Notation regarding polynomial ideals
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Suppose $F[x]$ is a field, and $p(x) in F[x]$ be a polynomial. We want to quotient $F[x]$ with the ideal generated by $p(x)$.
While the commonly used notation to denote that $F[x]/p(x)$, isn't the notation kinda misleading because $p(x)$ is not an ideal ?
I think $F[x]/p(x)F[x]$ should be the correct notation since $p(x)F[x]$ is an ideal.
abstract-algebra notation
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Suppose $F[x]$ is a field, and $p(x) in F[x]$ be a polynomial. We want to quotient $F[x]$ with the ideal generated by $p(x)$.
While the commonly used notation to denote that $F[x]/p(x)$, isn't the notation kinda misleading because $p(x)$ is not an ideal ?
I think $F[x]/p(x)F[x]$ should be the correct notation since $p(x)F[x]$ is an ideal.
abstract-algebra notation
5
You also can use $F[x]/bigl(p(x)bigr)$.
– Bernard
Jul 25 at 16:29
The same problem ("kinda misleading"?) is with $BbbZ/n$ or $BbbZ/(n)$ or $BbbZ/nBbbZ$. Even worse, $BbbZ_p$, which are also $p$-adic integers.
– Dietrich Burde
Jul 25 at 19:38
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $F[x]$ is a field, and $p(x) in F[x]$ be a polynomial. We want to quotient $F[x]$ with the ideal generated by $p(x)$.
While the commonly used notation to denote that $F[x]/p(x)$, isn't the notation kinda misleading because $p(x)$ is not an ideal ?
I think $F[x]/p(x)F[x]$ should be the correct notation since $p(x)F[x]$ is an ideal.
abstract-algebra notation
Suppose $F[x]$ is a field, and $p(x) in F[x]$ be a polynomial. We want to quotient $F[x]$ with the ideal generated by $p(x)$.
While the commonly used notation to denote that $F[x]/p(x)$, isn't the notation kinda misleading because $p(x)$ is not an ideal ?
I think $F[x]/p(x)F[x]$ should be the correct notation since $p(x)F[x]$ is an ideal.
abstract-algebra notation
asked Jul 25 at 16:26


alxchen
402220
402220
5
You also can use $F[x]/bigl(p(x)bigr)$.
– Bernard
Jul 25 at 16:29
The same problem ("kinda misleading"?) is with $BbbZ/n$ or $BbbZ/(n)$ or $BbbZ/nBbbZ$. Even worse, $BbbZ_p$, which are also $p$-adic integers.
– Dietrich Burde
Jul 25 at 19:38
add a comment |Â
5
You also can use $F[x]/bigl(p(x)bigr)$.
– Bernard
Jul 25 at 16:29
The same problem ("kinda misleading"?) is with $BbbZ/n$ or $BbbZ/(n)$ or $BbbZ/nBbbZ$. Even worse, $BbbZ_p$, which are also $p$-adic integers.
– Dietrich Burde
Jul 25 at 19:38
5
5
You also can use $F[x]/bigl(p(x)bigr)$.
– Bernard
Jul 25 at 16:29
You also can use $F[x]/bigl(p(x)bigr)$.
– Bernard
Jul 25 at 16:29
The same problem ("kinda misleading"?) is with $BbbZ/n$ or $BbbZ/(n)$ or $BbbZ/nBbbZ$. Even worse, $BbbZ_p$, which are also $p$-adic integers.
– Dietrich Burde
Jul 25 at 19:38
The same problem ("kinda misleading"?) is with $BbbZ/n$ or $BbbZ/(n)$ or $BbbZ/nBbbZ$. Even worse, $BbbZ_p$, which are also $p$-adic integers.
– Dietrich Burde
Jul 25 at 19:38
add a comment |Â
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5
You also can use $F[x]/bigl(p(x)bigr)$.
– Bernard
Jul 25 at 16:29
The same problem ("kinda misleading"?) is with $BbbZ/n$ or $BbbZ/(n)$ or $BbbZ/nBbbZ$. Even worse, $BbbZ_p$, which are also $p$-adic integers.
– Dietrich Burde
Jul 25 at 19:38