ gives the usual case of $mathbbQ_p$).
answered Jul 22 at 12:33
Gal Porat
718513
The valuation $v(f(t))$ is given by the number of times $t$ divides the power series $f(t)$. The norm is then defined accordingly, i.e. one may define it by setting for example $|f(t)| := e^-v(f(t))$. (More generally, this kind of construction whenever you take a Dedekind domain and consider some nonzero prime ideal of it. In this case the ring is $mathbbC[[t]]$ and the ideal is $(t)$. The case where the domain is $mathbbZ$ and the ideal is $(p)$ gives the usual case of $mathbbQ_p$).
answered Jul 22 at 12:33
Gal Porat
718513
answered Jul 22 at 12:33
Gal Porat
718513
answered Jul 22 at 12:33
Gal Porat
718513
answered Jul 22 at 12:33
Gal Porat
718513
718513
add a comment |Â
add a comment |Â
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