$F[x]$ is a domain?
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Given a field $F$, the ring of polynomials $F[x]$ is a popular example of ED. However, it seems that $F[x]$ does not even have to be a domain.
For example, take $F=mathbbF_2$. Then
$$x(x-1)equiv0$$
since LHS and RHS always have the same output for whatever input $x$. But of course we don't have $xequiv0$ nor $x-1equiv0$, which shows that zero divisor exists. How can we explain this?
abstract-algebra polynomials ring-theory
asked Sep 9 at 9:21
Easy
3,468817
add a comment |Â
up vote
3
down vote
favorite
Given a field $F$, the ring of polynomials $F[x]$ is a popular example of ED. However, it seems that $F[x]$ does not even have to be a domain.
For example, take $F=mathbbF_2$. Then
$$x(x-1)equiv0$$
since LHS and RHS always have the same output for whatever input $x$. But of course we don't have $xequiv0$ nor $x-1equiv0$, which shows that zero divisor exists. How can we explain this?
abstract-algebra polynomials ring-theory
asked Sep 9 at 9:21
Easy
3,468817
3
For an infinite field, there is a one-to-one correspondence between polynomials and polynomial functions. For finite fields, this does not hold.
– Wuestenfux
Sep 9 at 9:26
@Wuestenfux If so, then why people still regard $F[x]$ as a ED, as it is even NOT a domain! Or they consider this in a different sense?
– Easy
Sep 9 at 9:28
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Given a field $F$, the ring of polynomials $F[x]$ is a popular example of ED. However, it seems that $F[x]$ does not even have to be a domain.
For example, take $F=mathbbF_2$. Then
$$x(x-1)equiv0$$
since LHS and RHS always have the same output for whatever input $x$. But of course we don't have $xequiv0$ nor $x-1equiv0$, which shows that zero divisor exists. How can we explain this?
abstract-algebra polynomials ring-theory
asked Sep 9 at 9:21
Easy
3,468817
Given a field $F$, the ring of polynomials $F[x]$ is a popular example of ED. However, it seems that $F[x]$ does not even have to be a domain.
For example, take $F=mathbbF_2$. Then
$$x(x-1)equiv0$$
since LHS and RHS always have the same output for whatever input $x$. But of course we don't have $xequiv0$ nor $x-1equiv0$, which shows that zero divisor exists. How can we explain this?
abstract-algebra polynomials ring-theory
abstract-algebra polynomials ring-theory
asked Sep 9 at 9:21
Easy
3,468817
asked Sep 9 at 9:21
Easy
3,468817
asked Sep 9 at 9:21
Easy
3,468817
asked Sep 9 at 9:21
Easy
3,468817
asked Sep 9 at 9:21
Easy
3,468817
3,468817
3
For an infinite field, there is a one-to-one correspondence between polynomials and polynomial functions. For finite fields, this does not hold.
– Wuestenfux
Sep 9 at 9:26
@Wuestenfux If so, then why people still regard $F[x]$ as a ED, as it is even NOT a domain! Or they consider this in a different sense?
– Easy
Sep 9 at 9:28
add a comment |Â
3
For an infinite field, there is a one-to-one correspondence between polynomials and polynomial functions. For finite fields, this does not hold.
– Wuestenfux
Sep 9 at 9:26
@Wuestenfux If so, then why people still regard $F[x]$ as a ED, as it is even NOT a domain! Or they consider this in a different sense?
– Easy
Sep 9 at 9:28
3
3
For an infinite field, there is a one-to-one correspondence between polynomials and polynomial functions. For finite fields, this does not hold.
– Wuestenfux
Sep 9 at 9:26
For an infinite field, there is a one-to-one correspondence between polynomials and polynomial functions. For finite fields, this does not hold.
– Wuestenfux
Sep 9 at 9:26
@Wuestenfux If so, then why people still regard $F[x]$ as a ED, as it is even NOT a domain! Or they consider this in a different sense?
– Easy
Sep 9 at 9:28
@Wuestenfux If so, then why people still regard $F[x]$ as a ED, as it is even NOT a domain! Or they consider this in a different sense?
– Easy
Sep 9 at 9:28
add a comment |Â
1 Answer
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5
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The polynomial $x^2-x$ is identically $0$ on $F$, but it is not equal to $0$ in $F[x]$. Your issue is confusing equality of functions with equality of polynomials. Remember that two polynomials are equal if and only if all of their coefficients are equal. Whereas, two functions on the same sets are equal if and only if they agree everywhere.
answered Sep 9 at 9:28
eloiPrime
1,617519
Yes, we identify polynomials by their expressions as we define so. But in set of $mathbbF_2[x]$, do we really distinguish $x^2-x$ and $0$?
– Easy
Sep 9 at 9:36
We distinguish between $x^2−x$ and $0$ in $F[x]$ for precisely the same reason that we distinguish between $(0,0,0,0,0,…)$ and $(0,−1,1,0,0,…)$ in the set of all sequences of members of $F$.
– eloiPrime
Sep 9 at 9:50
One can check that all results of $F[x]$ about ED, PID, UFD still hold without this definition. I guess people make such a definition is because they can directly apply the results of ED, PID, UFD etc. Otherwise, this sounds very strange as it is just like $x^2-2x+1$ and $(x-1)^2$ in $mathbbQ[x]$ we call them different.
– Easy
Sep 9 at 10:05
@Easy: A polynomial in $x$ over $mathbbF$ is formally defined as a finitely-supported sequence $left(a_0, a_1, a_2, ldotsright)$ of elements of $mathbbF$; those elements are called its coefficients. Then, addition and multiplication and $mathbbF$-scalar multiplication of polynomials are defined. Then, $x$ is defined as the polynomial corresponding to the sequence $left(0, 1, 0, 0, 0, ldotsright)$ with exactly one $1$ between all the zeroes. Finally, it is ...
– darij grinberg
Sep 9 at 14:03
... shown that each polynomial $left(a_0, a_1, a_2, ldotsright)$ is equal to the sum $a_0 x^0 + a_1 x^1 + a_2 x^2 + cdots$ (which is a well-defined sum because all but finitely many $a_i$ are zero). This is the standard approach to defining polynomials, although it tends to fall through the cracks in undergraduate education. Both $x^2 - 2x + 1$ and $left(x-1right)^2$ are then equal to the polynomial $left(1,-2,1,0,0,0,ldotsright)$, by the rules for adding, subtracting and multiplying polynomials.
– darij grinberg
Sep 9 at 14:04
 |Â
show 15 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The polynomial $x^2-x$ is identically $0$ on $F$, but it is not equal to $0$ in $F[x]$. Your issue is confusing equality of functions with equality of polynomials. Remember that two polynomials are equal if and only if all of their coefficients are equal. Whereas, two functions on the same sets are equal if and only if they agree everywhere.
answered Sep 9 at 9:28
eloiPrime
1,617519
Yes, we identify polynomials by their expressions as we define so. But in set of $mathbbF_2[x]$, do we really distinguish $x^2-x$ and $0$?
– Easy
Sep 9 at 9:36
We distinguish between $x^2−x$ and $0$ in $F[x]$ for precisely the same reason that we distinguish between $(0,0,0,0,0,…)$ and $(0,−1,1,0,0,…)$ in the set of all sequences of members of $F$.
– eloiPrime
Sep 9 at 9:50
One can check that all results of $F[x]$ about ED, PID, UFD still hold without this definition. I guess people make such a definition is because they can directly apply the results of ED, PID, UFD etc. Otherwise, this sounds very strange as it is just like $x^2-2x+1$ and $(x-1)^2$ in $mathbbQ[x]$ we call them different.
– Easy
Sep 9 at 10:05
@Easy: A polynomial in $x$ over $mathbbF$ is formally defined as a finitely-supported sequence $left(a_0, a_1, a_2, ldotsright)$ of elements of $mathbbF$; those elements are called its coefficients. Then, addition and multiplication and $mathbbF$-scalar multiplication of polynomials are defined. Then, $x$ is defined as the polynomial corresponding to the sequence $left(0, 1, 0, 0, 0, ldotsright)$ with exactly one $1$ between all the zeroes. Finally, it is ...
– darij grinberg
Sep 9 at 14:03
... shown that each polynomial $left(a_0, a_1, a_2, ldotsright)$ is equal to the sum $a_0 x^0 + a_1 x^1 + a_2 x^2 + cdots$ (which is a well-defined sum because all but finitely many $a_i$ are zero). This is the standard approach to defining polynomials, although it tends to fall through the cracks in undergraduate education. Both $x^2 - 2x + 1$ and $left(x-1right)^2$ are then equal to the polynomial $left(1,-2,1,0,0,0,ldotsright)$, by the rules for adding, subtracting and multiplying polynomials.
– darij grinberg
Sep 9 at 14:04
 |Â
show 15 more comments
up vote
5
down vote
accepted
The polynomial $x^2-x$ is identically $0$ on $F$, but it is not equal to $0$ in $F[x]$. Your issue is confusing equality of functions with equality of polynomials. Remember that two polynomials are equal if and only if all of their coefficients are equal. Whereas, two functions on the same sets are equal if and only if they agree everywhere.
answered Sep 9 at 9:28
eloiPrime
1,617519
Yes, we identify polynomials by their expressions as we define so. But in set of $mathbbF_2[x]$, do we really distinguish $x^2-x$ and $0$?
– Easy
Sep 9 at 9:36
We distinguish between $x^2−x$ and $0$ in $F[x]$ for precisely the same reason that we distinguish between $(0,0,0,0,0,…)$ and $(0,−1,1,0,0,…)$ in the set of all sequences of members of $F$.
– eloiPrime
Sep 9 at 9:50
One can check that all results of $F[x]$ about ED, PID, UFD still hold without this definition. I guess people make such a definition is because they can directly apply the results of ED, PID, UFD etc. Otherwise, this sounds very strange as it is just like $x^2-2x+1$ and $(x-1)^2$ in $mathbbQ[x]$ we call them different.
– Easy
Sep 9 at 10:05
@Easy: A polynomial in $x$ over $mathbbF$ is formally defined as a finitely-supported sequence $left(a_0, a_1, a_2, ldotsright)$ of elements of $mathbbF$; those elements are called its coefficients. Then, addition and multiplication and $mathbbF$-scalar multiplication of polynomials are defined. Then, $x$ is defined as the polynomial corresponding to the sequence $left(0, 1, 0, 0, 0, ldotsright)$ with exactly one $1$ between all the zeroes. Finally, it is ...
– darij grinberg
Sep 9 at 14:03
... shown that each polynomial $left(a_0, a_1, a_2, ldotsright)$ is equal to the sum $a_0 x^0 + a_1 x^1 + a_2 x^2 + cdots$ (which is a well-defined sum because all but finitely many $a_i$ are zero). This is the standard approach to defining polynomials, although it tends to fall through the cracks in undergraduate education. Both $x^2 - 2x + 1$ and $left(x-1right)^2$ are then equal to the polynomial $left(1,-2,1,0,0,0,ldotsright)$, by the rules for adding, subtracting and multiplying polynomials.
– darij grinberg
Sep 9 at 14:04
 |Â
show 15 more comments
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The polynomial $x^2-x$ is identically $0$ on $F$, but it is not equal to $0$ in $F[x]$. Your issue is confusing equality of functions with equality of polynomials. Remember that two polynomials are equal if and only if all of their coefficients are equal. Whereas, two functions on the same sets are equal if and only if they agree everywhere.
answered Sep 9 at 9:28
eloiPrime
1,617519
The polynomial $x^2-x$ is identically $0$ on $F$, but it is not equal to $0$ in $F[x]$. Your issue is confusing equality of functions with equality of polynomials. Remember that two polynomials are equal if and only if all of their coefficients are equal. Whereas, two functions on the same sets are equal if and only if they agree everywhere.
answered Sep 9 at 9:28
eloiPrime
1,617519
edited Sep 9 at 9:34
answered Sep 9 at 9:28
eloiPrime
1,617519
answered Sep 9 at 9:28
eloiPrime
1,617519
answered Sep 9 at 9:28
eloiPrime
1,617519
1,617519
Yes, we identify polynomials by their expressions as we define so. But in set of $mathbbF_2[x]$, do we really distinguish $x^2-x$ and $0$?
– Easy
Sep 9 at 9:36
We distinguish between $x^2−x$ and $0$ in $F[x]$ for precisely the same reason that we distinguish between $(0,0,0,0,0,…)$ and $(0,−1,1,0,0,…)$ in the set of all sequences of members of $F$.
– eloiPrime
Sep 9 at 9:50
One can check that all results of $F[x]$ about ED, PID, UFD still hold without this definition. I guess people make such a definition is because they can directly apply the results of ED, PID, UFD etc. Otherwise, this sounds very strange as it is just like $x^2-2x+1$ and $(x-1)^2$ in $mathbbQ[x]$ we call them different.
– Easy
Sep 9 at 10:05
@Easy: A polynomial in $x$ over $mathbbF$ is formally defined as a finitely-supported sequence $left(a_0, a_1, a_2, ldotsright)$ of elements of $mathbbF$; those elements are called its coefficients. Then, addition and multiplication and $mathbbF$-scalar multiplication of polynomials are defined. Then, $x$ is defined as the polynomial corresponding to the sequence $left(0, 1, 0, 0, 0, ldotsright)$ with exactly one $1$ between all the zeroes. Finally, it is ...
– darij grinberg
Sep 9 at 14:03
... shown that each polynomial $left(a_0, a_1, a_2, ldotsright)$ is equal to the sum $a_0 x^0 + a_1 x^1 + a_2 x^2 + cdots$ (which is a well-defined sum because all but finitely many $a_i$ are zero). This is the standard approach to defining polynomials, although it tends to fall through the cracks in undergraduate education. Both $x^2 - 2x + 1$ and $left(x-1right)^2$ are then equal to the polynomial $left(1,-2,1,0,0,0,ldotsright)$, by the rules for adding, subtracting and multiplying polynomials.
– darij grinberg
Sep 9 at 14:04
 |Â
show 15 more comments
Yes, we identify polynomials by their expressions as we define so. But in set of $mathbbF_2[x]$, do we really distinguish $x^2-x$ and $0$?
– Easy
Sep 9 at 9:36
We distinguish between $x^2−x$ and $0$ in $F[x]$ for precisely the same reason that we distinguish between $(0,0,0,0,0,…)$ and $(0,−1,1,0,0,…)$ in the set of all sequences of members of $F$.
– eloiPrime
Sep 9 at 9:50
One can check that all results of $F[x]$ about ED, PID, UFD still hold without this definition. I guess people make such a definition is because they can directly apply the results of ED, PID, UFD etc. Otherwise, this sounds very strange as it is just like $x^2-2x+1$ and $(x-1)^2$ in $mathbbQ[x]$ we call them different.
– Easy
Sep 9 at 10:05
@Easy: A polynomial in $x$ over $mathbbF$ is formally defined as a finitely-supported sequence $left(a_0, a_1, a_2, ldotsright)$ of elements of $mathbbF$; those elements are called its coefficients. Then, addition and multiplication and $mathbbF$-scalar multiplication of polynomials are defined. Then, $x$ is defined as the polynomial corresponding to the sequence $left(0, 1, 0, 0, 0, ldotsright)$ with exactly one $1$ between all the zeroes. Finally, it is ...
– darij grinberg
Sep 9 at 14:03
... shown that each polynomial $left(a_0, a_1, a_2, ldotsright)$ is equal to the sum $a_0 x^0 + a_1 x^1 + a_2 x^2 + cdots$ (which is a well-defined sum because all but finitely many $a_i$ are zero). This is the standard approach to defining polynomials, although it tends to fall through the cracks in undergraduate education. Both $x^2 - 2x + 1$ and $left(x-1right)^2$ are then equal to the polynomial $left(1,-2,1,0,0,0,ldotsright)$, by the rules for adding, subtracting and multiplying polynomials.
– darij grinberg
Sep 9 at 14:04
Yes, we identify polynomials by their expressions as we define so. But in set of $mathbbF_2[x]$, do we really distinguish $x^2-x$ and $0$?
– Easy
Sep 9 at 9:36
Yes, we identify polynomials by their expressions as we define so. But in set of $mathbbF_2[x]$, do we really distinguish $x^2-x$ and $0$?
– Easy
Sep 9 at 9:36
We distinguish between $x^2−x$ and $0$ in $F[x]$ for precisely the same reason that we distinguish between $(0,0,0,0,0,…)$ and $(0,−1,1,0,0,…)$ in the set of all sequences of members of $F$.
– eloiPrime
Sep 9 at 9:50
We distinguish between $x^2−x$ and $0$ in $F[x]$ for precisely the same reason that we distinguish between $(0,0,0,0,0,…)$ and $(0,−1,1,0,0,…)$ in the set of all sequences of members of $F$.
– eloiPrime
Sep 9 at 9:50
One can check that all results of $F[x]$ about ED, PID, UFD still hold without this definition. I guess people make such a definition is because they can directly apply the results of ED, PID, UFD etc. Otherwise, this sounds very strange as it is just like $x^2-2x+1$ and $(x-1)^2$ in $mathbbQ[x]$ we call them different.
– Easy
Sep 9 at 10:05
One can check that all results of $F[x]$ about ED, PID, UFD still hold without this definition. I guess people make such a definition is because they can directly apply the results of ED, PID, UFD etc. Otherwise, this sounds very strange as it is just like $x^2-2x+1$ and $(x-1)^2$ in $mathbbQ[x]$ we call them different.
– Easy
Sep 9 at 10:05
@Easy: A polynomial in $x$ over $mathbbF$ is formally defined as a finitely-supported sequence $left(a_0, a_1, a_2, ldotsright)$ of elements of $mathbbF$; those elements are called its coefficients. Then, addition and multiplication and $mathbbF$-scalar multiplication of polynomials are defined. Then, $x$ is defined as the polynomial corresponding to the sequence $left(0, 1, 0, 0, 0, ldotsright)$ with exactly one $1$ between all the zeroes. Finally, it is ...
– darij grinberg
Sep 9 at 14:03
@Easy: A polynomial in $x$ over $mathbbF$ is formally defined as a finitely-supported sequence $left(a_0, a_1, a_2, ldotsright)$ of elements of $mathbbF$; those elements are called its coefficients. Then, addition and multiplication and $mathbbF$-scalar multiplication of polynomials are defined. Then, $x$ is defined as the polynomial corresponding to the sequence $left(0, 1, 0, 0, 0, ldotsright)$ with exactly one $1$ between all the zeroes. Finally, it is ...
– darij grinberg
Sep 9 at 14:03
... shown that each polynomial $left(a_0, a_1, a_2, ldotsright)$ is equal to the sum $a_0 x^0 + a_1 x^1 + a_2 x^2 + cdots$ (which is a well-defined sum because all but finitely many $a_i$ are zero). This is the standard approach to defining polynomials, although it tends to fall through the cracks in undergraduate education. Both $x^2 - 2x + 1$ and $left(x-1right)^2$ are then equal to the polynomial $left(1,-2,1,0,0,0,ldotsright)$, by the rules for adding, subtracting and multiplying polynomials.
– darij grinberg
Sep 9 at 14:04
... shown that each polynomial $left(a_0, a_1, a_2, ldotsright)$ is equal to the sum $a_0 x^0 + a_1 x^1 + a_2 x^2 + cdots$ (which is a well-defined sum because all but finitely many $a_i$ are zero). This is the standard approach to defining polynomials, although it tends to fall through the cracks in undergraduate education. Both $x^2 - 2x + 1$ and $left(x-1right)^2$ are then equal to the polynomial $left(1,-2,1,0,0,0,ldotsright)$, by the rules for adding, subtracting and multiplying polynomials.
– darij grinberg
Sep 9 at 14:04
 |Â
show 15 more comments
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3
For an infinite field, there is a one-to-one correspondence between polynomials and polynomial functions. For finite fields, this does not hold.
– Wuestenfux
Sep 9 at 9:26
@Wuestenfux If so, then why people still regard $F[x]$ as a ED, as it is even NOT a domain! Or they consider this in a different sense?
– Easy
Sep 9 at 9:28