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Minimal polynomial of $alpha=a+bsqrt d$ where $d$ is square free

How to describe when a simple extension $F(alpha)/F$ is Galois in terms of the minimal polynomial of $alpha$?finding the minimal polynomialMinimal Polynomial of $zeta+zeta^-1$Minimal polynomial of $sqrt2 + sqrt3$ over $BbbQ(sqrt6)$How do we find the minimal polynomial of $alpha = a + b sqrtd$ over $mathbbQ$?Minimal polynomial of $alpha=sqrt 2e^frac2pi i3$Minimal polynomial for $alpha=sqrt3-2sqrt2$ over $mathbbQ$Minimal polynomial of $alpha$ over $mathbbQ$ and $mathbbQ(sqrt2)$Minimal Polynomial of $mathbbQ(sqrt a +sqrt b)$Minimal polynomial of $sqrtsqrt[3]7-5$

3

1

$begingroup$

Here are two first steps that could be used to compute the integral closure of $mathbb Z[sqrt d]$:

(1) Reduce to the case when $d$ is square free.

(2) If $alpha=a+bsqrt d$ with $a,binmathbb Q$, then the minimal polynomial of $alpha$ is $x^2-2ax+(a^2-b^2d)$. Thus $alpha$ lies in the integral closure iff $2a,a^2-b^2dinmathbb Z$.

---

(1) How to do this? Assume $d$ is not square-free. Let's say $d=p^2q$. Then $mathbb Z[sqrt d]=mathbb Z[psqrt q]$. How to reduce to the case when $d$ is square free? I don't think the integral closure of $mathbb Z[sqrt d]$ equals the integral closure of $mathbb Z[sqrt q]$.

(2) Isn't the notion of a minimal polynomial defined over a ring? This is a minimal polynomial over which ring? If over $mathbb Z$, then $2a,a^2-b^2d$ are always integers, so I don't understand this step.

abstract-algebra number-theory elementary-number-theory

share|cite|improve this question

edited Mar 30 at 19:05

user419669

asked Mar 30 at 1:39

user419669user419669

271210

$endgroup$

add a comment | 

3

1

$begingroup$

Here are two first steps that could be used to compute the integral closure of $mathbb Z[sqrt d]$:

(1) Reduce to the case when $d$ is square free.

(2) If $alpha=a+bsqrt d$ with $a,binmathbb Q$, then the minimal polynomial of $alpha$ is $x^2-2ax+(a^2-b^2d)$. Thus $alpha$ lies in the integral closure iff $2a,a^2-b^2dinmathbb Z$.

---

(1) How to do this? Assume $d$ is not square-free. Let's say $d=p^2q$. Then $mathbb Z[sqrt d]=mathbb Z[psqrt q]$. How to reduce to the case when $d$ is square free? I don't think the integral closure of $mathbb Z[sqrt d]$ equals the integral closure of $mathbb Z[sqrt q]$.

(2) Isn't the notion of a minimal polynomial defined over a ring? This is a minimal polynomial over which ring? If over $mathbb Z$, then $2a,a^2-b^2d$ are always integers, so I don't understand this step.

abstract-algebra number-theory elementary-number-theory

share|cite|improve this question

edited Mar 30 at 19:05

user419669

asked Mar 30 at 1:39

user419669user419669

271210

$endgroup$

add a comment | 

$begingroup$

Here are two first steps that could be used to compute the integral closure of $mathbb Z[sqrt d]$:

(1) Reduce to the case when $d$ is square free.

(2) If $alpha=a+bsqrt d$ with $a,binmathbb Q$, then the minimal polynomial of $alpha$ is $x^2-2ax+(a^2-b^2d)$. Thus $alpha$ lies in the integral closure iff $2a,a^2-b^2dinmathbb Z$.

---

(1) How to do this? Assume $d$ is not square-free. Let's say $d=p^2q$. Then $mathbb Z[sqrt d]=mathbb Z[psqrt q]$. How to reduce to the case when $d$ is square free? I don't think the integral closure of $mathbb Z[sqrt d]$ equals the integral closure of $mathbb Z[sqrt q]$.

(2) Isn't the notion of a minimal polynomial defined over a ring? This is a minimal polynomial over which ring? If over $mathbb Z$, then $2a,a^2-b^2d$ are always integers, so I don't understand this step.

abstract-algebra number-theory elementary-number-theory

share|cite|improve this question

edited Mar 30 at 19:05

user419669

asked Mar 30 at 1:39

user419669user419669

271210

$endgroup$

share|cite|improve this question

edited Mar 30 at 19:05

user419669

asked Mar 30 at 1:39

user419669user419669

271210

share|cite|improve this question

edited Mar 30 at 19:05

user419669

asked Mar 30 at 1:39

user419669user419669

271210

asked Mar 30 at 1:39

user419669user419669

271210

asked Mar 30 at 1:39

user419669user419669

271210

1 Answer
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2

$begingroup$

Useful Fact : if $R_3$ is integral over $R_2$ and $R_2$ is integral over $R_1$ then $R_3$ is integral over $R_1$

1) Suppose $d$ is not square free then $d=s^2t$ where $t$ is square free. Observe that $mathbb Z[sqrt t]$ is integral over $mathbb Z[sqrt d] $ and hence the two have the same integral closure in $mathbb Q(sqrt d)$

2) $mathbb Z$ is integrally closed as it is a UFD. So if $alpha in$ integral closure of $mathbb Z[sqrt d]$ then all conjugates of $alpha$ over $mathbb Q$ are also integral and hence the coefficients of the minimal polynomial of $alpha$ are integral over $mathbb Z[sqrt d]$ and hence over $mathbb Z$. But the coefficients are in $mathbb Q$ so they are elements of $mathbb Q$ which are integral over $mathbb Z$ and hence $in mathbb Z$

share|cite|improve this answer

answered Mar 30 at 20:15

Soumik GhoshSoumik Ghosh

1,121112

$endgroup$

add a comment | 

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2

$begingroup$

Useful Fact : if $R_3$ is integral over $R_2$ and $R_2$ is integral over $R_1$ then $R_3$ is integral over $R_1$

1) Suppose $d$ is not square free then $d=s^2t$ where $t$ is square free. Observe that $mathbb Z[sqrt t]$ is integral over $mathbb Z[sqrt d] $ and hence the two have the same integral closure in $mathbb Q(sqrt d)$

2) $mathbb Z$ is integrally closed as it is a UFD. So if $alpha in$ integral closure of $mathbb Z[sqrt d]$ then all conjugates of $alpha$ over $mathbb Q$ are also integral and hence the coefficients of the minimal polynomial of $alpha$ are integral over $mathbb Z[sqrt d]$ and hence over $mathbb Z$. But the coefficients are in $mathbb Q$ so they are elements of $mathbb Q$ which are integral over $mathbb Z$ and hence $in mathbb Z$

share|cite|improve this answer

answered Mar 30 at 20:15

Soumik GhoshSoumik Ghosh

1,121112

$endgroup$

add a comment | 

2

$begingroup$

Useful Fact : if $R_3$ is integral over $R_2$ and $R_2$ is integral over $R_1$ then $R_3$ is integral over $R_1$

1) Suppose $d$ is not square free then $d=s^2t$ where $t$ is square free. Observe that $mathbb Z[sqrt t]$ is integral over $mathbb Z[sqrt d] $ and hence the two have the same integral closure in $mathbb Q(sqrt d)$

2) $mathbb Z$ is integrally closed as it is a UFD. So if $alpha in$ integral closure of $mathbb Z[sqrt d]$ then all conjugates of $alpha$ over $mathbb Q$ are also integral and hence the coefficients of the minimal polynomial of $alpha$ are integral over $mathbb Z[sqrt d]$ and hence over $mathbb Z$. But the coefficients are in $mathbb Q$ so they are elements of $mathbb Q$ which are integral over $mathbb Z$ and hence $in mathbb Z$

share|cite|improve this answer

answered Mar 30 at 20:15

Soumik GhoshSoumik Ghosh

1,121112

$endgroup$

add a comment | 

$begingroup$

Useful Fact : if $R_3$ is integral over $R_2$ and $R_2$ is integral over $R_1$ then $R_3$ is integral over $R_1$

1) Suppose $d$ is not square free then $d=s^2t$ where $t$ is square free. Observe that $mathbb Z[sqrt t]$ is integral over $mathbb Z[sqrt d] $ and hence the two have the same integral closure in $mathbb Q(sqrt d)$

2) $mathbb Z$ is integrally closed as it is a UFD. So if $alpha in$ integral closure of $mathbb Z[sqrt d]$ then all conjugates of $alpha$ over $mathbb Q$ are also integral and hence the coefficients of the minimal polynomial of $alpha$ are integral over $mathbb Z[sqrt d]$ and hence over $mathbb Z$. But the coefficients are in $mathbb Q$ so they are elements of $mathbb Q$ which are integral over $mathbb Z$ and hence $in mathbb Z$

share|cite|improve this answer

answered Mar 30 at 20:15

Soumik GhoshSoumik Ghosh

1,121112

$endgroup$

share|cite|improve this answer

answered Mar 30 at 20:15

Soumik GhoshSoumik Ghosh

1,121112

answered Mar 30 at 20:15

Soumik GhoshSoumik Ghosh

1,121112

answered Mar 30 at 20:15

Soumik GhoshSoumik Ghosh

1,121112