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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?













0












$begingroup$


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!










share|cite|improve this question











$endgroup$







  • 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















0












$begingroup$


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!










share|cite|improve this question











$endgroup$







  • 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













0












0








0





$begingroup$


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!










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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












  • 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










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