Mixing
Is $Bbb Q[alpha]$ a field ? where, $alpha in Bbb C$ such that $alpha + pi = {pi}^2 alpha $
$begingroup$
Is $Bbb Q[alpha]$ a field ? Here $alpha in Bbb C$ is such that $alpha + pi = {pi}^2 alpha $ .
I tried to argue by contradiction. Tried to come up with tricks commonly used for concluding that if $theta$ is transcendental, then so is $theta^2$, so is $sqrt{theta}$ and so on. But couldn't come up with a suitable one!
Any hint will be really appreciated.
abstract-algebra elementary-number-theory field-theory algebraic-number-theory transcendental-numbers
edited Jan 22 at 20:13
user26857
39.3k124183
asked Jan 21 at 13:49
CoherentCoherent
1,191623
$endgroup$
|
show 2 more comments
$begingroup$
Is $Bbb Q[alpha]$ a field ? Here $alpha in Bbb C$ is such that $alpha + pi = {pi}^2 alpha $ .
I tried to argue by contradiction. Tried to come up with tricks commonly used for concluding that if $theta$ is transcendental, then so is $theta^2$, so is $sqrt{theta}$ and so on. But couldn't come up with a suitable one!
Any hint will be really appreciated.
abstract-algebra elementary-number-theory field-theory algebraic-number-theory transcendental-numbers
edited Jan 22 at 20:13
user26857
39.3k124183
asked Jan 21 at 13:49
CoherentCoherent
1,191623
$endgroup$
3
$begingroup$
$alpha=frac{pi}{pi^2-1}$. Your ring is a field iff $alpha$ is algebraic. Thus, if $sum_k{a_kX^k}$ is a polynomial with degree $d$ and rational coefficients with $alpha$ as a root, then $sum_k{a_kX^k(X^2-1)^{d-k}}$ is a polynomial with rational coefficients, degree $2d$ if $a_0 neq 0$, and $pi$ is a root.
$endgroup$
– Mindlack
Jan 21 at 14:04
1
$begingroup$
What Mindlack said. Alternatively, the equation shows that $pi$ is algebraic over the field $Bbb{Q}(alpha)$. If $alpha$ were algebraic over $Bbb{Q}$ then ...
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:19
$begingroup$
@Mindlack Can I simply take any polynomial i.e. $a_n X^n+dots+a_0$ where $a_0 ne 0$ since $alpha notin Bbb Q$ to have $alpha$ as a root and thus, $a_n ({frac{pi}{{pi}^2 -1}})^n+dots+a_0=0 implies a_n {pi}^n+dots+a_0 {({pi}^2 -1)}^n=0$ and arrive at a contradiction!
$endgroup$
– Coherent
Jan 21 at 14:25
$begingroup$
Yes, but pay a closer attention to the powers of $pi^2-1$.
$endgroup$
– Mindlack
Jan 21 at 14:29
$begingroup$
@Mindlack Now,it's alright?
$endgroup$
– Coherent
Jan 21 at 14:31
|
show 2 more comments
$begingroup$
Is $Bbb Q[alpha]$ a field ? Here $alpha in Bbb C$ is such that $alpha + pi = {pi}^2 alpha $ .
I tried to argue by contradiction. Tried to come up with tricks commonly used for concluding that if $theta$ is transcendental, then so is $theta^2$, so is $sqrt{theta}$ and so on. But couldn't come up with a suitable one!
Any hint will be really appreciated.
abstract-algebra elementary-number-theory field-theory algebraic-number-theory transcendental-numbers
edited Jan 22 at 20:13
user26857
39.3k124183
asked Jan 21 at 13:49
CoherentCoherent
1,191623
$endgroup$
Is $Bbb Q[alpha]$ a field ? Here $alpha in Bbb C$ is such that $alpha + pi = {pi}^2 alpha $ .
I tried to argue by contradiction. Tried to come up with tricks commonly used for concluding that if $theta$ is transcendental, then so is $theta^2$, so is $sqrt{theta}$ and so on. But couldn't come up with a suitable one!
Any hint will be really appreciated.
abstract-algebra elementary-number-theory field-theory algebraic-number-theory transcendental-numbers
abstract-algebra elementary-number-theory field-theory algebraic-number-theory transcendental-numbers
edited Jan 22 at 20:13
user26857
39.3k124183
asked Jan 21 at 13:49
CoherentCoherent
1,191623
edited Jan 22 at 20:13
user26857
39.3k124183
asked Jan 21 at 13:49
CoherentCoherent
1,191623
edited Jan 22 at 20:13
user26857
39.3k124183
edited Jan 22 at 20:13
user26857
39.3k124183
edited Jan 22 at 20:13
user26857
39.3k124183
39.3k124183
asked Jan 21 at 13:49
CoherentCoherent
1,191623
asked Jan 21 at 13:49
CoherentCoherent
1,191623
asked Jan 21 at 13:49
CoherentCoherent
1,191623
1,191623
3
$begingroup$
$alpha=frac{pi}{pi^2-1}$. Your ring is a field iff $alpha$ is algebraic. Thus, if $sum_k{a_kX^k}$ is a polynomial with degree $d$ and rational coefficients with $alpha$ as a root, then $sum_k{a_kX^k(X^2-1)^{d-k}}$ is a polynomial with rational coefficients, degree $2d$ if $a_0 neq 0$, and $pi$ is a root.
$endgroup$
– Mindlack
Jan 21 at 14:04
1
$begingroup$
What Mindlack said. Alternatively, the equation shows that $pi$ is algebraic over the field $Bbb{Q}(alpha)$. If $alpha$ were algebraic over $Bbb{Q}$ then ...
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:19
$begingroup$
@Mindlack Can I simply take any polynomial i.e. $a_n X^n+dots+a_0$ where $a_0 ne 0$ since $alpha notin Bbb Q$ to have $alpha$ as a root and thus, $a_n ({frac{pi}{{pi}^2 -1}})^n+dots+a_0=0 implies a_n {pi}^n+dots+a_0 {({pi}^2 -1)}^n=0$ and arrive at a contradiction!
$endgroup$
– Coherent
Jan 21 at 14:25
$begingroup$
Yes, but pay a closer attention to the powers of $pi^2-1$.
$endgroup$
– Mindlack
Jan 21 at 14:29
$begingroup$
@Mindlack Now,it's alright?
$endgroup$
– Coherent
Jan 21 at 14:31
|
show 2 more comments
3
$begingroup$
$alpha=frac{pi}{pi^2-1}$. Your ring is a field iff $alpha$ is algebraic. Thus, if $sum_k{a_kX^k}$ is a polynomial with degree $d$ and rational coefficients with $alpha$ as a root, then $sum_k{a_kX^k(X^2-1)^{d-k}}$ is a polynomial with rational coefficients, degree $2d$ if $a_0 neq 0$, and $pi$ is a root.
$endgroup$
– Mindlack
Jan 21 at 14:04
1
$begingroup$
What Mindlack said. Alternatively, the equation shows that $pi$ is algebraic over the field $Bbb{Q}(alpha)$. If $alpha$ were algebraic over $Bbb{Q}$ then ...
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:19
$begingroup$
@Mindlack Can I simply take any polynomial i.e. $a_n X^n+dots+a_0$ where $a_0 ne 0$ since $alpha notin Bbb Q$ to have $alpha$ as a root and thus, $a_n ({frac{pi}{{pi}^2 -1}})^n+dots+a_0=0 implies a_n {pi}^n+dots+a_0 {({pi}^2 -1)}^n=0$ and arrive at a contradiction!
$endgroup$
– Coherent
Jan 21 at 14:25
$begingroup$
Yes, but pay a closer attention to the powers of $pi^2-1$.
$endgroup$
– Mindlack
Jan 21 at 14:29
$begingroup$
@Mindlack Now,it's alright?
$endgroup$
– Coherent
Jan 21 at 14:31
3
3
$begingroup$
$alpha=frac{pi}{pi^2-1}$. Your ring is a field iff $alpha$ is algebraic. Thus, if $sum_k{a_kX^k}$ is a polynomial with degree $d$ and rational coefficients with $alpha$ as a root, then $sum_k{a_kX^k(X^2-1)^{d-k}}$ is a polynomial with rational coefficients, degree $2d$ if $a_0 neq 0$, and $pi$ is a root.
$endgroup$
– Mindlack
Jan 21 at 14:04
$begingroup$
$alpha=frac{pi}{pi^2-1}$. Your ring is a field iff $alpha$ is algebraic. Thus, if $sum_k{a_kX^k}$ is a polynomial with degree $d$ and rational coefficients with $alpha$ as a root, then $sum_k{a_kX^k(X^2-1)^{d-k}}$ is a polynomial with rational coefficients, degree $2d$ if $a_0 neq 0$, and $pi$ is a root.
$endgroup$
– Mindlack
Jan 21 at 14:04
1
1
$begingroup$
What Mindlack said. Alternatively, the equation shows that $pi$ is algebraic over the field $Bbb{Q}(alpha)$. If $alpha$ were algebraic over $Bbb{Q}$ then ...
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:19
$begingroup$
What Mindlack said. Alternatively, the equation shows that $pi$ is algebraic over the field $Bbb{Q}(alpha)$. If $alpha$ were algebraic over $Bbb{Q}$ then ...
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:19
$begingroup$
@Mindlack Can I simply take any polynomial i.e. $a_n X^n+dots+a_0$ where $a_0 ne 0$ since $alpha notin Bbb Q$ to have $alpha$ as a root and thus, $a_n ({frac{pi}{{pi}^2 -1}})^n+dots+a_0=0 implies a_n {pi}^n+dots+a_0 {({pi}^2 -1)}^n=0$ and arrive at a contradiction!
$endgroup$
– Coherent
Jan 21 at 14:25
$begingroup$
@Mindlack Can I simply take any polynomial i.e. $a_n X^n+dots+a_0$ where $a_0 ne 0$ since $alpha notin Bbb Q$ to have $alpha$ as a root and thus, $a_n ({frac{pi}{{pi}^2 -1}})^n+dots+a_0=0 implies a_n {pi}^n+dots+a_0 {({pi}^2 -1)}^n=0$ and arrive at a contradiction!
$endgroup$
– Coherent
Jan 21 at 14:25
$begingroup$
Yes, but pay a closer attention to the powers of $pi^2-1$.
$endgroup$
– Mindlack
Jan 21 at 14:29
$begingroup$
Yes, but pay a closer attention to the powers of $pi^2-1$.
$endgroup$
– Mindlack
Jan 21 at 14:29
$begingroup$
@Mindlack Now,it's alright?
$endgroup$
– Coherent
Jan 21 at 14:31
$begingroup$
@Mindlack Now,it's alright?
$endgroup$
– Coherent
Jan 21 at 14:31
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Given that $alpha+pi=pi^2alpha$, by basic algebra $alpha=frac{pi}{pi^2-1}$. This is transcendental because $pi$ is. After all, if $alpha$ were algebraic then $f(alpha)=0$ for some $finBbb{Q}[X]$, but then $pi$ is a root of
$$(X^2-1)^nf(X)inBbb{Q}[X],$$
where $n:=deg f$, contradicting the fact that $pi$ is transcendental.
answered Jan 21 at 15:14
ServaesServaes
27.4k34098
$endgroup$
$begingroup$
...and since $alpha$ is transcendental, $mathbb{Q}[alpha]$ is not a field.
$endgroup$
– Levent
Jan 22 at 20:23
add a comment |
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$begingroup$
Given that $alpha+pi=pi^2alpha$, by basic algebra $alpha=frac{pi}{pi^2-1}$. This is transcendental because $pi$ is. After all, if $alpha$ were algebraic then $f(alpha)=0$ for some $finBbb{Q}[X]$, but then $pi$ is a root of
$$(X^2-1)^nf(X)inBbb{Q}[X],$$
where $n:=deg f$, contradicting the fact that $pi$ is transcendental.
answered Jan 21 at 15:14
ServaesServaes
27.4k34098
$endgroup$
$begingroup$
...and since $alpha$ is transcendental, $mathbb{Q}[alpha]$ is not a field.
$endgroup$
– Levent
Jan 22 at 20:23
add a comment |
$begingroup$
Given that $alpha+pi=pi^2alpha$, by basic algebra $alpha=frac{pi}{pi^2-1}$. This is transcendental because $pi$ is. After all, if $alpha$ were algebraic then $f(alpha)=0$ for some $finBbb{Q}[X]$, but then $pi$ is a root of
$$(X^2-1)^nf(X)inBbb{Q}[X],$$
where $n:=deg f$, contradicting the fact that $pi$ is transcendental.
answered Jan 21 at 15:14
ServaesServaes
27.4k34098
$endgroup$
$begingroup$
...and since $alpha$ is transcendental, $mathbb{Q}[alpha]$ is not a field.
$endgroup$
– Levent
Jan 22 at 20:23
add a comment |
$begingroup$
Given that $alpha+pi=pi^2alpha$, by basic algebra $alpha=frac{pi}{pi^2-1}$. This is transcendental because $pi$ is. After all, if $alpha$ were algebraic then $f(alpha)=0$ for some $finBbb{Q}[X]$, but then $pi$ is a root of
$$(X^2-1)^nf(X)inBbb{Q}[X],$$
where $n:=deg f$, contradicting the fact that $pi$ is transcendental.
answered Jan 21 at 15:14
ServaesServaes
27.4k34098
$endgroup$
Given that $alpha+pi=pi^2alpha$, by basic algebra $alpha=frac{pi}{pi^2-1}$. This is transcendental because $pi$ is. After all, if $alpha$ were algebraic then $f(alpha)=0$ for some $finBbb{Q}[X]$, but then $pi$ is a root of
$$(X^2-1)^nf(X)inBbb{Q}[X],$$
where $n:=deg f$, contradicting the fact that $pi$ is transcendental.
answered Jan 21 at 15:14
ServaesServaes
27.4k34098
answered Jan 21 at 15:14
ServaesServaes
27.4k34098
answered Jan 21 at 15:14
ServaesServaes
27.4k34098
answered Jan 21 at 15:14
ServaesServaes
27.4k34098
27.4k34098
$begingroup$
...and since $alpha$ is transcendental, $mathbb{Q}[alpha]$ is not a field.
$endgroup$
– Levent
Jan 22 at 20:23
add a comment |
$begingroup$
...and since $alpha$ is transcendental, $mathbb{Q}[alpha]$ is not a field.
$endgroup$
– Levent
Jan 22 at 20:23
$begingroup$
...and since $alpha$ is transcendental, $mathbb{Q}[alpha]$ is not a field.
$endgroup$
– Levent
Jan 22 at 20:23
$begingroup$
...and since $alpha$ is transcendental, $mathbb{Q}[alpha]$ is not a field.
$endgroup$
– Levent
Jan 22 at 20:23
add a comment |
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3
$begingroup$
$alpha=frac{pi}{pi^2-1}$. Your ring is a field iff $alpha$ is algebraic. Thus, if $sum_k{a_kX^k}$ is a polynomial with degree $d$ and rational coefficients with $alpha$ as a root, then $sum_k{a_kX^k(X^2-1)^{d-k}}$ is a polynomial with rational coefficients, degree $2d$ if $a_0 neq 0$, and $pi$ is a root.
$endgroup$
– Mindlack
Jan 21 at 14:04
1
$begingroup$
What Mindlack said. Alternatively, the equation shows that $pi$ is algebraic over the field $Bbb{Q}(alpha)$. If $alpha$ were algebraic over $Bbb{Q}$ then ...
$endgroup$
– Jyrki Lahtonen
Jan 21 at 14:19
$begingroup$
@Mindlack Can I simply take any polynomial i.e. $a_n X^n+dots+a_0$ where $a_0 ne 0$ since $alpha notin Bbb Q$ to have $alpha$ as a root and thus, $a_n ({frac{pi}{{pi}^2 -1}})^n+dots+a_0=0 implies a_n {pi}^n+dots+a_0 {({pi}^2 -1)}^n=0$ and arrive at a contradiction!
$endgroup$
– Coherent
Jan 21 at 14:25
$begingroup$
Yes, but pay a closer attention to the powers of $pi^2-1$.
$endgroup$
– Mindlack
Jan 21 at 14:29
$begingroup$
@Mindlack Now,it's alright?
$endgroup$
– Coherent
Jan 21 at 14:31