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What does $p(theta)$ mean in this theorem?
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Theorem Let $p(x) in F[x]$ be an irreducible polynomial of degree $n$. Then $K=F[x]/(p(x))$ is a field in which $theta=bar x=x+(p(x))$ that satisfies $p(theta)=0$. Furthermore, the elements $1, theta, theta^2, ..., theta^n-1$ form a basis of $K$ as a vector space over $F$. So $[K:F]=n$ and $K=F[theta]$.
This is a theorem from Lovett. "Abstract Algebra." p. 325. (Chapter 7. Field Extensions.) I know that $(p(x))$ means the smallest ideal in $F[x]$ containing $p(x)$. But if $theta=x+(p(x))$, then $p(theta)$ is $p(x+(p(x)))$, and I have no idea what this means. Please help me with this.
abstract-algebra field-theory
asked Jul 30 at 8:55
zxcv
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Theorem Let $p(x) in F[x]$ be an irreducible polynomial of degree $n$. Then $K=F[x]/(p(x))$ is a field in which $theta=bar x=x+(p(x))$ that satisfies $p(theta)=0$. Furthermore, the elements $1, theta, theta^2, ..., theta^n-1$ form a basis of $K$ as a vector space over $F$. So $[K:F]=n$ and $K=F[theta]$.
This is a theorem from Lovett. "Abstract Algebra." p. 325. (Chapter 7. Field Extensions.) I know that $(p(x))$ means the smallest ideal in $F[x]$ containing $p(x)$. But if $theta=x+(p(x))$, then $p(theta)$ is $p(x+(p(x)))$, and I have no idea what this means. Please help me with this.
abstract-algebra field-theory
asked Jul 30 at 8:55
zxcv
376
Do you know what $F[x]/p(x)$ means?
– Qiaochu Yuan
Jul 31 at 2:13
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Theorem Let $p(x) in F[x]$ be an irreducible polynomial of degree $n$. Then $K=F[x]/(p(x))$ is a field in which $theta=bar x=x+(p(x))$ that satisfies $p(theta)=0$. Furthermore, the elements $1, theta, theta^2, ..., theta^n-1$ form a basis of $K$ as a vector space over $F$. So $[K:F]=n$ and $K=F[theta]$.
This is a theorem from Lovett. "Abstract Algebra." p. 325. (Chapter 7. Field Extensions.) I know that $(p(x))$ means the smallest ideal in $F[x]$ containing $p(x)$. But if $theta=x+(p(x))$, then $p(theta)$ is $p(x+(p(x)))$, and I have no idea what this means. Please help me with this.
abstract-algebra field-theory
asked Jul 30 at 8:55
zxcv
376
Theorem Let $p(x) in F[x]$ be an irreducible polynomial of degree $n$. Then $K=F[x]/(p(x))$ is a field in which $theta=bar x=x+(p(x))$ that satisfies $p(theta)=0$. Furthermore, the elements $1, theta, theta^2, ..., theta^n-1$ form a basis of $K$ as a vector space over $F$. So $[K:F]=n$ and $K=F[theta]$.
This is a theorem from Lovett. "Abstract Algebra." p. 325. (Chapter 7. Field Extensions.) I know that $(p(x))$ means the smallest ideal in $F[x]$ containing $p(x)$. But if $theta=x+(p(x))$, then $p(theta)$ is $p(x+(p(x)))$, and I have no idea what this means. Please help me with this.
abstract-algebra field-theory
asked Jul 30 at 8:55
zxcv
376
edited Jul 31 at 2:11
asked Jul 30 at 8:55
zxcv
376
asked Jul 30 at 8:55
zxcv
376
asked Jul 30 at 8:55
zxcv
376
376
Do you know what $F[x]/p(x)$ means?
– Qiaochu Yuan
Jul 31 at 2:13
add a comment |Â
Do you know what $F[x]/p(x)$ means?
– Qiaochu Yuan
Jul 31 at 2:13
Do you know what $F[x]/p(x)$ means?
– Qiaochu Yuan
Jul 31 at 2:13
Do you know what $F[x]/p(x)$ means?
– Qiaochu Yuan
Jul 31 at 2:13
add a comment |Â
3 Answers
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There is an identification being made between the field $F$ and the field $tildeF := a + (p(x)) : a in F $, which is a subfield of $K$. We have that $F cong tildeF$ under the natural isomorphism $a mapsto a + (p(x))$.
(Henceforth, let $I = (p(x))$ for shorthand.)
This isomorphism induces an isomorphism between $F[x]$ and $tildeF[x]$, where the polynomial $$f(x) = a_0 + a_1 x + dots + a_n x^n in F[x]$$ is mapped to $$tildef(x) = (a_0 + I) + (a_1 + I)x + dots + (a_n + I)x^n in tildeF[x].$$
Rather than introducing many cumbersome notations, the author chooses to call the field $tildeF$ as $F$ itself, and the polynomial $tildef(x)$ as $f(x)$ itself.
For the sake of clarity, at least initially, let us retain the earlier notation with the $sim$'s. Now, it can be easily checked that for every $tildealpha = alpha + I in K$, $$tildef(tildealpha) = f(alpha) + I.$$
So, what does $p(theta)$ mean for $theta = x + I$? Clearly, the author is actually talking about $tildep(x + I)$. And, we have that $$tildep(x + I) = p(x) + I = 0 + I.$$
Thus, "$p(theta) = 0$" when making the appropriate identifications.
answered Jul 30 at 9:17
Brahadeesh
3,34331246
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Suppose $p(x)=sum_j=0^na_j x^j$ where the $a_jin K$.
You need to use the following facts about operating with cosets:
$$ overlinep_1(x)+p_2(x)=overlinep_1(x)+overlinep_2(x)$$
and
$$ overlinep_1(x)cdot p_2(x)=overlinep_1(x)cdotoverlinep_2(x)$$
and
$$ overlineccdot p_1(x)=ccdot overlinep_1(x).$$
Then calculate:
$$p(theta)=p(barx)=sum_j=0^na_j barx^j=overlinesum_j=0^na_j x^j=overlinep(x)=bar0.$$
answered Jul 30 at 9:10
ancientmathematician
4,0681312
add a comment |Â
up vote
0
down vote
Elements of the quotient set $K$ are equivalence classes by the relation $asim bLeftrightarrow a-bin(p(x))$. Polynomials conserve this equivalence, so if $asim b$ so $q(a) sim q(b)$ and we can then talk about $q(bar a)$, with $bar a$ is the equivalence class of $a$ and $b$.
As $p(x)-0in (p(x))$ so $p(bar x)=overlinep(x)=0$ that what means $p(theta)=0$.
answered Jul 30 at 9:13
Ahmed Lazhar
618
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
There is an identification being made between the field $F$ and the field $tildeF := a + (p(x)) : a in F $, which is a subfield of $K$. We have that $F cong tildeF$ under the natural isomorphism $a mapsto a + (p(x))$.
(Henceforth, let $I = (p(x))$ for shorthand.)
This isomorphism induces an isomorphism between $F[x]$ and $tildeF[x]$, where the polynomial $$f(x) = a_0 + a_1 x + dots + a_n x^n in F[x]$$ is mapped to $$tildef(x) = (a_0 + I) + (a_1 + I)x + dots + (a_n + I)x^n in tildeF[x].$$
Rather than introducing many cumbersome notations, the author chooses to call the field $tildeF$ as $F$ itself, and the polynomial $tildef(x)$ as $f(x)$ itself.
For the sake of clarity, at least initially, let us retain the earlier notation with the $sim$'s. Now, it can be easily checked that for every $tildealpha = alpha + I in K$, $$tildef(tildealpha) = f(alpha) + I.$$
So, what does $p(theta)$ mean for $theta = x + I$? Clearly, the author is actually talking about $tildep(x + I)$. And, we have that $$tildep(x + I) = p(x) + I = 0 + I.$$
Thus, "$p(theta) = 0$" when making the appropriate identifications.
answered Jul 30 at 9:17
Brahadeesh
3,34331246
add a comment |Â
up vote
0
down vote
accepted
There is an identification being made between the field $F$ and the field $tildeF := a + (p(x)) : a in F $, which is a subfield of $K$. We have that $F cong tildeF$ under the natural isomorphism $a mapsto a + (p(x))$.
(Henceforth, let $I = (p(x))$ for shorthand.)
This isomorphism induces an isomorphism between $F[x]$ and $tildeF[x]$, where the polynomial $$f(x) = a_0 + a_1 x + dots + a_n x^n in F[x]$$ is mapped to $$tildef(x) = (a_0 + I) + (a_1 + I)x + dots + (a_n + I)x^n in tildeF[x].$$
Rather than introducing many cumbersome notations, the author chooses to call the field $tildeF$ as $F$ itself, and the polynomial $tildef(x)$ as $f(x)$ itself.
For the sake of clarity, at least initially, let us retain the earlier notation with the $sim$'s. Now, it can be easily checked that for every $tildealpha = alpha + I in K$, $$tildef(tildealpha) = f(alpha) + I.$$
So, what does $p(theta)$ mean for $theta = x + I$? Clearly, the author is actually talking about $tildep(x + I)$. And, we have that $$tildep(x + I) = p(x) + I = 0 + I.$$
Thus, "$p(theta) = 0$" when making the appropriate identifications.
answered Jul 30 at 9:17
Brahadeesh
3,34331246
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
There is an identification being made between the field $F$ and the field $tildeF := a + (p(x)) : a in F $, which is a subfield of $K$. We have that $F cong tildeF$ under the natural isomorphism $a mapsto a + (p(x))$.
(Henceforth, let $I = (p(x))$ for shorthand.)
This isomorphism induces an isomorphism between $F[x]$ and $tildeF[x]$, where the polynomial $$f(x) = a_0 + a_1 x + dots + a_n x^n in F[x]$$ is mapped to $$tildef(x) = (a_0 + I) + (a_1 + I)x + dots + (a_n + I)x^n in tildeF[x].$$
Rather than introducing many cumbersome notations, the author chooses to call the field $tildeF$ as $F$ itself, and the polynomial $tildef(x)$ as $f(x)$ itself.
For the sake of clarity, at least initially, let us retain the earlier notation with the $sim$'s. Now, it can be easily checked that for every $tildealpha = alpha + I in K$, $$tildef(tildealpha) = f(alpha) + I.$$
So, what does $p(theta)$ mean for $theta = x + I$? Clearly, the author is actually talking about $tildep(x + I)$. And, we have that $$tildep(x + I) = p(x) + I = 0 + I.$$
Thus, "$p(theta) = 0$" when making the appropriate identifications.
answered Jul 30 at 9:17
Brahadeesh
3,34331246
There is an identification being made between the field $F$ and the field $tildeF := a + (p(x)) : a in F $, which is a subfield of $K$. We have that $F cong tildeF$ under the natural isomorphism $a mapsto a + (p(x))$.
(Henceforth, let $I = (p(x))$ for shorthand.)
This isomorphism induces an isomorphism between $F[x]$ and $tildeF[x]$, where the polynomial $$f(x) = a_0 + a_1 x + dots + a_n x^n in F[x]$$ is mapped to $$tildef(x) = (a_0 + I) + (a_1 + I)x + dots + (a_n + I)x^n in tildeF[x].$$
Rather than introducing many cumbersome notations, the author chooses to call the field $tildeF$ as $F$ itself, and the polynomial $tildef(x)$ as $f(x)$ itself.
For the sake of clarity, at least initially, let us retain the earlier notation with the $sim$'s. Now, it can be easily checked that for every $tildealpha = alpha + I in K$, $$tildef(tildealpha) = f(alpha) + I.$$
So, what does $p(theta)$ mean for $theta = x + I$? Clearly, the author is actually talking about $tildep(x + I)$. And, we have that $$tildep(x + I) = p(x) + I = 0 + I.$$
Thus, "$p(theta) = 0$" when making the appropriate identifications.
answered Jul 30 at 9:17
Brahadeesh
3,34331246
edited Jul 30 at 10:50
answered Jul 30 at 9:17
Brahadeesh
3,34331246
answered Jul 30 at 9:17
Brahadeesh
3,34331246
answered Jul 30 at 9:17
Brahadeesh
3,34331246
3,34331246
add a comment |Â
add a comment |Â
up vote
0
down vote
Suppose $p(x)=sum_j=0^na_j x^j$ where the $a_jin K$.
You need to use the following facts about operating with cosets:
$$ overlinep_1(x)+p_2(x)=overlinep_1(x)+overlinep_2(x)$$
and
$$ overlinep_1(x)cdot p_2(x)=overlinep_1(x)cdotoverlinep_2(x)$$
and
$$ overlineccdot p_1(x)=ccdot overlinep_1(x).$$
Then calculate:
$$p(theta)=p(barx)=sum_j=0^na_j barx^j=overlinesum_j=0^na_j x^j=overlinep(x)=bar0.$$
answered Jul 30 at 9:10
ancientmathematician
4,0681312
add a comment |Â
up vote
0
down vote
Suppose $p(x)=sum_j=0^na_j x^j$ where the $a_jin K$.
You need to use the following facts about operating with cosets:
$$ overlinep_1(x)+p_2(x)=overlinep_1(x)+overlinep_2(x)$$
and
$$ overlinep_1(x)cdot p_2(x)=overlinep_1(x)cdotoverlinep_2(x)$$
and
$$ overlineccdot p_1(x)=ccdot overlinep_1(x).$$
Then calculate:
$$p(theta)=p(barx)=sum_j=0^na_j barx^j=overlinesum_j=0^na_j x^j=overlinep(x)=bar0.$$
answered Jul 30 at 9:10
ancientmathematician
4,0681312
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Suppose $p(x)=sum_j=0^na_j x^j$ where the $a_jin K$.
You need to use the following facts about operating with cosets:
$$ overlinep_1(x)+p_2(x)=overlinep_1(x)+overlinep_2(x)$$
and
$$ overlinep_1(x)cdot p_2(x)=overlinep_1(x)cdotoverlinep_2(x)$$
and
$$ overlineccdot p_1(x)=ccdot overlinep_1(x).$$
Then calculate:
$$p(theta)=p(barx)=sum_j=0^na_j barx^j=overlinesum_j=0^na_j x^j=overlinep(x)=bar0.$$
answered Jul 30 at 9:10
ancientmathematician
4,0681312
Suppose $p(x)=sum_j=0^na_j x^j$ where the $a_jin K$.
You need to use the following facts about operating with cosets:
$$ overlinep_1(x)+p_2(x)=overlinep_1(x)+overlinep_2(x)$$
and
$$ overlinep_1(x)cdot p_2(x)=overlinep_1(x)cdotoverlinep_2(x)$$
and
$$ overlineccdot p_1(x)=ccdot overlinep_1(x).$$
Then calculate:
$$p(theta)=p(barx)=sum_j=0^na_j barx^j=overlinesum_j=0^na_j x^j=overlinep(x)=bar0.$$
answered Jul 30 at 9:10
ancientmathematician
4,0681312
answered Jul 30 at 9:10
ancientmathematician
4,0681312
answered Jul 30 at 9:10
ancientmathematician
4,0681312
answered Jul 30 at 9:10
ancientmathematician
4,0681312
4,0681312
add a comment |Â
add a comment |Â
up vote
0
down vote
Elements of the quotient set $K$ are equivalence classes by the relation $asim bLeftrightarrow a-bin(p(x))$. Polynomials conserve this equivalence, so if $asim b$ so $q(a) sim q(b)$ and we can then talk about $q(bar a)$, with $bar a$ is the equivalence class of $a$ and $b$.
As $p(x)-0in (p(x))$ so $p(bar x)=overlinep(x)=0$ that what means $p(theta)=0$.
answered Jul 30 at 9:13
Ahmed Lazhar
618
add a comment |Â
up vote
0
down vote
Elements of the quotient set $K$ are equivalence classes by the relation $asim bLeftrightarrow a-bin(p(x))$. Polynomials conserve this equivalence, so if $asim b$ so $q(a) sim q(b)$ and we can then talk about $q(bar a)$, with $bar a$ is the equivalence class of $a$ and $b$.
As $p(x)-0in (p(x))$ so $p(bar x)=overlinep(x)=0$ that what means $p(theta)=0$.
answered Jul 30 at 9:13
Ahmed Lazhar
618
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Elements of the quotient set $K$ are equivalence classes by the relation $asim bLeftrightarrow a-bin(p(x))$. Polynomials conserve this equivalence, so if $asim b$ so $q(a) sim q(b)$ and we can then talk about $q(bar a)$, with $bar a$ is the equivalence class of $a$ and $b$.
As $p(x)-0in (p(x))$ so $p(bar x)=overlinep(x)=0$ that what means $p(theta)=0$.
answered Jul 30 at 9:13
Ahmed Lazhar
618
Elements of the quotient set $K$ are equivalence classes by the relation $asim bLeftrightarrow a-bin(p(x))$. Polynomials conserve this equivalence, so if $asim b$ so $q(a) sim q(b)$ and we can then talk about $q(bar a)$, with $bar a$ is the equivalence class of $a$ and $b$.
As $p(x)-0in (p(x))$ so $p(bar x)=overlinep(x)=0$ that what means $p(theta)=0$.
answered Jul 30 at 9:13
Ahmed Lazhar
618
edited Jul 30 at 9:19
answered Jul 30 at 9:13
Ahmed Lazhar
618
answered Jul 30 at 9:13
Ahmed Lazhar
618
answered Jul 30 at 9:13
Ahmed Lazhar
618
618
add a comment |Â
add a comment |Â
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Do you know what $F[x]/p(x)$ means?
– Qiaochu Yuan
Jul 31 at 2:13