Mixing
Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the...
$begingroup$
Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the monic irreducible polynomials?
In Algebra by Artin, he says this proposition is proven analogously to:
Here, he shows that if $n$ is prime, then $mathbb Z/(n)$ is a field. Then we use the fact that $R/I$ is a field iff $I$ is maximal, and he concludes that $(n)$ is maximal.
The analogous proof would be that if $f(x)$ is monic irreducible, then $F[x]/(f)$ is a field. The only problem is that he has not proven that $F[x]$ modulo a monic irreducible polynomial is a field.
abstract-algebra proof-verification ring-theory field-theory maximal-and-prime-ideals
asked Jan 31 at 22:48
Al JebrAl Jebr
4,42143478
$endgroup$
add a comment |
$begingroup$
Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the monic irreducible polynomials?
In Algebra by Artin, he says this proposition is proven analogously to:
Here, he shows that if $n$ is prime, then $mathbb Z/(n)$ is a field. Then we use the fact that $R/I$ is a field iff $I$ is maximal, and he concludes that $(n)$ is maximal.
The analogous proof would be that if $f(x)$ is monic irreducible, then $F[x]/(f)$ is a field. The only problem is that he has not proven that $F[x]$ modulo a monic irreducible polynomial is a field.
abstract-algebra proof-verification ring-theory field-theory maximal-and-prime-ideals
asked Jan 31 at 22:48
Al JebrAl Jebr
4,42143478
$endgroup$
$begingroup$
Related : math.stackexchange.com/q/350054
$endgroup$
– Jean Marie
Jan 31 at 23:55
$begingroup$
$F[x]/(f)$ is a field because $(f)$ is a maximal ideal and $(f)$ is a maximal ideal because $(f)subseteq (g)iff gmid fiff g=f$ or $g=1 iff (f)=(g)$ or $(g)=F[x]$. In the step $gmid f iff g=f$ or $g=1$ I am using that $F[x]$ is a UFD (and hence irreducible elements are prime elements).
$endgroup$
– yamete kudasai
Feb 1 at 0:14
add a comment |
$begingroup$
Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the monic irreducible polynomials?
In Algebra by Artin, he says this proposition is proven analogously to:
Here, he shows that if $n$ is prime, then $mathbb Z/(n)$ is a field. Then we use the fact that $R/I$ is a field iff $I$ is maximal, and he concludes that $(n)$ is maximal.
The analogous proof would be that if $f(x)$ is monic irreducible, then $F[x]/(f)$ is a field. The only problem is that he has not proven that $F[x]$ modulo a monic irreducible polynomial is a field.
abstract-algebra proof-verification ring-theory field-theory maximal-and-prime-ideals
asked Jan 31 at 22:48
Al JebrAl Jebr
4,42143478
$endgroup$
Let $F$ be a field. How do we show that maximal ideals of $F[x]$ are the principal ideals generated by the monic irreducible polynomials?
In Algebra by Artin, he says this proposition is proven analogously to:
Here, he shows that if $n$ is prime, then $mathbb Z/(n)$ is a field. Then we use the fact that $R/I$ is a field iff $I$ is maximal, and he concludes that $(n)$ is maximal.
The analogous proof would be that if $f(x)$ is monic irreducible, then $F[x]/(f)$ is a field. The only problem is that he has not proven that $F[x]$ modulo a monic irreducible polynomial is a field.
abstract-algebra proof-verification ring-theory field-theory maximal-and-prime-ideals
abstract-algebra proof-verification ring-theory field-theory maximal-and-prime-ideals
asked Jan 31 at 22:48
Al JebrAl Jebr
4,42143478
asked Jan 31 at 22:48
Al JebrAl Jebr
4,42143478
edited Feb 1 at 1:46
Al Jebr
asked Jan 31 at 22:48
Al JebrAl Jebr
4,42143478
asked Jan 31 at 22:48
Al JebrAl Jebr
4,42143478
asked Jan 31 at 22:48
Al JebrAl Jebr
4,42143478
4,42143478
$begingroup$
Related : math.stackexchange.com/q/350054
$endgroup$
– Jean Marie
Jan 31 at 23:55
$begingroup$
$F[x]/(f)$ is a field because $(f)$ is a maximal ideal and $(f)$ is a maximal ideal because $(f)subseteq (g)iff gmid fiff g=f$ or $g=1 iff (f)=(g)$ or $(g)=F[x]$. In the step $gmid f iff g=f$ or $g=1$ I am using that $F[x]$ is a UFD (and hence irreducible elements are prime elements).
$endgroup$
– yamete kudasai
Feb 1 at 0:14
add a comment |
$begingroup$
Related : math.stackexchange.com/q/350054
$endgroup$
– Jean Marie
Jan 31 at 23:55
$begingroup$
$F[x]/(f)$ is a field because $(f)$ is a maximal ideal and $(f)$ is a maximal ideal because $(f)subseteq (g)iff gmid fiff g=f$ or $g=1 iff (f)=(g)$ or $(g)=F[x]$. In the step $gmid f iff g=f$ or $g=1$ I am using that $F[x]$ is a UFD (and hence irreducible elements are prime elements).
$endgroup$
– yamete kudasai
Feb 1 at 0:14
$begingroup$
Related : math.stackexchange.com/q/350054
$endgroup$
– Jean Marie
Jan 31 at 23:55
$begingroup$
Related : math.stackexchange.com/q/350054
$endgroup$
– Jean Marie
Jan 31 at 23:55
$begingroup$
$F[x]/(f)$ is a field because $(f)$ is a maximal ideal and $(f)$ is a maximal ideal because $(f)subseteq (g)iff gmid fiff g=f$ or $g=1 iff (f)=(g)$ or $(g)=F[x]$. In the step $gmid f iff g=f$ or $g=1$ I am using that $F[x]$ is a UFD (and hence irreducible elements are prime elements).
$endgroup$
– yamete kudasai
Feb 1 at 0:14
$begingroup$
$F[x]/(f)$ is a field because $(f)$ is a maximal ideal and $(f)$ is a maximal ideal because $(f)subseteq (g)iff gmid fiff g=f$ or $g=1 iff (f)=(g)$ or $(g)=F[x]$. In the step $gmid f iff g=f$ or $g=1$ I am using that $F[x]$ is a UFD (and hence irreducible elements are prime elements).
$endgroup$
– yamete kudasai
Feb 1 at 0:14
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In general we have $F(x)/(f)$ is a field iff f(x) is irreducible.
If reducible, then we have a zero divisor, so it can't be a field.
If irreducible, then all polynomial can be subjected to Euclidean algorithm which gives you a multiplicative inverse.
The remaining field axioms follow from the fact that we have a ring quotient.
answered Jan 31 at 22:59
AlexandrosAlexandros
1,0151413
$endgroup$
add a comment |
$begingroup$
There''s a classic result in commutative algebra that you can apply:
Let $B$ be an integral domain, $A$ be a subring such that $B$ is
integral over $A$. Then $B$ is a field if and only if $A$ is a field.
answered Jan 31 at 22:54
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
For any field $F$, $F[x]$ is a principal ideal domain; this is a very well-known and oft-quoted result, which I will accept here.
Now let
$M subset F[x] tag 1$
be a maximal ideal; since $F[x]$ is a principal ideal domain, we have
$M = (m(x)) tag 2$
for some
$m(x) in F[x]; tag 3$
we may clearly take $m(x)$ to be monic, since the leading coefficient $mu$ of $m(x)$, satisfying as it does $mu ne 0$, is a unit; thus $mu^{-1} m(x)$ is monic and
$(mu^{-1} m(x)) = (m(x)); tag 4$
now if $m(x)$ were reducible in $F[x]$, we would have
$m(x) = p(x)q(x), ; p(x), q(x) in F[x], ; deg p(x), deg q(x) ge 1; tag 5$
consider the ideal
$(p(x)) subsetneq F[x]; tag 6$
it is clearly proper: since $deg p(x) ge 1$, $(p(x))$ contains no polynomials of degree $0$, that is, contains no elements of $F$. Also,
$(m(x)) = (p(x)q(x)) subset (p(x)), tag 7$
which shows that $(m(x))$ is not a maximal ideal in $F[x]$; this contradiction implies that $m(x)$ is irreducible in $F[x]$. Finis.
answered Jan 31 at 23:48
Robert LewisRobert Lewis
48.9k23168
$endgroup$
1
$begingroup$
In (4), did you want $(m(x))$ on the right-hand side?
$endgroup$
– J. W. Tanner
Feb 1 at 1:13
$begingroup$
@J.W.Tanner: 'deed I did! Thanks, will fix!
$endgroup$
– Robert Lewis
Feb 1 at 1:15
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In general we have $F(x)/(f)$ is a field iff f(x) is irreducible.
If reducible, then we have a zero divisor, so it can't be a field.
If irreducible, then all polynomial can be subjected to Euclidean algorithm which gives you a multiplicative inverse.
The remaining field axioms follow from the fact that we have a ring quotient.
answered Jan 31 at 22:59
AlexandrosAlexandros
1,0151413
$endgroup$
add a comment |
$begingroup$
In general we have $F(x)/(f)$ is a field iff f(x) is irreducible.
If reducible, then we have a zero divisor, so it can't be a field.
If irreducible, then all polynomial can be subjected to Euclidean algorithm which gives you a multiplicative inverse.
The remaining field axioms follow from the fact that we have a ring quotient.
answered Jan 31 at 22:59
AlexandrosAlexandros
1,0151413
$endgroup$
add a comment |
$begingroup$
In general we have $F(x)/(f)$ is a field iff f(x) is irreducible.
If reducible, then we have a zero divisor, so it can't be a field.
If irreducible, then all polynomial can be subjected to Euclidean algorithm which gives you a multiplicative inverse.
The remaining field axioms follow from the fact that we have a ring quotient.
answered Jan 31 at 22:59
AlexandrosAlexandros
1,0151413
$endgroup$
In general we have $F(x)/(f)$ is a field iff f(x) is irreducible.
If reducible, then we have a zero divisor, so it can't be a field.
If irreducible, then all polynomial can be subjected to Euclidean algorithm which gives you a multiplicative inverse.
The remaining field axioms follow from the fact that we have a ring quotient.
answered Jan 31 at 22:59
AlexandrosAlexandros
1,0151413
answered Jan 31 at 22:59
AlexandrosAlexandros
1,0151413
answered Jan 31 at 22:59
AlexandrosAlexandros
1,0151413
answered Jan 31 at 22:59
AlexandrosAlexandros
1,0151413
1,0151413
add a comment |
add a comment |
$begingroup$
There''s a classic result in commutative algebra that you can apply:
Let $B$ be an integral domain, $A$ be a subring such that $B$ is
integral over $A$. Then $B$ is a field if and only if $A$ is a field.
answered Jan 31 at 22:54
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
There''s a classic result in commutative algebra that you can apply:
Let $B$ be an integral domain, $A$ be a subring such that $B$ is
integral over $A$. Then $B$ is a field if and only if $A$ is a field.
answered Jan 31 at 22:54
BernardBernard
124k741117
$endgroup$
add a comment |
$begingroup$
There''s a classic result in commutative algebra that you can apply:
Let $B$ be an integral domain, $A$ be a subring such that $B$ is
integral over $A$. Then $B$ is a field if and only if $A$ is a field.
answered Jan 31 at 22:54
BernardBernard
124k741117
$endgroup$
There''s a classic result in commutative algebra that you can apply:
Let $B$ be an integral domain, $A$ be a subring such that $B$ is
integral over $A$. Then $B$ is a field if and only if $A$ is a field.
answered Jan 31 at 22:54
BernardBernard
124k741117
edited Jan 31 at 23:07
answered Jan 31 at 22:54
BernardBernard
124k741117
answered Jan 31 at 22:54
BernardBernard
124k741117
answered Jan 31 at 22:54
BernardBernard
124k741117
124k741117
add a comment |
add a comment |
$begingroup$
For any field $F$, $F[x]$ is a principal ideal domain; this is a very well-known and oft-quoted result, which I will accept here.
Now let
$M subset F[x] tag 1$
be a maximal ideal; since $F[x]$ is a principal ideal domain, we have
$M = (m(x)) tag 2$
for some
$m(x) in F[x]; tag 3$
we may clearly take $m(x)$ to be monic, since the leading coefficient $mu$ of $m(x)$, satisfying as it does $mu ne 0$, is a unit; thus $mu^{-1} m(x)$ is monic and
$(mu^{-1} m(x)) = (m(x)); tag 4$
now if $m(x)$ were reducible in $F[x]$, we would have
$m(x) = p(x)q(x), ; p(x), q(x) in F[x], ; deg p(x), deg q(x) ge 1; tag 5$
consider the ideal
$(p(x)) subsetneq F[x]; tag 6$
it is clearly proper: since $deg p(x) ge 1$, $(p(x))$ contains no polynomials of degree $0$, that is, contains no elements of $F$. Also,
$(m(x)) = (p(x)q(x)) subset (p(x)), tag 7$
which shows that $(m(x))$ is not a maximal ideal in $F[x]$; this contradiction implies that $m(x)$ is irreducible in $F[x]$. Finis.
answered Jan 31 at 23:48
Robert LewisRobert Lewis
48.9k23168
$endgroup$
1
$begingroup$
In (4), did you want $(m(x))$ on the right-hand side?
$endgroup$
– J. W. Tanner
Feb 1 at 1:13
$begingroup$
@J.W.Tanner: 'deed I did! Thanks, will fix!
$endgroup$
– Robert Lewis
Feb 1 at 1:15
add a comment |
$begingroup$
For any field $F$, $F[x]$ is a principal ideal domain; this is a very well-known and oft-quoted result, which I will accept here.
Now let
$M subset F[x] tag 1$
be a maximal ideal; since $F[x]$ is a principal ideal domain, we have
$M = (m(x)) tag 2$
for some
$m(x) in F[x]; tag 3$
we may clearly take $m(x)$ to be monic, since the leading coefficient $mu$ of $m(x)$, satisfying as it does $mu ne 0$, is a unit; thus $mu^{-1} m(x)$ is monic and
$(mu^{-1} m(x)) = (m(x)); tag 4$
now if $m(x)$ were reducible in $F[x]$, we would have
$m(x) = p(x)q(x), ; p(x), q(x) in F[x], ; deg p(x), deg q(x) ge 1; tag 5$
consider the ideal
$(p(x)) subsetneq F[x]; tag 6$
it is clearly proper: since $deg p(x) ge 1$, $(p(x))$ contains no polynomials of degree $0$, that is, contains no elements of $F$. Also,
$(m(x)) = (p(x)q(x)) subset (p(x)), tag 7$
which shows that $(m(x))$ is not a maximal ideal in $F[x]$; this contradiction implies that $m(x)$ is irreducible in $F[x]$. Finis.
answered Jan 31 at 23:48
Robert LewisRobert Lewis
48.9k23168
$endgroup$
1
$begingroup$
In (4), did you want $(m(x))$ on the right-hand side?
$endgroup$
– J. W. Tanner
Feb 1 at 1:13
$begingroup$
@J.W.Tanner: 'deed I did! Thanks, will fix!
$endgroup$
– Robert Lewis
Feb 1 at 1:15
add a comment |
$begingroup$
For any field $F$, $F[x]$ is a principal ideal domain; this is a very well-known and oft-quoted result, which I will accept here.
Now let
$M subset F[x] tag 1$
be a maximal ideal; since $F[x]$ is a principal ideal domain, we have
$M = (m(x)) tag 2$
for some
$m(x) in F[x]; tag 3$
we may clearly take $m(x)$ to be monic, since the leading coefficient $mu$ of $m(x)$, satisfying as it does $mu ne 0$, is a unit; thus $mu^{-1} m(x)$ is monic and
$(mu^{-1} m(x)) = (m(x)); tag 4$
now if $m(x)$ were reducible in $F[x]$, we would have
$m(x) = p(x)q(x), ; p(x), q(x) in F[x], ; deg p(x), deg q(x) ge 1; tag 5$
consider the ideal
$(p(x)) subsetneq F[x]; tag 6$
it is clearly proper: since $deg p(x) ge 1$, $(p(x))$ contains no polynomials of degree $0$, that is, contains no elements of $F$. Also,
$(m(x)) = (p(x)q(x)) subset (p(x)), tag 7$
which shows that $(m(x))$ is not a maximal ideal in $F[x]$; this contradiction implies that $m(x)$ is irreducible in $F[x]$. Finis.
answered Jan 31 at 23:48
Robert LewisRobert Lewis
48.9k23168
$endgroup$
For any field $F$, $F[x]$ is a principal ideal domain; this is a very well-known and oft-quoted result, which I will accept here.
Now let
$M subset F[x] tag 1$
be a maximal ideal; since $F[x]$ is a principal ideal domain, we have
$M = (m(x)) tag 2$
for some
$m(x) in F[x]; tag 3$
we may clearly take $m(x)$ to be monic, since the leading coefficient $mu$ of $m(x)$, satisfying as it does $mu ne 0$, is a unit; thus $mu^{-1} m(x)$ is monic and
$(mu^{-1} m(x)) = (m(x)); tag 4$
now if $m(x)$ were reducible in $F[x]$, we would have
$m(x) = p(x)q(x), ; p(x), q(x) in F[x], ; deg p(x), deg q(x) ge 1; tag 5$
consider the ideal
$(p(x)) subsetneq F[x]; tag 6$
it is clearly proper: since $deg p(x) ge 1$, $(p(x))$ contains no polynomials of degree $0$, that is, contains no elements of $F$. Also,
$(m(x)) = (p(x)q(x)) subset (p(x)), tag 7$
which shows that $(m(x))$ is not a maximal ideal in $F[x]$; this contradiction implies that $m(x)$ is irreducible in $F[x]$. Finis.
answered Jan 31 at 23:48
Robert LewisRobert Lewis
48.9k23168
edited Feb 1 at 1:16
answered Jan 31 at 23:48
Robert LewisRobert Lewis
48.9k23168
answered Jan 31 at 23:48
Robert LewisRobert Lewis
48.9k23168
answered Jan 31 at 23:48
Robert LewisRobert Lewis
48.9k23168
48.9k23168
1
$begingroup$
In (4), did you want $(m(x))$ on the right-hand side?
$endgroup$
– J. W. Tanner
Feb 1 at 1:13
$begingroup$
@J.W.Tanner: 'deed I did! Thanks, will fix!
$endgroup$
– Robert Lewis
Feb 1 at 1:15
add a comment |
1
$begingroup$
In (4), did you want $(m(x))$ on the right-hand side?
$endgroup$
– J. W. Tanner
Feb 1 at 1:13
$begingroup$
@J.W.Tanner: 'deed I did! Thanks, will fix!
$endgroup$
– Robert Lewis
Feb 1 at 1:15
1
1
$begingroup$
In (4), did you want $(m(x))$ on the right-hand side?
$endgroup$
– J. W. Tanner
Feb 1 at 1:13
$begingroup$
In (4), did you want $(m(x))$ on the right-hand side?
$endgroup$
– J. W. Tanner
Feb 1 at 1:13
$begingroup$
@J.W.Tanner: 'deed I did! Thanks, will fix!
$endgroup$
– Robert Lewis
Feb 1 at 1:15
$begingroup$
@J.W.Tanner: 'deed I did! Thanks, will fix!
$endgroup$
– Robert Lewis
Feb 1 at 1:15
add a comment |
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$begingroup$
Related : math.stackexchange.com/q/350054
$endgroup$
– Jean Marie
Jan 31 at 23:55
$begingroup$
$F[x]/(f)$ is a field because $(f)$ is a maximal ideal and $(f)$ is a maximal ideal because $(f)subseteq (g)iff gmid fiff g=f$ or $g=1 iff (f)=(g)$ or $(g)=F[x]$. In the step $gmid f iff g=f$ or $g=1$ I am using that $F[x]$ is a UFD (and hence irreducible elements are prime elements).
$endgroup$
– yamete kudasai
Feb 1 at 0:14