Why does the integral domain “being trapped between a finite field extension” implies that it is a field?Linear map $f:Vrightarrow V$ injective $Longleftrightarrow$ surjectiveDoes this morphism necessarily give rise to a finite extension of residue fields?Points lying over a closed point in a separable extension of the base field are rationnalWhat kind of points are there in a finite type $k$-scheme?Quotient of ring is flat gives an identity of idealsWhen is the tensor product of a separable field extension with itself a domain?Why is the residue field of a $k$-scheme an extension of $k$?An example of normalization of schemeFinite fiber property and integral extension.Characterize integral extension of rings by maximal idealsMaximal ideal of $K[x_1,cdots,x_n]$ such that the quotient field equals to $K$
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Why does the integral domain "being trapped between a finite field extension" implies that it is a field?
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Why does the integral domain “being trapped between a finite field extension” implies that it is a field?
Linear map $f:Vrightarrow V$ injective $Longleftrightarrow$ surjectiveDoes this morphism necessarily give rise to a finite extension of residue fields?Points lying over a closed point in a separable extension of the base field are rationnalWhat kind of points are there in a finite type $k$-scheme?Quotient of ring is flat gives an identity of idealsWhen is the tensor product of a separable field extension with itself a domain?Why is the residue field of a $k$-scheme an extension of $k$?An example of normalization of schemeFinite fiber property and integral extension.Characterize integral extension of rings by maximal idealsMaximal ideal of $K[x_1,cdots,x_n]$ such that the quotient field equals to $K$
$begingroup$
The following is an exercise from Qing Liu's Algebraic Geometry and Arithmetic Curves.
Exercise 1.2.
Let ϕ : A → B be a homomorphism of finitely generated algebras over a field. Show that the image of a closed point under Spec ϕ is a closed point.
The following is the solution from Cihan Bahran. http://www-users.math.umn.edu/~bahra004/alg-geo/liu-soln.pdf.
Write $k$ for the underlying field. Let’s parse the statement. A closed point in $operatornameSpec B$ means a maximal ideal $n$ of $B$. And $operatornameSpec(ϕ)(n) = ϕ^−1(n)$. So we want to show that $p := ϕ−1(n)$ is a maximal ideal in $A$. First of all, $p$ is definitely a prime ideal of $A$ and $ϕ$ descends to an injective $k$-algebra homomorphism $ψ : A/p to B/n$. But the map $k to B/n$ defines a finite field extension of $k$ by Corollary 1.12. So the integral domain $A/p$ is trapped between a finite field extension. Such domains are necessarily fields, thus $p$ is maximal in $A$.
In the second last sentence, the writer says that the integral domain $A/p$ is trapped between a finite field extension. I don't exactly know what it means, but I think it means that there are two injective ring homomorphisms $f:kto A/p$ and $g:A/pto B/n$ such that $gcirc f$ makes $B/n$ a finite field extension of $k$. But why does it imply that $A/p$ is a field?
abstract-algebra algebraic-geometry commutative-algebra
$endgroup$
add a comment |
$begingroup$
The following is an exercise from Qing Liu's Algebraic Geometry and Arithmetic Curves.
Exercise 1.2.
Let ϕ : A → B be a homomorphism of finitely generated algebras over a field. Show that the image of a closed point under Spec ϕ is a closed point.
The following is the solution from Cihan Bahran. http://www-users.math.umn.edu/~bahra004/alg-geo/liu-soln.pdf.
Write $k$ for the underlying field. Let’s parse the statement. A closed point in $operatornameSpec B$ means a maximal ideal $n$ of $B$. And $operatornameSpec(ϕ)(n) = ϕ^−1(n)$. So we want to show that $p := ϕ−1(n)$ is a maximal ideal in $A$. First of all, $p$ is definitely a prime ideal of $A$ and $ϕ$ descends to an injective $k$-algebra homomorphism $ψ : A/p to B/n$. But the map $k to B/n$ defines a finite field extension of $k$ by Corollary 1.12. So the integral domain $A/p$ is trapped between a finite field extension. Such domains are necessarily fields, thus $p$ is maximal in $A$.
In the second last sentence, the writer says that the integral domain $A/p$ is trapped between a finite field extension. I don't exactly know what it means, but I think it means that there are two injective ring homomorphisms $f:kto A/p$ and $g:A/pto B/n$ such that $gcirc f$ makes $B/n$ a finite field extension of $k$. But why does it imply that $A/p$ is a field?
abstract-algebra algebraic-geometry commutative-algebra
$endgroup$
add a comment |
$begingroup$
The following is an exercise from Qing Liu's Algebraic Geometry and Arithmetic Curves.
Exercise 1.2.
Let ϕ : A → B be a homomorphism of finitely generated algebras over a field. Show that the image of a closed point under Spec ϕ is a closed point.
The following is the solution from Cihan Bahran. http://www-users.math.umn.edu/~bahra004/alg-geo/liu-soln.pdf.
Write $k$ for the underlying field. Let’s parse the statement. A closed point in $operatornameSpec B$ means a maximal ideal $n$ of $B$. And $operatornameSpec(ϕ)(n) = ϕ^−1(n)$. So we want to show that $p := ϕ−1(n)$ is a maximal ideal in $A$. First of all, $p$ is definitely a prime ideal of $A$ and $ϕ$ descends to an injective $k$-algebra homomorphism $ψ : A/p to B/n$. But the map $k to B/n$ defines a finite field extension of $k$ by Corollary 1.12. So the integral domain $A/p$ is trapped between a finite field extension. Such domains are necessarily fields, thus $p$ is maximal in $A$.
In the second last sentence, the writer says that the integral domain $A/p$ is trapped between a finite field extension. I don't exactly know what it means, but I think it means that there are two injective ring homomorphisms $f:kto A/p$ and $g:A/pto B/n$ such that $gcirc f$ makes $B/n$ a finite field extension of $k$. But why does it imply that $A/p$ is a field?
abstract-algebra algebraic-geometry commutative-algebra
$endgroup$
The following is an exercise from Qing Liu's Algebraic Geometry and Arithmetic Curves.
Exercise 1.2.
Let ϕ : A → B be a homomorphism of finitely generated algebras over a field. Show that the image of a closed point under Spec ϕ is a closed point.
The following is the solution from Cihan Bahran. http://www-users.math.umn.edu/~bahra004/alg-geo/liu-soln.pdf.
Write $k$ for the underlying field. Let’s parse the statement. A closed point in $operatornameSpec B$ means a maximal ideal $n$ of $B$. And $operatornameSpec(ϕ)(n) = ϕ^−1(n)$. So we want to show that $p := ϕ−1(n)$ is a maximal ideal in $A$. First of all, $p$ is definitely a prime ideal of $A$ and $ϕ$ descends to an injective $k$-algebra homomorphism $ψ : A/p to B/n$. But the map $k to B/n$ defines a finite field extension of $k$ by Corollary 1.12. So the integral domain $A/p$ is trapped between a finite field extension. Such domains are necessarily fields, thus $p$ is maximal in $A$.
In the second last sentence, the writer says that the integral domain $A/p$ is trapped between a finite field extension. I don't exactly know what it means, but I think it means that there are two injective ring homomorphisms $f:kto A/p$ and $g:A/pto B/n$ such that $gcirc f$ makes $B/n$ a finite field extension of $k$. But why does it imply that $A/p$ is a field?
abstract-algebra algebraic-geometry commutative-algebra
abstract-algebra algebraic-geometry commutative-algebra
asked 1 hour ago
zxcvzxcv
1609
1609
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Theorem 1. Let $K$ be a field. Let $R$ and $L$ be two $K$-algebras such that $L$ is a finite-dimensional $K$-vector space and $R$ is an integral domain. Let $g : R to L$ be an injective $K$-linear map. Then, $R$ is a field.
Proof of Theorem 1. Since the $K$-linear map $g : R to L$ is injective, we have $dim R leq dim L$, where "$dim$" refers to the dimension of a $K$-vector space. But $dim L < infty$, since $L$ is finite-dimensional. Hence, $dim R leq dim L < infty$; thus, $R$ is a finite-dimensional $K$-vector space. Therefore, any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces (according to a well-known fact from linear algebra).
Now, let $a in R$ be nonzero. Let $M_a$ denote the map $R to R, r mapsto ar$. This map $M_a : R to R$ is $K$-linear and has kernel $0$ (because every $r in R$ satisfying $ar = 0$ must satisfy $r = 0$ (since $R$ is an integral domain and $a$ is nonzero)); thus, it is injective. Hence, it is an isomorphism of $K$-vector spaces (since any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces). Thus, it is surjective. Therefore, there exists some $s in R$ such that $M_aleft(sright) = 1$. Consider this $s$. Now, the definition of $M_a$ yields $M_aleft(sright) = as$, so that $as = M_aleft(sright) = 1$. In other words, $s$ is a (multiplicative) inverse of $a$. Hence, $a$ has an inverse.
We have thus proven that every nonzero $a in R$ has an inverse. In other words, the ring $R$ is an integral domain. This proves Theorem 1. $blacksquare$
In your situation, you should apply Theorem 1 to $K = k$, $R = A/p$, $L = B/n$ and $g = psi$.
$endgroup$
add a comment |
$begingroup$
Suppose $F$ is any field, $E$ is a finite extension field of $F$, and $D$ is an integral domain such that
$F subset D subset E; tag 1$
since
$[E:F] = n < infty, tag 2$
every element of $D$ is algebraic over $F$; thus
$0 ne d in D tag 3$
satisfies some
$p(x) in F[x]; tag 4$
that is,
$p(d) = 0; tag 5$
we may write
$p(x) = displaystyle sum_0^deg p p_j x^j, ; p_j in F; tag 6$
then
$displaystyle sum_0^deg p p_j d^j = p(d) = 0; tag 7$
furthermore, we may assume $p(x)$ is of minimal degree of all polynomials in $F[x]$ satisfied by $d$. In this case, we must have
$p_0 ne 0; tag 8$
if not, then
$p(x) = displaystyle sum_1^deg p p_jx^j = x sum_1^deg p p_j x^j - 1; tag 9$
thus
$d displaystyle sum_1^deg p p_j d^j - 1 = 0, tag10$
and via (4) this forces
$displaystyle sum_1^deg p p_j d^j - 1 = 0, tag11$
since $D$ is an integral domain; but this asserts that $d$ satisfies the polynomial
$displaystyle sum_1^deg p p_ x^j - 1 in F[x] tag12$
of degree $deg p - 1$, which contradicts the minimality of the degree of $p(x)$; therefore (8) binds and we may write
$displaystyle sum_1^deg pp_j d^j = -p_0, tag13$
or
$d left( -p_0^-1displaystyle sum_1^deg p p_j d^j- 1 right ) = 1, tag14$
which shows that
$d^-1 = -p_0^-1displaystyle sum_1^deg p p_j d^j- 1 in D; tag15$
since every $0 ne d in D$ has in iverse in $D$ by (15), $D$ is indeed a field.
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
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votes
$begingroup$
Theorem 1. Let $K$ be a field. Let $R$ and $L$ be two $K$-algebras such that $L$ is a finite-dimensional $K$-vector space and $R$ is an integral domain. Let $g : R to L$ be an injective $K$-linear map. Then, $R$ is a field.
Proof of Theorem 1. Since the $K$-linear map $g : R to L$ is injective, we have $dim R leq dim L$, where "$dim$" refers to the dimension of a $K$-vector space. But $dim L < infty$, since $L$ is finite-dimensional. Hence, $dim R leq dim L < infty$; thus, $R$ is a finite-dimensional $K$-vector space. Therefore, any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces (according to a well-known fact from linear algebra).
Now, let $a in R$ be nonzero. Let $M_a$ denote the map $R to R, r mapsto ar$. This map $M_a : R to R$ is $K$-linear and has kernel $0$ (because every $r in R$ satisfying $ar = 0$ must satisfy $r = 0$ (since $R$ is an integral domain and $a$ is nonzero)); thus, it is injective. Hence, it is an isomorphism of $K$-vector spaces (since any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces). Thus, it is surjective. Therefore, there exists some $s in R$ such that $M_aleft(sright) = 1$. Consider this $s$. Now, the definition of $M_a$ yields $M_aleft(sright) = as$, so that $as = M_aleft(sright) = 1$. In other words, $s$ is a (multiplicative) inverse of $a$. Hence, $a$ has an inverse.
We have thus proven that every nonzero $a in R$ has an inverse. In other words, the ring $R$ is an integral domain. This proves Theorem 1. $blacksquare$
In your situation, you should apply Theorem 1 to $K = k$, $R = A/p$, $L = B/n$ and $g = psi$.
$endgroup$
add a comment |
$begingroup$
Theorem 1. Let $K$ be a field. Let $R$ and $L$ be two $K$-algebras such that $L$ is a finite-dimensional $K$-vector space and $R$ is an integral domain. Let $g : R to L$ be an injective $K$-linear map. Then, $R$ is a field.
Proof of Theorem 1. Since the $K$-linear map $g : R to L$ is injective, we have $dim R leq dim L$, where "$dim$" refers to the dimension of a $K$-vector space. But $dim L < infty$, since $L$ is finite-dimensional. Hence, $dim R leq dim L < infty$; thus, $R$ is a finite-dimensional $K$-vector space. Therefore, any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces (according to a well-known fact from linear algebra).
Now, let $a in R$ be nonzero. Let $M_a$ denote the map $R to R, r mapsto ar$. This map $M_a : R to R$ is $K$-linear and has kernel $0$ (because every $r in R$ satisfying $ar = 0$ must satisfy $r = 0$ (since $R$ is an integral domain and $a$ is nonzero)); thus, it is injective. Hence, it is an isomorphism of $K$-vector spaces (since any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces). Thus, it is surjective. Therefore, there exists some $s in R$ such that $M_aleft(sright) = 1$. Consider this $s$. Now, the definition of $M_a$ yields $M_aleft(sright) = as$, so that $as = M_aleft(sright) = 1$. In other words, $s$ is a (multiplicative) inverse of $a$. Hence, $a$ has an inverse.
We have thus proven that every nonzero $a in R$ has an inverse. In other words, the ring $R$ is an integral domain. This proves Theorem 1. $blacksquare$
In your situation, you should apply Theorem 1 to $K = k$, $R = A/p$, $L = B/n$ and $g = psi$.
$endgroup$
add a comment |
$begingroup$
Theorem 1. Let $K$ be a field. Let $R$ and $L$ be two $K$-algebras such that $L$ is a finite-dimensional $K$-vector space and $R$ is an integral domain. Let $g : R to L$ be an injective $K$-linear map. Then, $R$ is a field.
Proof of Theorem 1. Since the $K$-linear map $g : R to L$ is injective, we have $dim R leq dim L$, where "$dim$" refers to the dimension of a $K$-vector space. But $dim L < infty$, since $L$ is finite-dimensional. Hence, $dim R leq dim L < infty$; thus, $R$ is a finite-dimensional $K$-vector space. Therefore, any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces (according to a well-known fact from linear algebra).
Now, let $a in R$ be nonzero. Let $M_a$ denote the map $R to R, r mapsto ar$. This map $M_a : R to R$ is $K$-linear and has kernel $0$ (because every $r in R$ satisfying $ar = 0$ must satisfy $r = 0$ (since $R$ is an integral domain and $a$ is nonzero)); thus, it is injective. Hence, it is an isomorphism of $K$-vector spaces (since any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces). Thus, it is surjective. Therefore, there exists some $s in R$ such that $M_aleft(sright) = 1$. Consider this $s$. Now, the definition of $M_a$ yields $M_aleft(sright) = as$, so that $as = M_aleft(sright) = 1$. In other words, $s$ is a (multiplicative) inverse of $a$. Hence, $a$ has an inverse.
We have thus proven that every nonzero $a in R$ has an inverse. In other words, the ring $R$ is an integral domain. This proves Theorem 1. $blacksquare$
In your situation, you should apply Theorem 1 to $K = k$, $R = A/p$, $L = B/n$ and $g = psi$.
$endgroup$
Theorem 1. Let $K$ be a field. Let $R$ and $L$ be two $K$-algebras such that $L$ is a finite-dimensional $K$-vector space and $R$ is an integral domain. Let $g : R to L$ be an injective $K$-linear map. Then, $R$ is a field.
Proof of Theorem 1. Since the $K$-linear map $g : R to L$ is injective, we have $dim R leq dim L$, where "$dim$" refers to the dimension of a $K$-vector space. But $dim L < infty$, since $L$ is finite-dimensional. Hence, $dim R leq dim L < infty$; thus, $R$ is a finite-dimensional $K$-vector space. Therefore, any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces (according to a well-known fact from linear algebra).
Now, let $a in R$ be nonzero. Let $M_a$ denote the map $R to R, r mapsto ar$. This map $M_a : R to R$ is $K$-linear and has kernel $0$ (because every $r in R$ satisfying $ar = 0$ must satisfy $r = 0$ (since $R$ is an integral domain and $a$ is nonzero)); thus, it is injective. Hence, it is an isomorphism of $K$-vector spaces (since any injective $K$-linear map $f : R to R$ is an isomorphism of $K$-vector spaces). Thus, it is surjective. Therefore, there exists some $s in R$ such that $M_aleft(sright) = 1$. Consider this $s$. Now, the definition of $M_a$ yields $M_aleft(sright) = as$, so that $as = M_aleft(sright) = 1$. In other words, $s$ is a (multiplicative) inverse of $a$. Hence, $a$ has an inverse.
We have thus proven that every nonzero $a in R$ has an inverse. In other words, the ring $R$ is an integral domain. This proves Theorem 1. $blacksquare$
In your situation, you should apply Theorem 1 to $K = k$, $R = A/p$, $L = B/n$ and $g = psi$.
answered 35 mins ago
darij grinbergdarij grinberg
11.2k33167
11.2k33167
add a comment |
add a comment |
$begingroup$
Suppose $F$ is any field, $E$ is a finite extension field of $F$, and $D$ is an integral domain such that
$F subset D subset E; tag 1$
since
$[E:F] = n < infty, tag 2$
every element of $D$ is algebraic over $F$; thus
$0 ne d in D tag 3$
satisfies some
$p(x) in F[x]; tag 4$
that is,
$p(d) = 0; tag 5$
we may write
$p(x) = displaystyle sum_0^deg p p_j x^j, ; p_j in F; tag 6$
then
$displaystyle sum_0^deg p p_j d^j = p(d) = 0; tag 7$
furthermore, we may assume $p(x)$ is of minimal degree of all polynomials in $F[x]$ satisfied by $d$. In this case, we must have
$p_0 ne 0; tag 8$
if not, then
$p(x) = displaystyle sum_1^deg p p_jx^j = x sum_1^deg p p_j x^j - 1; tag 9$
thus
$d displaystyle sum_1^deg p p_j d^j - 1 = 0, tag10$
and via (4) this forces
$displaystyle sum_1^deg p p_j d^j - 1 = 0, tag11$
since $D$ is an integral domain; but this asserts that $d$ satisfies the polynomial
$displaystyle sum_1^deg p p_ x^j - 1 in F[x] tag12$
of degree $deg p - 1$, which contradicts the minimality of the degree of $p(x)$; therefore (8) binds and we may write
$displaystyle sum_1^deg pp_j d^j = -p_0, tag13$
or
$d left( -p_0^-1displaystyle sum_1^deg p p_j d^j- 1 right ) = 1, tag14$
which shows that
$d^-1 = -p_0^-1displaystyle sum_1^deg p p_j d^j- 1 in D; tag15$
since every $0 ne d in D$ has in iverse in $D$ by (15), $D$ is indeed a field.
$endgroup$
add a comment |
$begingroup$
Suppose $F$ is any field, $E$ is a finite extension field of $F$, and $D$ is an integral domain such that
$F subset D subset E; tag 1$
since
$[E:F] = n < infty, tag 2$
every element of $D$ is algebraic over $F$; thus
$0 ne d in D tag 3$
satisfies some
$p(x) in F[x]; tag 4$
that is,
$p(d) = 0; tag 5$
we may write
$p(x) = displaystyle sum_0^deg p p_j x^j, ; p_j in F; tag 6$
then
$displaystyle sum_0^deg p p_j d^j = p(d) = 0; tag 7$
furthermore, we may assume $p(x)$ is of minimal degree of all polynomials in $F[x]$ satisfied by $d$. In this case, we must have
$p_0 ne 0; tag 8$
if not, then
$p(x) = displaystyle sum_1^deg p p_jx^j = x sum_1^deg p p_j x^j - 1; tag 9$
thus
$d displaystyle sum_1^deg p p_j d^j - 1 = 0, tag10$
and via (4) this forces
$displaystyle sum_1^deg p p_j d^j - 1 = 0, tag11$
since $D$ is an integral domain; but this asserts that $d$ satisfies the polynomial
$displaystyle sum_1^deg p p_ x^j - 1 in F[x] tag12$
of degree $deg p - 1$, which contradicts the minimality of the degree of $p(x)$; therefore (8) binds and we may write
$displaystyle sum_1^deg pp_j d^j = -p_0, tag13$
or
$d left( -p_0^-1displaystyle sum_1^deg p p_j d^j- 1 right ) = 1, tag14$
which shows that
$d^-1 = -p_0^-1displaystyle sum_1^deg p p_j d^j- 1 in D; tag15$
since every $0 ne d in D$ has in iverse in $D$ by (15), $D$ is indeed a field.
$endgroup$
add a comment |
$begingroup$
Suppose $F$ is any field, $E$ is a finite extension field of $F$, and $D$ is an integral domain such that
$F subset D subset E; tag 1$
since
$[E:F] = n < infty, tag 2$
every element of $D$ is algebraic over $F$; thus
$0 ne d in D tag 3$
satisfies some
$p(x) in F[x]; tag 4$
that is,
$p(d) = 0; tag 5$
we may write
$p(x) = displaystyle sum_0^deg p p_j x^j, ; p_j in F; tag 6$
then
$displaystyle sum_0^deg p p_j d^j = p(d) = 0; tag 7$
furthermore, we may assume $p(x)$ is of minimal degree of all polynomials in $F[x]$ satisfied by $d$. In this case, we must have
$p_0 ne 0; tag 8$
if not, then
$p(x) = displaystyle sum_1^deg p p_jx^j = x sum_1^deg p p_j x^j - 1; tag 9$
thus
$d displaystyle sum_1^deg p p_j d^j - 1 = 0, tag10$
and via (4) this forces
$displaystyle sum_1^deg p p_j d^j - 1 = 0, tag11$
since $D$ is an integral domain; but this asserts that $d$ satisfies the polynomial
$displaystyle sum_1^deg p p_ x^j - 1 in F[x] tag12$
of degree $deg p - 1$, which contradicts the minimality of the degree of $p(x)$; therefore (8) binds and we may write
$displaystyle sum_1^deg pp_j d^j = -p_0, tag13$
or
$d left( -p_0^-1displaystyle sum_1^deg p p_j d^j- 1 right ) = 1, tag14$
which shows that
$d^-1 = -p_0^-1displaystyle sum_1^deg p p_j d^j- 1 in D; tag15$
since every $0 ne d in D$ has in iverse in $D$ by (15), $D$ is indeed a field.
$endgroup$
Suppose $F$ is any field, $E$ is a finite extension field of $F$, and $D$ is an integral domain such that
$F subset D subset E; tag 1$
since
$[E:F] = n < infty, tag 2$
every element of $D$ is algebraic over $F$; thus
$0 ne d in D tag 3$
satisfies some
$p(x) in F[x]; tag 4$
that is,
$p(d) = 0; tag 5$
we may write
$p(x) = displaystyle sum_0^deg p p_j x^j, ; p_j in F; tag 6$
then
$displaystyle sum_0^deg p p_j d^j = p(d) = 0; tag 7$
furthermore, we may assume $p(x)$ is of minimal degree of all polynomials in $F[x]$ satisfied by $d$. In this case, we must have
$p_0 ne 0; tag 8$
if not, then
$p(x) = displaystyle sum_1^deg p p_jx^j = x sum_1^deg p p_j x^j - 1; tag 9$
thus
$d displaystyle sum_1^deg p p_j d^j - 1 = 0, tag10$
and via (4) this forces
$displaystyle sum_1^deg p p_j d^j - 1 = 0, tag11$
since $D$ is an integral domain; but this asserts that $d$ satisfies the polynomial
$displaystyle sum_1^deg p p_ x^j - 1 in F[x] tag12$
of degree $deg p - 1$, which contradicts the minimality of the degree of $p(x)$; therefore (8) binds and we may write
$displaystyle sum_1^deg pp_j d^j = -p_0, tag13$
or
$d left( -p_0^-1displaystyle sum_1^deg p p_j d^j- 1 right ) = 1, tag14$
which shows that
$d^-1 = -p_0^-1displaystyle sum_1^deg p p_j d^j- 1 in D; tag15$
since every $0 ne d in D$ has in iverse in $D$ by (15), $D$ is indeed a field.
answered 31 mins ago
Robert LewisRobert Lewis
48.3k23167
48.3k23167
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