Maximal Ideal in a polynomial ring
$begingroup$
Let $G$ be a field, $c in G$, and let $H = {(x - c)k, k in G[X]}$ be an ideal of the polynomial ring $G[X]$. How to show that $H$ is a maximal ideal?
Ideas:
I know that $H$ is maximal if and only if $H = <p(x)>$ for some irreducible polynomial $p(x)$ in $G[X]$. I also know that it suffices to show that $G[X]/H$ is a field however, I'm having trouble coming up with a coherent proof. Any help would be much appreciated.
abstract-algebra ring-theory
asked Dec 12 '18 at 0:33
J. MolinaJ. Molina
82
$endgroup$
add a comment |
$begingroup$
Let $G$ be a field, $c in G$, and let $H = {(x - c)k, k in G[X]}$ be an ideal of the polynomial ring $G[X]$. How to show that $H$ is a maximal ideal?
Ideas:
I know that $H$ is maximal if and only if $H = <p(x)>$ for some irreducible polynomial $p(x)$ in $G[X]$. I also know that it suffices to show that $G[X]/H$ is a field however, I'm having trouble coming up with a coherent proof. Any help would be much appreciated.
abstract-algebra ring-theory
asked Dec 12 '18 at 0:33
J. MolinaJ. Molina
82
$endgroup$
$begingroup$
Your $H$ is $langle x-crangle$. Can you show that $x-c$ is irreducible?
$endgroup$
– user9077
Dec 12 '18 at 0:37
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Is ${(x-c)k, kin G[X]}$ of the form $langle p(x)rangle$?
$endgroup$
– Ben
Dec 12 '18 at 0:38
add a comment |
$begingroup$
Let $G$ be a field, $c in G$, and let $H = {(x - c)k, k in G[X]}$ be an ideal of the polynomial ring $G[X]$. How to show that $H$ is a maximal ideal?
Ideas:
I know that $H$ is maximal if and only if $H = <p(x)>$ for some irreducible polynomial $p(x)$ in $G[X]$. I also know that it suffices to show that $G[X]/H$ is a field however, I'm having trouble coming up with a coherent proof. Any help would be much appreciated.
abstract-algebra ring-theory
asked Dec 12 '18 at 0:33
J. MolinaJ. Molina
82
$endgroup$
Let $G$ be a field, $c in G$, and let $H = {(x - c)k, k in G[X]}$ be an ideal of the polynomial ring $G[X]$. How to show that $H$ is a maximal ideal?
Ideas:
I know that $H$ is maximal if and only if $H = <p(x)>$ for some irreducible polynomial $p(x)$ in $G[X]$. I also know that it suffices to show that $G[X]/H$ is a field however, I'm having trouble coming up with a coherent proof. Any help would be much appreciated.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Dec 12 '18 at 0:33
J. MolinaJ. Molina
82
asked Dec 12 '18 at 0:33
J. MolinaJ. Molina
82
asked Dec 12 '18 at 0:33
J. MolinaJ. Molina
82
asked Dec 12 '18 at 0:33
J. MolinaJ. Molina
82
asked Dec 12 '18 at 0:33
J. MolinaJ. Molina
82
82
$begingroup$
Your $H$ is $langle x-crangle$. Can you show that $x-c$ is irreducible?
$endgroup$
– user9077
Dec 12 '18 at 0:37
$begingroup$
Is ${(x-c)k, kin G[X]}$ of the form $langle p(x)rangle$?
$endgroup$
– Ben
Dec 12 '18 at 0:38
add a comment |
$begingroup$
Your $H$ is $langle x-crangle$. Can you show that $x-c$ is irreducible?
$endgroup$
– user9077
Dec 12 '18 at 0:37
$begingroup$
Is ${(x-c)k, kin G[X]}$ of the form $langle p(x)rangle$?
$endgroup$
– Ben
Dec 12 '18 at 0:38
$begingroup$
Your $H$ is $langle x-crangle$. Can you show that $x-c$ is irreducible?
$endgroup$
– user9077
Dec 12 '18 at 0:37
$begingroup$
Your $H$ is $langle x-crangle$. Can you show that $x-c$ is irreducible?
$endgroup$
– user9077
Dec 12 '18 at 0:37
$begingroup$
Is ${(x-c)k, kin G[X]}$ of the form $langle p(x)rangle$?
$endgroup$
– Ben
Dec 12 '18 at 0:38
$begingroup$
Is ${(x-c)k, kin G[X]}$ of the form $langle p(x)rangle$?
$endgroup$
– Ben
Dec 12 '18 at 0:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Both approaches work. One way is to see that $X - c$ is irreducible: if it weren't then we would have $X-c = fg$ with $f,g$ non units and non-zero (because $X -c neq 0$). Thus both $f$ and $g$ must have degree at least $1$, which is absurd since
$$
1 = deg (X-c) = deg fg = deg f + deg g geq 2
$$
The other approach is proving that $G[X]/langle X-crangle$ is a field. Here's a hint: consider the unital ring morphism $G[X] to G$ that sends $X$ to $c$, and recall the statement of the first isomorphism theorem.
answered Dec 12 '18 at 0:43
Guido A.Guido A.
7,3601730
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Both approaches work. One way is to see that $X - c$ is irreducible: if it weren't then we would have $X-c = fg$ with $f,g$ non units and non-zero (because $X -c neq 0$). Thus both $f$ and $g$ must have degree at least $1$, which is absurd since
$$
1 = deg (X-c) = deg fg = deg f + deg g geq 2
$$
The other approach is proving that $G[X]/langle X-crangle$ is a field. Here's a hint: consider the unital ring morphism $G[X] to G$ that sends $X$ to $c$, and recall the statement of the first isomorphism theorem.
answered Dec 12 '18 at 0:43
Guido A.Guido A.
7,3601730
$endgroup$
add a comment |
$begingroup$
Both approaches work. One way is to see that $X - c$ is irreducible: if it weren't then we would have $X-c = fg$ with $f,g$ non units and non-zero (because $X -c neq 0$). Thus both $f$ and $g$ must have degree at least $1$, which is absurd since
$$
1 = deg (X-c) = deg fg = deg f + deg g geq 2
$$
The other approach is proving that $G[X]/langle X-crangle$ is a field. Here's a hint: consider the unital ring morphism $G[X] to G$ that sends $X$ to $c$, and recall the statement of the first isomorphism theorem.
answered Dec 12 '18 at 0:43
Guido A.Guido A.
7,3601730
$endgroup$
add a comment |
$begingroup$
Both approaches work. One way is to see that $X - c$ is irreducible: if it weren't then we would have $X-c = fg$ with $f,g$ non units and non-zero (because $X -c neq 0$). Thus both $f$ and $g$ must have degree at least $1$, which is absurd since
$$
1 = deg (X-c) = deg fg = deg f + deg g geq 2
$$
The other approach is proving that $G[X]/langle X-crangle$ is a field. Here's a hint: consider the unital ring morphism $G[X] to G$ that sends $X$ to $c$, and recall the statement of the first isomorphism theorem.
answered Dec 12 '18 at 0:43
Guido A.Guido A.
7,3601730
$endgroup$
Both approaches work. One way is to see that $X - c$ is irreducible: if it weren't then we would have $X-c = fg$ with $f,g$ non units and non-zero (because $X -c neq 0$). Thus both $f$ and $g$ must have degree at least $1$, which is absurd since
$$
1 = deg (X-c) = deg fg = deg f + deg g geq 2
$$
The other approach is proving that $G[X]/langle X-crangle$ is a field. Here's a hint: consider the unital ring morphism $G[X] to G$ that sends $X$ to $c$, and recall the statement of the first isomorphism theorem.
answered Dec 12 '18 at 0:43
Guido A.Guido A.
7,3601730
answered Dec 12 '18 at 0:43
Guido A.Guido A.
7,3601730
answered Dec 12 '18 at 0:43
Guido A.Guido A.
7,3601730
answered Dec 12 '18 at 0:43
Guido A.Guido A.
7,3601730
7,3601730
add a comment |
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$begingroup$
Your $H$ is $langle x-crangle$. Can you show that $x-c$ is irreducible?
$endgroup$
– user9077
Dec 12 '18 at 0:37
$begingroup$
Is ${(x-c)k, kin G[X]}$ of the form $langle p(x)rangle$?
$endgroup$
– Ben
Dec 12 '18 at 0:38