a question for the irreducibility test for degree 2 or 3 polynomialThe ring $ℤ/nℤ$ is a field if and only if $n$ is prime'shifted' polynomial preserves validity of Eisenstein irreducibility criterion of original, in finite fields?About irreducibility of a particular class of bihomogeneous polynomialsProving irreducibility of polynomialIrreducibility of a polynomial with degree $4$question on testing irreducibility of a polynomialDetermine the irreducibility of polynomial.Is there a way to find one irreducible polynomial of degree n in the field Z2Mod $p$ irreducibility test proof questionFor every $n$, there exists a polynomial of degree $n$ with Galois Group $S_n$Why is $p_3(x) = 9x^2-3=3(xsqrt3-1)(xsqrt3+1)$ reducible over integers?
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a question for the irreducibility test for degree 2 or 3 polynomial
The ring $ℤ/nℤ$ is a field if and only if $n$ is prime'shifted' polynomial preserves validity of Eisenstein irreducibility criterion of original, in finite fields?About irreducibility of a particular class of bihomogeneous polynomialsProving irreducibility of polynomialIrreducibility of a polynomial with degree $4$question on testing irreducibility of a polynomialDetermine the irreducibility of polynomial.Is there a way to find one irreducible polynomial of degree n in the field Z2Mod $p$ irreducibility test proof questionFor every $n$, there exists a polynomial of degree $n$ with Galois Group $S_n$Why is $p_3(x) = 9x^2-3=3(xsqrt3-1)(xsqrt3+1)$ reducible over integers?
$begingroup$
I have a question for the irreducibility test for degree 2 or 3 polynomial. The test states: "Let F be a field, f(x) ∈ F[x], and degf(x) = 2 or 3. Then f(x) is reducible over F if and only if f(x) has a zero in F."
In the textbook, it states the test can be applied to polynomials in Z2[x] or Z3[x]. However, I don't think Z2 and Z3 are fields, just like Z is not field. Can anyone help? Thanks!
abstract-algebra polynomials irreducible-polynomials
$endgroup$
add a comment |
$begingroup$
I have a question for the irreducibility test for degree 2 or 3 polynomial. The test states: "Let F be a field, f(x) ∈ F[x], and degf(x) = 2 or 3. Then f(x) is reducible over F if and only if f(x) has a zero in F."
In the textbook, it states the test can be applied to polynomials in Z2[x] or Z3[x]. However, I don't think Z2 and Z3 are fields, just like Z is not field. Can anyone help? Thanks!
abstract-algebra polynomials irreducible-polynomials
$endgroup$
3
$begingroup$
Wrong! Z2 and Z3 are fields, if this means $mathbb Z/2mathbb Z$, respectively $mathbb Z/3mathbb Z$.
$endgroup$
– user26857
Mar 3 at 16:34
2
$begingroup$
For a prime $p,$ the ring $mathbb Z / p mathbb Z$ is a field. The reason there are multiplicative inverses: if $n$ is an integer not divisible by $p,$ there are integers $x,y$ such that $nx-py = 1,$ or $nx= 1+py,$ finally $nx equiv 1 pmod p$
$endgroup$
– Will Jagy
Mar 3 at 16:37
$begingroup$
I see! Thanks guys!
$endgroup$
– RandomThinker
Mar 3 at 16:49
$begingroup$
See this duplicate about $Bbb Z/n$.
$endgroup$
– Dietrich Burde
Mar 3 at 17:49
add a comment |
$begingroup$
I have a question for the irreducibility test for degree 2 or 3 polynomial. The test states: "Let F be a field, f(x) ∈ F[x], and degf(x) = 2 or 3. Then f(x) is reducible over F if and only if f(x) has a zero in F."
In the textbook, it states the test can be applied to polynomials in Z2[x] or Z3[x]. However, I don't think Z2 and Z3 are fields, just like Z is not field. Can anyone help? Thanks!
abstract-algebra polynomials irreducible-polynomials
$endgroup$
I have a question for the irreducibility test for degree 2 or 3 polynomial. The test states: "Let F be a field, f(x) ∈ F[x], and degf(x) = 2 or 3. Then f(x) is reducible over F if and only if f(x) has a zero in F."
In the textbook, it states the test can be applied to polynomials in Z2[x] or Z3[x]. However, I don't think Z2 and Z3 are fields, just like Z is not field. Can anyone help? Thanks!
abstract-algebra polynomials irreducible-polynomials
abstract-algebra polynomials irreducible-polynomials
edited 2 days ago
Sil
5,16421643
5,16421643
asked Mar 3 at 16:33
RandomThinkerRandomThinker
474
474
3
$begingroup$
Wrong! Z2 and Z3 are fields, if this means $mathbb Z/2mathbb Z$, respectively $mathbb Z/3mathbb Z$.
$endgroup$
– user26857
Mar 3 at 16:34
2
$begingroup$
For a prime $p,$ the ring $mathbb Z / p mathbb Z$ is a field. The reason there are multiplicative inverses: if $n$ is an integer not divisible by $p,$ there are integers $x,y$ such that $nx-py = 1,$ or $nx= 1+py,$ finally $nx equiv 1 pmod p$
$endgroup$
– Will Jagy
Mar 3 at 16:37
$begingroup$
I see! Thanks guys!
$endgroup$
– RandomThinker
Mar 3 at 16:49
$begingroup$
See this duplicate about $Bbb Z/n$.
$endgroup$
– Dietrich Burde
Mar 3 at 17:49
add a comment |
3
$begingroup$
Wrong! Z2 and Z3 are fields, if this means $mathbb Z/2mathbb Z$, respectively $mathbb Z/3mathbb Z$.
$endgroup$
– user26857
Mar 3 at 16:34
2
$begingroup$
For a prime $p,$ the ring $mathbb Z / p mathbb Z$ is a field. The reason there are multiplicative inverses: if $n$ is an integer not divisible by $p,$ there are integers $x,y$ such that $nx-py = 1,$ or $nx= 1+py,$ finally $nx equiv 1 pmod p$
$endgroup$
– Will Jagy
Mar 3 at 16:37
$begingroup$
I see! Thanks guys!
$endgroup$
– RandomThinker
Mar 3 at 16:49
$begingroup$
See this duplicate about $Bbb Z/n$.
$endgroup$
– Dietrich Burde
Mar 3 at 17:49
3
3
$begingroup$
Wrong! Z2 and Z3 are fields, if this means $mathbb Z/2mathbb Z$, respectively $mathbb Z/3mathbb Z$.
$endgroup$
– user26857
Mar 3 at 16:34
$begingroup$
Wrong! Z2 and Z3 are fields, if this means $mathbb Z/2mathbb Z$, respectively $mathbb Z/3mathbb Z$.
$endgroup$
– user26857
Mar 3 at 16:34
2
2
$begingroup$
For a prime $p,$ the ring $mathbb Z / p mathbb Z$ is a field. The reason there are multiplicative inverses: if $n$ is an integer not divisible by $p,$ there are integers $x,y$ such that $nx-py = 1,$ or $nx= 1+py,$ finally $nx equiv 1 pmod p$
$endgroup$
– Will Jagy
Mar 3 at 16:37
$begingroup$
For a prime $p,$ the ring $mathbb Z / p mathbb Z$ is a field. The reason there are multiplicative inverses: if $n$ is an integer not divisible by $p,$ there are integers $x,y$ such that $nx-py = 1,$ or $nx= 1+py,$ finally $nx equiv 1 pmod p$
$endgroup$
– Will Jagy
Mar 3 at 16:37
$begingroup$
I see! Thanks guys!
$endgroup$
– RandomThinker
Mar 3 at 16:49
$begingroup$
I see! Thanks guys!
$endgroup$
– RandomThinker
Mar 3 at 16:49
$begingroup$
See this duplicate about $Bbb Z/n$.
$endgroup$
– Dietrich Burde
Mar 3 at 17:49
$begingroup$
See this duplicate about $Bbb Z/n$.
$endgroup$
– Dietrich Burde
Mar 3 at 17:49
add a comment |
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3
$begingroup$
Wrong! Z2 and Z3 are fields, if this means $mathbb Z/2mathbb Z$, respectively $mathbb Z/3mathbb Z$.
$endgroup$
– user26857
Mar 3 at 16:34
2
$begingroup$
For a prime $p,$ the ring $mathbb Z / p mathbb Z$ is a field. The reason there are multiplicative inverses: if $n$ is an integer not divisible by $p,$ there are integers $x,y$ such that $nx-py = 1,$ or $nx= 1+py,$ finally $nx equiv 1 pmod p$
$endgroup$
– Will Jagy
Mar 3 at 16:37
$begingroup$
I see! Thanks guys!
$endgroup$
– RandomThinker
Mar 3 at 16:49
$begingroup$
See this duplicate about $Bbb Z/n$.
$endgroup$
– Dietrich Burde
Mar 3 at 17:49