Ring Homomorphism: $mathbb{Z}[x] / (f(x)) to mathbb{Q}$
up vote
2
down vote
favorite
Let $f(x) in mathbb{Z}[x]$. Prove that $f(x)$ has a root in $mathbb{Q}$ iff there is a ring homomorphism from $mathbb{Z}[x]/(f(x)) rightarrow mathbb{Q}$.
I tried using a homomorphism from $mathbb{Z}[x] rightarrow mathbb{Q}$ defined by $varphi(f(x)) = f(q)$ for a fixed $q in mathbb{Q}$. When $q$ is a root this could be useful, but that's all I've managed to come up with, and I'm unclear how to proceed.
abstract-algebra ring-theory ring-homomorphism
add a comment |
up vote
2
down vote
favorite
Let $f(x) in mathbb{Z}[x]$. Prove that $f(x)$ has a root in $mathbb{Q}$ iff there is a ring homomorphism from $mathbb{Z}[x]/(f(x)) rightarrow mathbb{Q}$.
I tried using a homomorphism from $mathbb{Z}[x] rightarrow mathbb{Q}$ defined by $varphi(f(x)) = f(q)$ for a fixed $q in mathbb{Q}$. When $q$ is a root this could be useful, but that's all I've managed to come up with, and I'm unclear how to proceed.
abstract-algebra ring-theory ring-homomorphism
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f(x) in mathbb{Z}[x]$. Prove that $f(x)$ has a root in $mathbb{Q}$ iff there is a ring homomorphism from $mathbb{Z}[x]/(f(x)) rightarrow mathbb{Q}$.
I tried using a homomorphism from $mathbb{Z}[x] rightarrow mathbb{Q}$ defined by $varphi(f(x)) = f(q)$ for a fixed $q in mathbb{Q}$. When $q$ is a root this could be useful, but that's all I've managed to come up with, and I'm unclear how to proceed.
abstract-algebra ring-theory ring-homomorphism
Let $f(x) in mathbb{Z}[x]$. Prove that $f(x)$ has a root in $mathbb{Q}$ iff there is a ring homomorphism from $mathbb{Z}[x]/(f(x)) rightarrow mathbb{Q}$.
I tried using a homomorphism from $mathbb{Z}[x] rightarrow mathbb{Q}$ defined by $varphi(f(x)) = f(q)$ for a fixed $q in mathbb{Q}$. When $q$ is a root this could be useful, but that's all I've managed to come up with, and I'm unclear how to proceed.
abstract-algebra ring-theory ring-homomorphism
abstract-algebra ring-theory ring-homomorphism
edited Nov 16 at 7:30
Shubham
1,3171518
1,3171518
asked Nov 16 at 5:46
Physics Blader
413
413
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
The following lemma will be useful.
Lemma. Let $varphi: A to B$ be a ring homomorphism and $I$ be an ideal of $A$. Then $varphi$ descends to a homomorphism $overline{varphi}: A/I to B$ iff $I subseteq ker(varphi)$.
You've defined the map
begin{align*}
varphi: mathbb{Z}[x] &to mathbb{Q}\
g(x) &mapsto g(q)
end{align*}
Can you see why you can apply the lemma to get a map on the quotient? (As a note, you shouldn't use $f(x)$ to refer to two different things.)
add a comment |
up vote
1
down vote
A ring homomorphism $Bbb Z[X]/(f(X))toBbb Q$ is essentially a ring homomorphism
$phi:Bbb Z[X]toBbb Q$ with the property that $phi(f(X))=0$.
The ring homomorphisms $Bbb Z[X]toBbb Q$ all have the form $phi_q:g(X)mapsto
g(q)$ for $qinBbb Q$, as you say. This map induces a homomorphism
$Bbb Z[X]/(f(X))toBbb Q$ iff $phi_q(f(X))=0$. But $phi_q(f(X))=f(q)$.
add a comment |
up vote
0
down vote
Recall that a map between unital rings $mathbb{Z}[X] xrightarrow{q} A$ is determined by $q(X)$. Moreover, if $q_x := q(X)$, then $q(f) = f(q_x)$. This comes from writing $f$ as a sum of monomials, expanding and using that $f(X) = q_x$. Hence all morphisms $mathbb{Z}[X] to A$ are an evaluation.
So, take $ev_x$ an evaluation map from $mathbb{Z}[X]$ to $mathbb{Q} ni x$. If $f(x) = ev_x(f) = 0$, then $(f) subset ker(ev_x)$ and so $ev_x$ factors through $mathbb{Z}[X]/(f)$. To see this you can appeal to the first isomorphism theorem.
Reciprocally, if you have a morphism $q : mathbb{Z}[X]/(f) to mathbb{Q}$, then you have a morphism $g = qpi$ defined as the following composition,
$$
mathbb{Z}[X] xrightarrow{pi} mathbb{Z}[X]/(f) xrightarrow{q} mathbb{Q}.
$$
We have proved that $g equiv ev_x$ for some rational $x$. Thus,
$$
f(x) = ev_x(f) = g(f) = qpi(f) = q(0) = 0,
$$
which concludes the proof.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The following lemma will be useful.
Lemma. Let $varphi: A to B$ be a ring homomorphism and $I$ be an ideal of $A$. Then $varphi$ descends to a homomorphism $overline{varphi}: A/I to B$ iff $I subseteq ker(varphi)$.
You've defined the map
begin{align*}
varphi: mathbb{Z}[x] &to mathbb{Q}\
g(x) &mapsto g(q)
end{align*}
Can you see why you can apply the lemma to get a map on the quotient? (As a note, you shouldn't use $f(x)$ to refer to two different things.)
add a comment |
up vote
1
down vote
The following lemma will be useful.
Lemma. Let $varphi: A to B$ be a ring homomorphism and $I$ be an ideal of $A$. Then $varphi$ descends to a homomorphism $overline{varphi}: A/I to B$ iff $I subseteq ker(varphi)$.
You've defined the map
begin{align*}
varphi: mathbb{Z}[x] &to mathbb{Q}\
g(x) &mapsto g(q)
end{align*}
Can you see why you can apply the lemma to get a map on the quotient? (As a note, you shouldn't use $f(x)$ to refer to two different things.)
add a comment |
up vote
1
down vote
up vote
1
down vote
The following lemma will be useful.
Lemma. Let $varphi: A to B$ be a ring homomorphism and $I$ be an ideal of $A$. Then $varphi$ descends to a homomorphism $overline{varphi}: A/I to B$ iff $I subseteq ker(varphi)$.
You've defined the map
begin{align*}
varphi: mathbb{Z}[x] &to mathbb{Q}\
g(x) &mapsto g(q)
end{align*}
Can you see why you can apply the lemma to get a map on the quotient? (As a note, you shouldn't use $f(x)$ to refer to two different things.)
The following lemma will be useful.
Lemma. Let $varphi: A to B$ be a ring homomorphism and $I$ be an ideal of $A$. Then $varphi$ descends to a homomorphism $overline{varphi}: A/I to B$ iff $I subseteq ker(varphi)$.
You've defined the map
begin{align*}
varphi: mathbb{Z}[x] &to mathbb{Q}\
g(x) &mapsto g(q)
end{align*}
Can you see why you can apply the lemma to get a map on the quotient? (As a note, you shouldn't use $f(x)$ to refer to two different things.)
answered Nov 16 at 5:57
André 3000
12.1k22041
12.1k22041
add a comment |
add a comment |
up vote
1
down vote
A ring homomorphism $Bbb Z[X]/(f(X))toBbb Q$ is essentially a ring homomorphism
$phi:Bbb Z[X]toBbb Q$ with the property that $phi(f(X))=0$.
The ring homomorphisms $Bbb Z[X]toBbb Q$ all have the form $phi_q:g(X)mapsto
g(q)$ for $qinBbb Q$, as you say. This map induces a homomorphism
$Bbb Z[X]/(f(X))toBbb Q$ iff $phi_q(f(X))=0$. But $phi_q(f(X))=f(q)$.
add a comment |
up vote
1
down vote
A ring homomorphism $Bbb Z[X]/(f(X))toBbb Q$ is essentially a ring homomorphism
$phi:Bbb Z[X]toBbb Q$ with the property that $phi(f(X))=0$.
The ring homomorphisms $Bbb Z[X]toBbb Q$ all have the form $phi_q:g(X)mapsto
g(q)$ for $qinBbb Q$, as you say. This map induces a homomorphism
$Bbb Z[X]/(f(X))toBbb Q$ iff $phi_q(f(X))=0$. But $phi_q(f(X))=f(q)$.
add a comment |
up vote
1
down vote
up vote
1
down vote
A ring homomorphism $Bbb Z[X]/(f(X))toBbb Q$ is essentially a ring homomorphism
$phi:Bbb Z[X]toBbb Q$ with the property that $phi(f(X))=0$.
The ring homomorphisms $Bbb Z[X]toBbb Q$ all have the form $phi_q:g(X)mapsto
g(q)$ for $qinBbb Q$, as you say. This map induces a homomorphism
$Bbb Z[X]/(f(X))toBbb Q$ iff $phi_q(f(X))=0$. But $phi_q(f(X))=f(q)$.
A ring homomorphism $Bbb Z[X]/(f(X))toBbb Q$ is essentially a ring homomorphism
$phi:Bbb Z[X]toBbb Q$ with the property that $phi(f(X))=0$.
The ring homomorphisms $Bbb Z[X]toBbb Q$ all have the form $phi_q:g(X)mapsto
g(q)$ for $qinBbb Q$, as you say. This map induces a homomorphism
$Bbb Z[X]/(f(X))toBbb Q$ iff $phi_q(f(X))=0$. But $phi_q(f(X))=f(q)$.
answered Nov 16 at 5:58
Lord Shark the Unknown
97.4k958128
97.4k958128
add a comment |
add a comment |
up vote
0
down vote
Recall that a map between unital rings $mathbb{Z}[X] xrightarrow{q} A$ is determined by $q(X)$. Moreover, if $q_x := q(X)$, then $q(f) = f(q_x)$. This comes from writing $f$ as a sum of monomials, expanding and using that $f(X) = q_x$. Hence all morphisms $mathbb{Z}[X] to A$ are an evaluation.
So, take $ev_x$ an evaluation map from $mathbb{Z}[X]$ to $mathbb{Q} ni x$. If $f(x) = ev_x(f) = 0$, then $(f) subset ker(ev_x)$ and so $ev_x$ factors through $mathbb{Z}[X]/(f)$. To see this you can appeal to the first isomorphism theorem.
Reciprocally, if you have a morphism $q : mathbb{Z}[X]/(f) to mathbb{Q}$, then you have a morphism $g = qpi$ defined as the following composition,
$$
mathbb{Z}[X] xrightarrow{pi} mathbb{Z}[X]/(f) xrightarrow{q} mathbb{Q}.
$$
We have proved that $g equiv ev_x$ for some rational $x$. Thus,
$$
f(x) = ev_x(f) = g(f) = qpi(f) = q(0) = 0,
$$
which concludes the proof.
add a comment |
up vote
0
down vote
Recall that a map between unital rings $mathbb{Z}[X] xrightarrow{q} A$ is determined by $q(X)$. Moreover, if $q_x := q(X)$, then $q(f) = f(q_x)$. This comes from writing $f$ as a sum of monomials, expanding and using that $f(X) = q_x$. Hence all morphisms $mathbb{Z}[X] to A$ are an evaluation.
So, take $ev_x$ an evaluation map from $mathbb{Z}[X]$ to $mathbb{Q} ni x$. If $f(x) = ev_x(f) = 0$, then $(f) subset ker(ev_x)$ and so $ev_x$ factors through $mathbb{Z}[X]/(f)$. To see this you can appeal to the first isomorphism theorem.
Reciprocally, if you have a morphism $q : mathbb{Z}[X]/(f) to mathbb{Q}$, then you have a morphism $g = qpi$ defined as the following composition,
$$
mathbb{Z}[X] xrightarrow{pi} mathbb{Z}[X]/(f) xrightarrow{q} mathbb{Q}.
$$
We have proved that $g equiv ev_x$ for some rational $x$. Thus,
$$
f(x) = ev_x(f) = g(f) = qpi(f) = q(0) = 0,
$$
which concludes the proof.
add a comment |
up vote
0
down vote
up vote
0
down vote
Recall that a map between unital rings $mathbb{Z}[X] xrightarrow{q} A$ is determined by $q(X)$. Moreover, if $q_x := q(X)$, then $q(f) = f(q_x)$. This comes from writing $f$ as a sum of monomials, expanding and using that $f(X) = q_x$. Hence all morphisms $mathbb{Z}[X] to A$ are an evaluation.
So, take $ev_x$ an evaluation map from $mathbb{Z}[X]$ to $mathbb{Q} ni x$. If $f(x) = ev_x(f) = 0$, then $(f) subset ker(ev_x)$ and so $ev_x$ factors through $mathbb{Z}[X]/(f)$. To see this you can appeal to the first isomorphism theorem.
Reciprocally, if you have a morphism $q : mathbb{Z}[X]/(f) to mathbb{Q}$, then you have a morphism $g = qpi$ defined as the following composition,
$$
mathbb{Z}[X] xrightarrow{pi} mathbb{Z}[X]/(f) xrightarrow{q} mathbb{Q}.
$$
We have proved that $g equiv ev_x$ for some rational $x$. Thus,
$$
f(x) = ev_x(f) = g(f) = qpi(f) = q(0) = 0,
$$
which concludes the proof.
Recall that a map between unital rings $mathbb{Z}[X] xrightarrow{q} A$ is determined by $q(X)$. Moreover, if $q_x := q(X)$, then $q(f) = f(q_x)$. This comes from writing $f$ as a sum of monomials, expanding and using that $f(X) = q_x$. Hence all morphisms $mathbb{Z}[X] to A$ are an evaluation.
So, take $ev_x$ an evaluation map from $mathbb{Z}[X]$ to $mathbb{Q} ni x$. If $f(x) = ev_x(f) = 0$, then $(f) subset ker(ev_x)$ and so $ev_x$ factors through $mathbb{Z}[X]/(f)$. To see this you can appeal to the first isomorphism theorem.
Reciprocally, if you have a morphism $q : mathbb{Z}[X]/(f) to mathbb{Q}$, then you have a morphism $g = qpi$ defined as the following composition,
$$
mathbb{Z}[X] xrightarrow{pi} mathbb{Z}[X]/(f) xrightarrow{q} mathbb{Q}.
$$
We have proved that $g equiv ev_x$ for some rational $x$. Thus,
$$
f(x) = ev_x(f) = g(f) = qpi(f) = q(0) = 0,
$$
which concludes the proof.
answered Nov 16 at 6:19
Guido A.
6,6321730
6,6321730
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000767%2fring-homomorphism-mathbbzx-fx-to-mathbbq%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown