Mixing
Constructing a field with exactly 81 elements
Clash Royale CLAN TAG#URR8PPP
up vote
4
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I was thinking $fracmathbbZ_3[x](x^3+x+1) times fracmathbbZ_3[x](x^3+x+1)$.
$(x^3+x+1)$ is irreducible in $mathbbZ_3[x]$ so the quotient will be a field, and a field cross a field looks like it should be a field to me!
How would I do this if working over $mathbbZ_9$ and $mathbbZ_2$? in $mathbbZ_9$ i would have to find an irreducible quadratic, and then quotient out by that, correct? Thanks in advance!!
abstract-algebra number-theory field-theory ideals
asked 1 hour ago
Michael Vaughan
50611
 |Â
show 2 more comments
up vote
4
down vote
favorite
I was thinking $fracmathbbZ_3[x](x^3+x+1) times fracmathbbZ_3[x](x^3+x+1)$.
$(x^3+x+1)$ is irreducible in $mathbbZ_3[x]$ so the quotient will be a field, and a field cross a field looks like it should be a field to me!
How would I do this if working over $mathbbZ_9$ and $mathbbZ_2$? in $mathbbZ_9$ i would have to find an irreducible quadratic, and then quotient out by that, correct? Thanks in advance!!
abstract-algebra number-theory field-theory ideals
asked 1 hour ago
Michael Vaughan
50611
5
A "field cross a field" is not a field. It can have zero divisors.
– David Peterson
1 hour ago
1
That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbbZ_9$ either, because that has zero divisors.
– Arturo Magidin
1 hour ago
1
Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
– Arturo Magidin
1 hour ago
Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
– Michael Vaughan
1 hour ago
$mathbb Z_9$ is not a field.
– Thomas Andrews
1 hour ago
 |Â
show 2 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I was thinking $fracmathbbZ_3[x](x^3+x+1) times fracmathbbZ_3[x](x^3+x+1)$.
$(x^3+x+1)$ is irreducible in $mathbbZ_3[x]$ so the quotient will be a field, and a field cross a field looks like it should be a field to me!
How would I do this if working over $mathbbZ_9$ and $mathbbZ_2$? in $mathbbZ_9$ i would have to find an irreducible quadratic, and then quotient out by that, correct? Thanks in advance!!
abstract-algebra number-theory field-theory ideals
asked 1 hour ago
Michael Vaughan
50611
I was thinking $fracmathbbZ_3[x](x^3+x+1) times fracmathbbZ_3[x](x^3+x+1)$.
$(x^3+x+1)$ is irreducible in $mathbbZ_3[x]$ so the quotient will be a field, and a field cross a field looks like it should be a field to me!
How would I do this if working over $mathbbZ_9$ and $mathbbZ_2$? in $mathbbZ_9$ i would have to find an irreducible quadratic, and then quotient out by that, correct? Thanks in advance!!
abstract-algebra number-theory field-theory ideals
abstract-algebra number-theory field-theory ideals
asked 1 hour ago
Michael Vaughan
50611
asked 1 hour ago
Michael Vaughan
50611
asked 1 hour ago
Michael Vaughan
50611
asked 1 hour ago
Michael Vaughan
50611
asked 1 hour ago
Michael Vaughan
50611
50611
5
A "field cross a field" is not a field. It can have zero divisors.
– David Peterson
1 hour ago
1
That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbbZ_9$ either, because that has zero divisors.
– Arturo Magidin
1 hour ago
1
Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
– Arturo Magidin
1 hour ago
Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
– Michael Vaughan
1 hour ago
$mathbb Z_9$ is not a field.
– Thomas Andrews
1 hour ago
 |Â
show 2 more comments
5
A "field cross a field" is not a field. It can have zero divisors.
– David Peterson
1 hour ago
1
That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbbZ_9$ either, because that has zero divisors.
– Arturo Magidin
1 hour ago
1
Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
– Arturo Magidin
1 hour ago
Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
– Michael Vaughan
1 hour ago
$mathbb Z_9$ is not a field.
– Thomas Andrews
1 hour ago
5
5
A "field cross a field" is not a field. It can have zero divisors.
– David Peterson
1 hour ago
A "field cross a field" is not a field. It can have zero divisors.
– David Peterson
1 hour ago
1
1
That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbbZ_9$ either, because that has zero divisors.
– Arturo Magidin
1 hour ago
That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbbZ_9$ either, because that has zero divisors.
– Arturo Magidin
1 hour ago
1
1
Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
– Arturo Magidin
1 hour ago
Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
– Arturo Magidin
1 hour ago
Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
– Michael Vaughan
1 hour ago
Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
– Michael Vaughan
1 hour ago
$mathbb Z_9$ is not a field.
– Thomas Andrews
1 hour ago
$mathbb Z_9$ is not a field.
– Thomas Andrews
1 hour ago
 |Â
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
2
down vote
No you can't do $F times F$ because $(a,0) times (0,b) = (0,0)$ would break the field axioms.
You need to give an irreducible polynomial $P$ of degree $4$ over $mathbbZ_3$ so you can give representatives of $mathbbZ_3[x]/(P)$ by polynomials of degree $leq 3$. The coefficients of $1$, $x$ $x^2$ and $x^3$ are free to choose so there are a total of $3^4$ elements of the field as desired.
Getting $P$ is more complicated. If you have a guess for an irreducible polynomial check it with Rabin's test
answered 1 hour ago
AHusain
2,237714
Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbbZ_3$ because checking for roots won't be enough... got any tips?
– Michael Vaughan
1 hour ago
6
@MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbbZ_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
– Arturo Magidin
1 hour ago
3
The field extensions of degree 2, 3, 4, .. of $mathbbZ_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $fracx^5-1x-1=x^4+x^3+x^2+x+1$ must work.
– random
1 hour ago
add a comment |Â
up vote
1
down vote
A field with $81$ elements must be the splitting field of $x^81-x$. It is also an extension of $mathbb F_3$ of degree $4$. Therefore, take any irreducible quartic factor of $x^81-x$, for instance, $x^4+x+2$. WA tells you all of them.
answered 1 hour ago
lhf
158k9161375
While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbbZ_9$ or $mathbbZ_2$ to build the field.
– Morgan Rodgers
42 secs ago
add a comment |Â
up vote
0
down vote
Let's see. What are the irreducible quadratics: $x^2+x+2,,x^2+2x+2,,x^2+1,,2x^2+x+1$.
I count $4$ (check me on this).
If so, there are $4choose 2+4=10$ combinations to check.
But there are $2cdot 3^4=162$ quartics to choose from...
answered 54 mins ago
Chris Custer
7,1522622
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
No you can't do $F times F$ because $(a,0) times (0,b) = (0,0)$ would break the field axioms.
You need to give an irreducible polynomial $P$ of degree $4$ over $mathbbZ_3$ so you can give representatives of $mathbbZ_3[x]/(P)$ by polynomials of degree $leq 3$. The coefficients of $1$, $x$ $x^2$ and $x^3$ are free to choose so there are a total of $3^4$ elements of the field as desired.
Getting $P$ is more complicated. If you have a guess for an irreducible polynomial check it with Rabin's test
answered 1 hour ago
AHusain
2,237714
Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbbZ_3$ because checking for roots won't be enough... got any tips?
– Michael Vaughan
1 hour ago
6
@MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbbZ_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
– Arturo Magidin
1 hour ago
3
The field extensions of degree 2, 3, 4, .. of $mathbbZ_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $fracx^5-1x-1=x^4+x^3+x^2+x+1$ must work.
– random
1 hour ago
add a comment |Â
up vote
2
down vote
No you can't do $F times F$ because $(a,0) times (0,b) = (0,0)$ would break the field axioms.
You need to give an irreducible polynomial $P$ of degree $4$ over $mathbbZ_3$ so you can give representatives of $mathbbZ_3[x]/(P)$ by polynomials of degree $leq 3$. The coefficients of $1$, $x$ $x^2$ and $x^3$ are free to choose so there are a total of $3^4$ elements of the field as desired.
Getting $P$ is more complicated. If you have a guess for an irreducible polynomial check it with Rabin's test
answered 1 hour ago
AHusain
2,237714
Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbbZ_3$ because checking for roots won't be enough... got any tips?
– Michael Vaughan
1 hour ago
6
@MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbbZ_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
– Arturo Magidin
1 hour ago
3
The field extensions of degree 2, 3, 4, .. of $mathbbZ_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $fracx^5-1x-1=x^4+x^3+x^2+x+1$ must work.
– random
1 hour ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
No you can't do $F times F$ because $(a,0) times (0,b) = (0,0)$ would break the field axioms.
You need to give an irreducible polynomial $P$ of degree $4$ over $mathbbZ_3$ so you can give representatives of $mathbbZ_3[x]/(P)$ by polynomials of degree $leq 3$. The coefficients of $1$, $x$ $x^2$ and $x^3$ are free to choose so there are a total of $3^4$ elements of the field as desired.
Getting $P$ is more complicated. If you have a guess for an irreducible polynomial check it with Rabin's test
answered 1 hour ago
AHusain
2,237714
No you can't do $F times F$ because $(a,0) times (0,b) = (0,0)$ would break the field axioms.
You need to give an irreducible polynomial $P$ of degree $4$ over $mathbbZ_3$ so you can give representatives of $mathbbZ_3[x]/(P)$ by polynomials of degree $leq 3$. The coefficients of $1$, $x$ $x^2$ and $x^3$ are free to choose so there are a total of $3^4$ elements of the field as desired.
Getting $P$ is more complicated. If you have a guess for an irreducible polynomial check it with Rabin's test
answered 1 hour ago
AHusain
2,237714
edited 1 hour ago
answered 1 hour ago
AHusain
2,237714
answered 1 hour ago
AHusain
2,237714
answered 1 hour ago
AHusain
2,237714
2,237714
Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbbZ_3$ because checking for roots won't be enough... got any tips?
– Michael Vaughan
1 hour ago
6
@MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbbZ_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
– Arturo Magidin
1 hour ago
3
The field extensions of degree 2, 3, 4, .. of $mathbbZ_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $fracx^5-1x-1=x^4+x^3+x^2+x+1$ must work.
– random
1 hour ago
add a comment |Â
Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbbZ_3$ because checking for roots won't be enough... got any tips?
– Michael Vaughan
1 hour ago
6
@MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbbZ_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
– Arturo Magidin
1 hour ago
3
The field extensions of degree 2, 3, 4, .. of $mathbbZ_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $fracx^5-1x-1=x^4+x^3+x^2+x+1$ must work.
– random
1 hour ago
Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbbZ_3$ because checking for roots won't be enough... got any tips?
– Michael Vaughan
1 hour ago
Seems like it would be tough to find a degree 4 irreducible polynomial over $mathbbZ_3$ because checking for roots won't be enough... got any tips?
– Michael Vaughan
1 hour ago
6
6
@MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbbZ_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
– Arturo Magidin
1 hour ago
@MichaelVaughan: Checking for roots isn’t enough; but then, if it is of degree $4$ and has no roots in $mathbbZ_3$, then the only way it can fail to be irreducible is if it is the product of two irreducible quadratics. There aren’t that many irreducible quadratics, so you can check what all the products of two of them are.
– Arturo Magidin
1 hour ago
3
3
The field extensions of degree 2, 3, 4, .. of $mathbbZ_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $fracx^5-1x-1=x^4+x^3+x^2+x+1$ must work.
– random
1 hour ago
The field extensions of degree 2, 3, 4, .. of $mathbbZ_3$ have 8, 26, 80, ... non-zero elements, so the degree 4 extension is the smallest one with elements of order 5 and the polynomial $fracx^5-1x-1=x^4+x^3+x^2+x+1$ must work.
– random
1 hour ago
add a comment |Â
up vote
1
down vote
A field with $81$ elements must be the splitting field of $x^81-x$. It is also an extension of $mathbb F_3$ of degree $4$. Therefore, take any irreducible quartic factor of $x^81-x$, for instance, $x^4+x+2$. WA tells you all of them.
answered 1 hour ago
lhf
158k9161375
While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbbZ_9$ or $mathbbZ_2$ to build the field.
– Morgan Rodgers
42 secs ago
add a comment |Â
up vote
1
down vote
A field with $81$ elements must be the splitting field of $x^81-x$. It is also an extension of $mathbb F_3$ of degree $4$. Therefore, take any irreducible quartic factor of $x^81-x$, for instance, $x^4+x+2$. WA tells you all of them.
answered 1 hour ago
lhf
158k9161375
While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbbZ_9$ or $mathbbZ_2$ to build the field.
– Morgan Rodgers
42 secs ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A field with $81$ elements must be the splitting field of $x^81-x$. It is also an extension of $mathbb F_3$ of degree $4$. Therefore, take any irreducible quartic factor of $x^81-x$, for instance, $x^4+x+2$. WA tells you all of them.
answered 1 hour ago
lhf
158k9161375
A field with $81$ elements must be the splitting field of $x^81-x$. It is also an extension of $mathbb F_3$ of degree $4$. Therefore, take any irreducible quartic factor of $x^81-x$, for instance, $x^4+x+2$. WA tells you all of them.
answered 1 hour ago
lhf
158k9161375
answered 1 hour ago
lhf
158k9161375
answered 1 hour ago
lhf
158k9161375
answered 1 hour ago
lhf
158k9161375
158k9161375
While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbbZ_9$ or $mathbbZ_2$ to build the field.
– Morgan Rodgers
42 secs ago
add a comment |Â
While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbbZ_9$ or $mathbbZ_2$ to build the field.
– Morgan Rodgers
42 secs ago
While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbbZ_9$ or $mathbbZ_2$ to build the field.
– Morgan Rodgers
42 secs ago
While this answer is technically correct, it feels incomplete without addressing the confusion about using $mathbbZ_9$ or $mathbbZ_2$ to build the field.
– Morgan Rodgers
42 secs ago
add a comment |Â
up vote
0
down vote
Let's see. What are the irreducible quadratics: $x^2+x+2,,x^2+2x+2,,x^2+1,,2x^2+x+1$.
I count $4$ (check me on this).
If so, there are $4choose 2+4=10$ combinations to check.
But there are $2cdot 3^4=162$ quartics to choose from...
answered 54 mins ago
Chris Custer
7,1522622
add a comment |Â
up vote
0
down vote
Let's see. What are the irreducible quadratics: $x^2+x+2,,x^2+2x+2,,x^2+1,,2x^2+x+1$.
I count $4$ (check me on this).
If so, there are $4choose 2+4=10$ combinations to check.
But there are $2cdot 3^4=162$ quartics to choose from...
answered 54 mins ago
Chris Custer
7,1522622
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let's see. What are the irreducible quadratics: $x^2+x+2,,x^2+2x+2,,x^2+1,,2x^2+x+1$.
I count $4$ (check me on this).
If so, there are $4choose 2+4=10$ combinations to check.
But there are $2cdot 3^4=162$ quartics to choose from...
answered 54 mins ago
Chris Custer
7,1522622
Let's see. What are the irreducible quadratics: $x^2+x+2,,x^2+2x+2,,x^2+1,,2x^2+x+1$.
I count $4$ (check me on this).
If so, there are $4choose 2+4=10$ combinations to check.
But there are $2cdot 3^4=162$ quartics to choose from...
answered 54 mins ago
Chris Custer
7,1522622
edited 25 mins ago
answered 54 mins ago
Chris Custer
7,1522622
answered 54 mins ago
Chris Custer
7,1522622
answered 54 mins ago
Chris Custer
7,1522622
7,1522622
add a comment |Â
add a comment |Â
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5
A "field cross a field" is not a field. It can have zero divisors.
– David Peterson
1 hour ago
1
That can’t be a field: a direct product of two rings always has zero divisors: $(a,0)*(0,a) = (0,0)$, even when $aneq 0$. And you can’t do it working over $mathbbZ_9$ either, because that has zero divisors.
– Arturo Magidin
1 hour ago
1
Since $81 = 3^4$, try to find an irreducible polynomial of degree 4. Also, notice that $Rtimes S$ has $|R|*|S|$ elements; your cross product has $27times 27 = 729$ elements, not 81.
– Arturo Magidin
1 hour ago
Is there a process for finding an irreducible polynomial over a given modulus or do you just try things ?? Seems like it would be really tough to find an a degree 4 irreducible polynomial since checking for roots won't be enough
– Michael Vaughan
1 hour ago
$mathbb Z_9$ is not a field.
– Thomas Andrews
1 hour ago