Mixing
Determine if an ideal is a prime ideal over $mathbb{Z}_5[x]$
$begingroup$
Conside the ring $<mathbb{Z}_5[x],+,.>$.
Is the ideal $ langle1 + 3x + 3x^2 + x^3rangle $ a prime ideal? If we have a ring of integers then we can simply check that if product of $a$ and $b$ belongs to the ideal $P$ then either $a$ or $b$ should belong to $P$.
How can I check this for polynomials?
abstract-algebra polynomials finite-fields
edited Jan 28 at 6:33
stressed out
6,5831939
asked Jan 28 at 5:59
MathVictimMathVictim
303
$endgroup$
add a comment |
$begingroup$
Conside the ring $<mathbb{Z}_5[x],+,.>$.
Is the ideal $ langle1 + 3x + 3x^2 + x^3rangle $ a prime ideal? If we have a ring of integers then we can simply check that if product of $a$ and $b$ belongs to the ideal $P$ then either $a$ or $b$ should belong to $P$.
How can I check this for polynomials?
abstract-algebra polynomials finite-fields
edited Jan 28 at 6:33
stressed out
6,5831939
asked Jan 28 at 5:59
MathVictimMathVictim
303
$endgroup$
$begingroup$
how about showing that the quotient is an Integral domain then the ideal will be a prime ideal. Can you somehow do that?
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 6:02
add a comment |
$begingroup$
Conside the ring $<mathbb{Z}_5[x],+,.>$.
Is the ideal $ langle1 + 3x + 3x^2 + x^3rangle $ a prime ideal? If we have a ring of integers then we can simply check that if product of $a$ and $b$ belongs to the ideal $P$ then either $a$ or $b$ should belong to $P$.
How can I check this for polynomials?
abstract-algebra polynomials finite-fields
edited Jan 28 at 6:33
stressed out
6,5831939
asked Jan 28 at 5:59
MathVictimMathVictim
303
$endgroup$
Conside the ring $<mathbb{Z}_5[x],+,.>$.
Is the ideal $ langle1 + 3x + 3x^2 + x^3rangle $ a prime ideal? If we have a ring of integers then we can simply check that if product of $a$ and $b$ belongs to the ideal $P$ then either $a$ or $b$ should belong to $P$.
How can I check this for polynomials?
abstract-algebra polynomials finite-fields
abstract-algebra polynomials finite-fields
edited Jan 28 at 6:33
stressed out
6,5831939
asked Jan 28 at 5:59
MathVictimMathVictim
303
edited Jan 28 at 6:33
stressed out
6,5831939
asked Jan 28 at 5:59
MathVictimMathVictim
303
edited Jan 28 at 6:33
stressed out
6,5831939
edited Jan 28 at 6:33
stressed out
6,5831939
edited Jan 28 at 6:33
stressed out
6,5831939
6,5831939
asked Jan 28 at 5:59
MathVictimMathVictim
303
asked Jan 28 at 5:59
MathVictimMathVictim
303
asked Jan 28 at 5:59
MathVictimMathVictim
303
303
$begingroup$
how about showing that the quotient is an Integral domain then the ideal will be a prime ideal. Can you somehow do that?
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 6:02
add a comment |
$begingroup$
how about showing that the quotient is an Integral domain then the ideal will be a prime ideal. Can you somehow do that?
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 6:02
$begingroup$
how about showing that the quotient is an Integral domain then the ideal will be a prime ideal. Can you somehow do that?
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 6:02
$begingroup$
how about showing that the quotient is an Integral domain then the ideal will be a prime ideal. Can you somehow do that?
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 6:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $(1+x)^3 = 1+3x+3x^2+x^3 in (1+3x+3x^2+x^3)$, but $1+x notin (1+3x+3x^2+x^3)$ because all non-zero multiples of $1+3x+3x^2+x^3$ must be of degree at least $3$. Thus, $(1+3x+3x^2+x^3)$ cannot be prime.
In general, if $F$ is a field then $F[x]$ is a principal ideal domain, and in particular every non-zero prime ideal is maximal. Therefore, one can merely check for reducibility of $p(x)$ in $F$ to see if $(p(x))$ is prime or not. If $deg p leq 3$, then in fact it is enough to check if $p$ has a root in $F$ or not i.e. if $exists x in F,p(x) =0$.
In our case, with $p(x) = x^3+3x^2+3x+1$,we may consider the element $4 in mathbb Z_5$, and see that $p(4) = 0$. So $p$ is reducible mod $5$, and hence we have that it cannot be prime.
If $F$ is not a field, then $F[x]$ need not be a principal ideal domain, as shown by the example $(2,x)$ being non-principal over $mathbb Z[x]$. Here, prime ideals are more difficult to characterize, because one cannot say what every ideal looks like in such a field(it may not be principal, but also may not have other nice properties), so it is more difficult to deduce conclusions from something being an ideal in this case.
answered Jan 28 at 6:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.9k33477
$endgroup$
add a comment |
$begingroup$
Hint:
Generally speaking, the strategy for determining the prime ideals of a polynomial ring is a bit different and more complicated. You need to know some facts about the polynomial rings first. So, I'm going to sketch the general idea for you and I'll let you check the details and convince yourself. If you had any question, don't hesitate to ask it in the comment section.
$mathbb{Z}_5$ is a field. So, $mathbb{Z}_5[X]$ is a principal ideal domain because of the Euclidean algorithm for division which works over any field. You can check whether $p(x)=1+3x+3x^2+3x^3$ is reducible or not. Since $deg(p(x))<4$, it suffices to check whether it has a root or not.
Note that prime elements in an integral domain are irreducible. Since $mathbb{Z}_5$ is a field, $mathbb{Z}_5[X]$ is an integral domain. So, if some element is reducible, then it is NOT prime and the principal ideal generated by it is not prime either. On the other hand, if $p(x)$ turns out to be irreducible (which is not true in our case), then it generates a maximal ideal and we know that maximal ideals are prime.
answered Jan 28 at 6:20
stressed outstressed out
6,5831939
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note that $(1+x)^3 = 1+3x+3x^2+x^3 in (1+3x+3x^2+x^3)$, but $1+x notin (1+3x+3x^2+x^3)$ because all non-zero multiples of $1+3x+3x^2+x^3$ must be of degree at least $3$. Thus, $(1+3x+3x^2+x^3)$ cannot be prime.
In general, if $F$ is a field then $F[x]$ is a principal ideal domain, and in particular every non-zero prime ideal is maximal. Therefore, one can merely check for reducibility of $p(x)$ in $F$ to see if $(p(x))$ is prime or not. If $deg p leq 3$, then in fact it is enough to check if $p$ has a root in $F$ or not i.e. if $exists x in F,p(x) =0$.
In our case, with $p(x) = x^3+3x^2+3x+1$,we may consider the element $4 in mathbb Z_5$, and see that $p(4) = 0$. So $p$ is reducible mod $5$, and hence we have that it cannot be prime.
If $F$ is not a field, then $F[x]$ need not be a principal ideal domain, as shown by the example $(2,x)$ being non-principal over $mathbb Z[x]$. Here, prime ideals are more difficult to characterize, because one cannot say what every ideal looks like in such a field(it may not be principal, but also may not have other nice properties), so it is more difficult to deduce conclusions from something being an ideal in this case.
answered Jan 28 at 6:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.9k33477
$endgroup$
add a comment |
$begingroup$
Note that $(1+x)^3 = 1+3x+3x^2+x^3 in (1+3x+3x^2+x^3)$, but $1+x notin (1+3x+3x^2+x^3)$ because all non-zero multiples of $1+3x+3x^2+x^3$ must be of degree at least $3$. Thus, $(1+3x+3x^2+x^3)$ cannot be prime.
In general, if $F$ is a field then $F[x]$ is a principal ideal domain, and in particular every non-zero prime ideal is maximal. Therefore, one can merely check for reducibility of $p(x)$ in $F$ to see if $(p(x))$ is prime or not. If $deg p leq 3$, then in fact it is enough to check if $p$ has a root in $F$ or not i.e. if $exists x in F,p(x) =0$.
In our case, with $p(x) = x^3+3x^2+3x+1$,we may consider the element $4 in mathbb Z_5$, and see that $p(4) = 0$. So $p$ is reducible mod $5$, and hence we have that it cannot be prime.
If $F$ is not a field, then $F[x]$ need not be a principal ideal domain, as shown by the example $(2,x)$ being non-principal over $mathbb Z[x]$. Here, prime ideals are more difficult to characterize, because one cannot say what every ideal looks like in such a field(it may not be principal, but also may not have other nice properties), so it is more difficult to deduce conclusions from something being an ideal in this case.
answered Jan 28 at 6:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.9k33477
$endgroup$
add a comment |
$begingroup$
Note that $(1+x)^3 = 1+3x+3x^2+x^3 in (1+3x+3x^2+x^3)$, but $1+x notin (1+3x+3x^2+x^3)$ because all non-zero multiples of $1+3x+3x^2+x^3$ must be of degree at least $3$. Thus, $(1+3x+3x^2+x^3)$ cannot be prime.
In general, if $F$ is a field then $F[x]$ is a principal ideal domain, and in particular every non-zero prime ideal is maximal. Therefore, one can merely check for reducibility of $p(x)$ in $F$ to see if $(p(x))$ is prime or not. If $deg p leq 3$, then in fact it is enough to check if $p$ has a root in $F$ or not i.e. if $exists x in F,p(x) =0$.
In our case, with $p(x) = x^3+3x^2+3x+1$,we may consider the element $4 in mathbb Z_5$, and see that $p(4) = 0$. So $p$ is reducible mod $5$, and hence we have that it cannot be prime.
If $F$ is not a field, then $F[x]$ need not be a principal ideal domain, as shown by the example $(2,x)$ being non-principal over $mathbb Z[x]$. Here, prime ideals are more difficult to characterize, because one cannot say what every ideal looks like in such a field(it may not be principal, but also may not have other nice properties), so it is more difficult to deduce conclusions from something being an ideal in this case.
answered Jan 28 at 6:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.9k33477
$endgroup$
Note that $(1+x)^3 = 1+3x+3x^2+x^3 in (1+3x+3x^2+x^3)$, but $1+x notin (1+3x+3x^2+x^3)$ because all non-zero multiples of $1+3x+3x^2+x^3$ must be of degree at least $3$. Thus, $(1+3x+3x^2+x^3)$ cannot be prime.
In general, if $F$ is a field then $F[x]$ is a principal ideal domain, and in particular every non-zero prime ideal is maximal. Therefore, one can merely check for reducibility of $p(x)$ in $F$ to see if $(p(x))$ is prime or not. If $deg p leq 3$, then in fact it is enough to check if $p$ has a root in $F$ or not i.e. if $exists x in F,p(x) =0$.
In our case, with $p(x) = x^3+3x^2+3x+1$,we may consider the element $4 in mathbb Z_5$, and see that $p(4) = 0$. So $p$ is reducible mod $5$, and hence we have that it cannot be prime.
If $F$ is not a field, then $F[x]$ need not be a principal ideal domain, as shown by the example $(2,x)$ being non-principal over $mathbb Z[x]$. Here, prime ideals are more difficult to characterize, because one cannot say what every ideal looks like in such a field(it may not be principal, but also may not have other nice properties), so it is more difficult to deduce conclusions from something being an ideal in this case.
answered Jan 28 at 6:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.9k33477
edited Jan 28 at 10:18
answered Jan 28 at 6:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.9k33477
answered Jan 28 at 6:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.9k33477
answered Jan 28 at 6:19
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.9k33477
39.9k33477
add a comment |
add a comment |
$begingroup$
Hint:
Generally speaking, the strategy for determining the prime ideals of a polynomial ring is a bit different and more complicated. You need to know some facts about the polynomial rings first. So, I'm going to sketch the general idea for you and I'll let you check the details and convince yourself. If you had any question, don't hesitate to ask it in the comment section.
$mathbb{Z}_5$ is a field. So, $mathbb{Z}_5[X]$ is a principal ideal domain because of the Euclidean algorithm for division which works over any field. You can check whether $p(x)=1+3x+3x^2+3x^3$ is reducible or not. Since $deg(p(x))<4$, it suffices to check whether it has a root or not.
Note that prime elements in an integral domain are irreducible. Since $mathbb{Z}_5$ is a field, $mathbb{Z}_5[X]$ is an integral domain. So, if some element is reducible, then it is NOT prime and the principal ideal generated by it is not prime either. On the other hand, if $p(x)$ turns out to be irreducible (which is not true in our case), then it generates a maximal ideal and we know that maximal ideals are prime.
answered Jan 28 at 6:20
stressed outstressed out
6,5831939
$endgroup$
add a comment |
$begingroup$
Hint:
Generally speaking, the strategy for determining the prime ideals of a polynomial ring is a bit different and more complicated. You need to know some facts about the polynomial rings first. So, I'm going to sketch the general idea for you and I'll let you check the details and convince yourself. If you had any question, don't hesitate to ask it in the comment section.
$mathbb{Z}_5$ is a field. So, $mathbb{Z}_5[X]$ is a principal ideal domain because of the Euclidean algorithm for division which works over any field. You can check whether $p(x)=1+3x+3x^2+3x^3$ is reducible or not. Since $deg(p(x))<4$, it suffices to check whether it has a root or not.
Note that prime elements in an integral domain are irreducible. Since $mathbb{Z}_5$ is a field, $mathbb{Z}_5[X]$ is an integral domain. So, if some element is reducible, then it is NOT prime and the principal ideal generated by it is not prime either. On the other hand, if $p(x)$ turns out to be irreducible (which is not true in our case), then it generates a maximal ideal and we know that maximal ideals are prime.
answered Jan 28 at 6:20
stressed outstressed out
6,5831939
$endgroup$
add a comment |
$begingroup$
Hint:
Generally speaking, the strategy for determining the prime ideals of a polynomial ring is a bit different and more complicated. You need to know some facts about the polynomial rings first. So, I'm going to sketch the general idea for you and I'll let you check the details and convince yourself. If you had any question, don't hesitate to ask it in the comment section.
$mathbb{Z}_5$ is a field. So, $mathbb{Z}_5[X]$ is a principal ideal domain because of the Euclidean algorithm for division which works over any field. You can check whether $p(x)=1+3x+3x^2+3x^3$ is reducible or not. Since $deg(p(x))<4$, it suffices to check whether it has a root or not.
Note that prime elements in an integral domain are irreducible. Since $mathbb{Z}_5$ is a field, $mathbb{Z}_5[X]$ is an integral domain. So, if some element is reducible, then it is NOT prime and the principal ideal generated by it is not prime either. On the other hand, if $p(x)$ turns out to be irreducible (which is not true in our case), then it generates a maximal ideal and we know that maximal ideals are prime.
answered Jan 28 at 6:20
stressed outstressed out
6,5831939
$endgroup$
Hint:
Generally speaking, the strategy for determining the prime ideals of a polynomial ring is a bit different and more complicated. You need to know some facts about the polynomial rings first. So, I'm going to sketch the general idea for you and I'll let you check the details and convince yourself. If you had any question, don't hesitate to ask it in the comment section.
$mathbb{Z}_5$ is a field. So, $mathbb{Z}_5[X]$ is a principal ideal domain because of the Euclidean algorithm for division which works over any field. You can check whether $p(x)=1+3x+3x^2+3x^3$ is reducible or not. Since $deg(p(x))<4$, it suffices to check whether it has a root or not.
Note that prime elements in an integral domain are irreducible. Since $mathbb{Z}_5$ is a field, $mathbb{Z}_5[X]$ is an integral domain. So, if some element is reducible, then it is NOT prime and the principal ideal generated by it is not prime either. On the other hand, if $p(x)$ turns out to be irreducible (which is not true in our case), then it generates a maximal ideal and we know that maximal ideals are prime.
answered Jan 28 at 6:20
stressed outstressed out
6,5831939
edited Jan 28 at 6:52
answered Jan 28 at 6:20
stressed outstressed out
6,5831939
answered Jan 28 at 6:20
stressed outstressed out
6,5831939
answered Jan 28 at 6:20
stressed outstressed out
6,5831939
6,5831939
add a comment |
add a comment |
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$begingroup$
how about showing that the quotient is an Integral domain then the ideal will be a prime ideal. Can you somehow do that?
$endgroup$
– Sujit Bhattacharyya
Jan 28 at 6:02