Is $sqrtX-1$ an element of the completed ring $mathbbC[X]_mathfrakp$ with $mathfrakp = (X+1)$?
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Let's say we complete the ring $mathbbC[X]$ respect to the ideal $mathfrakp = (X+1)$, then we have the ring $mathbbC[X]_mathfrakp$. Can this ring have square roots of other polynomials? E.g.
$$ sqrtX-a in mathbbC[X]_(X+1) $$
for some $a in mathbbC$ ? This might be somewhat straightforward since we have the fundamental theorem of calculus available.
commutative-algebra maximal-and-prime-ideals
asked Sep 3 at 11:59
cactus314
15.2k41862
add a comment |Â
up vote
1
down vote
favorite
Let's say we complete the ring $mathbbC[X]$ respect to the ideal $mathfrakp = (X+1)$, then we have the ring $mathbbC[X]_mathfrakp$. Can this ring have square roots of other polynomials? E.g.
$$ sqrtX-a in mathbbC[X]_(X+1) $$
for some $a in mathbbC$ ? This might be somewhat straightforward since we have the fundamental theorem of calculus available.
commutative-algebra maximal-and-prime-ideals
asked Sep 3 at 11:59
cactus314
15.2k41862
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let's say we complete the ring $mathbbC[X]$ respect to the ideal $mathfrakp = (X+1)$, then we have the ring $mathbbC[X]_mathfrakp$. Can this ring have square roots of other polynomials? E.g.
$$ sqrtX-a in mathbbC[X]_(X+1) $$
for some $a in mathbbC$ ? This might be somewhat straightforward since we have the fundamental theorem of calculus available.
commutative-algebra maximal-and-prime-ideals
asked Sep 3 at 11:59
cactus314
15.2k41862
Let's say we complete the ring $mathbbC[X]$ respect to the ideal $mathfrakp = (X+1)$, then we have the ring $mathbbC[X]_mathfrakp$. Can this ring have square roots of other polynomials? E.g.
$$ sqrtX-a in mathbbC[X]_(X+1) $$
for some $a in mathbbC$ ? This might be somewhat straightforward since we have the fundamental theorem of calculus available.
commutative-algebra maximal-and-prime-ideals
commutative-algebra maximal-and-prime-ideals
asked Sep 3 at 11:59
cactus314
15.2k41862
asked Sep 3 at 11:59
cactus314
15.2k41862
asked Sep 3 at 11:59
cactus314
15.2k41862
asked Sep 3 at 11:59
cactus314
15.2k41862
asked Sep 3 at 11:59
cactus314
15.2k41862
15.2k41862
add a comment |Â
add a comment |Â
1 Answer
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active
oldest
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up vote
0
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accepted
The $mathbbC$-algebra morphism $Tmapsto X+1$ defines an isomorphism $mathbbC[T]cong mathbbC[X]$ identifying the ideal $(T)$ with the ideal $(X+1)$. In particular, we obtain an induced isomorphism after completing, i.e.
$$mathbbC[![ T ]!] cong (X+1)-textadic completion of mathbbC[X].$$ (Here I am using that the $(T)$-adic completion of $mathbbC[T]$ is the ring of formal power series). Now $X+a$ has a square root on the right-hand side if and only if $T+(a-1)$ has a square root on the left-hand side.
You can use the binomial formula to deduce that $T+b$ has a square root in $mathbbC[![ T ]!]$ if and only if $bneq 0$. Consequently, $X+a$ has a square root in your ring of interest if and only if $aneq 1$.
answered Sep 3 at 12:41
user363120
856310
OK. Binomial Theorem $(1 + x)^1/2 = sum binomfrac12k x^k$ thanks!
– cactus314
Sep 3 at 20:13
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The $mathbbC$-algebra morphism $Tmapsto X+1$ defines an isomorphism $mathbbC[T]cong mathbbC[X]$ identifying the ideal $(T)$ with the ideal $(X+1)$. In particular, we obtain an induced isomorphism after completing, i.e.
$$mathbbC[![ T ]!] cong (X+1)-textadic completion of mathbbC[X].$$ (Here I am using that the $(T)$-adic completion of $mathbbC[T]$ is the ring of formal power series). Now $X+a$ has a square root on the right-hand side if and only if $T+(a-1)$ has a square root on the left-hand side.
You can use the binomial formula to deduce that $T+b$ has a square root in $mathbbC[![ T ]!]$ if and only if $bneq 0$. Consequently, $X+a$ has a square root in your ring of interest if and only if $aneq 1$.
answered Sep 3 at 12:41
user363120
856310
OK. Binomial Theorem $(1 + x)^1/2 = sum binomfrac12k x^k$ thanks!
– cactus314
Sep 3 at 20:13
add a comment |Â
up vote
0
down vote
accepted
The $mathbbC$-algebra morphism $Tmapsto X+1$ defines an isomorphism $mathbbC[T]cong mathbbC[X]$ identifying the ideal $(T)$ with the ideal $(X+1)$. In particular, we obtain an induced isomorphism after completing, i.e.
$$mathbbC[![ T ]!] cong (X+1)-textadic completion of mathbbC[X].$$ (Here I am using that the $(T)$-adic completion of $mathbbC[T]$ is the ring of formal power series). Now $X+a$ has a square root on the right-hand side if and only if $T+(a-1)$ has a square root on the left-hand side.
You can use the binomial formula to deduce that $T+b$ has a square root in $mathbbC[![ T ]!]$ if and only if $bneq 0$. Consequently, $X+a$ has a square root in your ring of interest if and only if $aneq 1$.
answered Sep 3 at 12:41
user363120
856310
OK. Binomial Theorem $(1 + x)^1/2 = sum binomfrac12k x^k$ thanks!
– cactus314
Sep 3 at 20:13
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The $mathbbC$-algebra morphism $Tmapsto X+1$ defines an isomorphism $mathbbC[T]cong mathbbC[X]$ identifying the ideal $(T)$ with the ideal $(X+1)$. In particular, we obtain an induced isomorphism after completing, i.e.
$$mathbbC[![ T ]!] cong (X+1)-textadic completion of mathbbC[X].$$ (Here I am using that the $(T)$-adic completion of $mathbbC[T]$ is the ring of formal power series). Now $X+a$ has a square root on the right-hand side if and only if $T+(a-1)$ has a square root on the left-hand side.
You can use the binomial formula to deduce that $T+b$ has a square root in $mathbbC[![ T ]!]$ if and only if $bneq 0$. Consequently, $X+a$ has a square root in your ring of interest if and only if $aneq 1$.
answered Sep 3 at 12:41
user363120
856310
The $mathbbC$-algebra morphism $Tmapsto X+1$ defines an isomorphism $mathbbC[T]cong mathbbC[X]$ identifying the ideal $(T)$ with the ideal $(X+1)$. In particular, we obtain an induced isomorphism after completing, i.e.
$$mathbbC[![ T ]!] cong (X+1)-textadic completion of mathbbC[X].$$ (Here I am using that the $(T)$-adic completion of $mathbbC[T]$ is the ring of formal power series). Now $X+a$ has a square root on the right-hand side if and only if $T+(a-1)$ has a square root on the left-hand side.
You can use the binomial formula to deduce that $T+b$ has a square root in $mathbbC[![ T ]!]$ if and only if $bneq 0$. Consequently, $X+a$ has a square root in your ring of interest if and only if $aneq 1$.
answered Sep 3 at 12:41
user363120
856310
answered Sep 3 at 12:41
user363120
856310
answered Sep 3 at 12:41
user363120
856310
answered Sep 3 at 12:41
user363120
856310
856310
OK. Binomial Theorem $(1 + x)^1/2 = sum binomfrac12k x^k$ thanks!
– cactus314
Sep 3 at 20:13
add a comment |Â
OK. Binomial Theorem $(1 + x)^1/2 = sum binomfrac12k x^k$ thanks!
– cactus314
Sep 3 at 20:13
OK. Binomial Theorem $(1 + x)^1/2 = sum binomfrac12k x^k$ thanks!
– cactus314
Sep 3 at 20:13
OK. Binomial Theorem $(1 + x)^1/2 = sum binomfrac12k x^k$ thanks!
– cactus314
Sep 3 at 20:13
add a comment |Â
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