Is $F(x)/$ isomorphic to a subfield of the splitting field of $f(x)$ over $F$?
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$F$ is a field. $F[x]$ denotes the ring which consists of all the polynomials with coefficients in $F$. Suppose $f(x)in F[x]$ and $f(x)$ is irreducible over $F$, $a$ is a zero of $f(x)$ in some extension field $E$ of $F$. $F(a)$ denote the smallest subfield of $E$ that contains both $F$ and $a$. It is well known that all such $F(a)$ is isomorphic to
$F(x)/<f(x)>$, and all splitting fields of $f(x)$ over $F$ are isomorphic. I am quite curious that is it the case that $F(x)/<f(x)>$ is isomorphic to a subfield of the splitting fields? Some enlightening would be very helpful!
abstract-algebra
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$F$ is a field. $F[x]$ denotes the ring which consists of all the polynomials with coefficients in $F$. Suppose $f(x)in F[x]$ and $f(x)$ is irreducible over $F$, $a$ is a zero of $f(x)$ in some extension field $E$ of $F$. $F(a)$ denote the smallest subfield of $E$ that contains both $F$ and $a$. It is well known that all such $F(a)$ is isomorphic to
$F(x)/<f(x)>$, and all splitting fields of $f(x)$ over $F$ are isomorphic. I am quite curious that is it the case that $F(x)/<f(x)>$ is isomorphic to a subfield of the splitting fields? Some enlightening would be very helpful!
abstract-algebra
YouâÂÂre forgetting the assumption that $f$ is irreducible.
â egreg
Sep 1 at 7:13
Thank you, editing now.
â qiang heng
Sep 1 at 7:14
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$F$ is a field. $F[x]$ denotes the ring which consists of all the polynomials with coefficients in $F$. Suppose $f(x)in F[x]$ and $f(x)$ is irreducible over $F$, $a$ is a zero of $f(x)$ in some extension field $E$ of $F$. $F(a)$ denote the smallest subfield of $E$ that contains both $F$ and $a$. It is well known that all such $F(a)$ is isomorphic to
$F(x)/<f(x)>$, and all splitting fields of $f(x)$ over $F$ are isomorphic. I am quite curious that is it the case that $F(x)/<f(x)>$ is isomorphic to a subfield of the splitting fields? Some enlightening would be very helpful!
abstract-algebra
$F$ is a field. $F[x]$ denotes the ring which consists of all the polynomials with coefficients in $F$. Suppose $f(x)in F[x]$ and $f(x)$ is irreducible over $F$, $a$ is a zero of $f(x)$ in some extension field $E$ of $F$. $F(a)$ denote the smallest subfield of $E$ that contains both $F$ and $a$. It is well known that all such $F(a)$ is isomorphic to
$F(x)/<f(x)>$, and all splitting fields of $f(x)$ over $F$ are isomorphic. I am quite curious that is it the case that $F(x)/<f(x)>$ is isomorphic to a subfield of the splitting fields? Some enlightening would be very helpful!
abstract-algebra
abstract-algebra
edited Sep 1 at 7:25
asked Sep 1 at 7:12
qiang heng
17715
17715
YouâÂÂre forgetting the assumption that $f$ is irreducible.
â egreg
Sep 1 at 7:13
Thank you, editing now.
â qiang heng
Sep 1 at 7:14
add a comment |Â
YouâÂÂre forgetting the assumption that $f$ is irreducible.
â egreg
Sep 1 at 7:13
Thank you, editing now.
â qiang heng
Sep 1 at 7:14
YouâÂÂre forgetting the assumption that $f$ is irreducible.
â egreg
Sep 1 at 7:13
YouâÂÂre forgetting the assumption that $f$ is irreducible.
â egreg
Sep 1 at 7:13
Thank you, editing now.
â qiang heng
Sep 1 at 7:14
Thank you, editing now.
â qiang heng
Sep 1 at 7:14
add a comment |Â
1 Answer
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The splitting field $K$ of $f$ by definition contains a root $b$ of $f$ (actually all roots). Since $F(b)cong F[x]/langle f(x)rangle$, youâÂÂre done.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The splitting field $K$ of $f$ by definition contains a root $b$ of $f$ (actually all roots). Since $F(b)cong F[x]/langle f(x)rangle$, youâÂÂre done.
add a comment |Â
up vote
1
down vote
accepted
The splitting field $K$ of $f$ by definition contains a root $b$ of $f$ (actually all roots). Since $F(b)cong F[x]/langle f(x)rangle$, youâÂÂre done.
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The splitting field $K$ of $f$ by definition contains a root $b$ of $f$ (actually all roots). Since $F(b)cong F[x]/langle f(x)rangle$, youâÂÂre done.
The splitting field $K$ of $f$ by definition contains a root $b$ of $f$ (actually all roots). Since $F(b)cong F[x]/langle f(x)rangle$, youâÂÂre done.
answered Sep 1 at 7:28
egreg
166k1180189
166k1180189
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YouâÂÂre forgetting the assumption that $f$ is irreducible.
â egreg
Sep 1 at 7:13
Thank you, editing now.
â qiang heng
Sep 1 at 7:14