Why does the integral domain “being trapped between a finite field extension” implies that it is a field?












3












$begingroup$


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 $operatorname{Spec} B$ means a maximal ideal $n$ of $B$. And $operatorname{Spec}(ϕ)(n) = ϕ^{−1}(n)$. 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?










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  • $begingroup$
    This is close to being a duplicate of this old thread. But, it is not clear cut, so I won't use my dupehammer privilege to force my opinion.
    $endgroup$
    – Jyrki Lahtonen
    40 secs ago
















3












$begingroup$


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 $operatorname{Spec} B$ means a maximal ideal $n$ of $B$. And $operatorname{Spec}(ϕ)(n) = ϕ^{−1}(n)$. 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?










share|cite|improve this question









$endgroup$












  • $begingroup$
    This is close to being a duplicate of this old thread. But, it is not clear cut, so I won't use my dupehammer privilege to force my opinion.
    $endgroup$
    – Jyrki Lahtonen
    40 secs ago














3












3








3


1



$begingroup$


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 $operatorname{Spec} B$ means a maximal ideal $n$ of $B$. And $operatorname{Spec}(ϕ)(n) = ϕ^{−1}(n)$. 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?










share|cite|improve this question









$endgroup$




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 $operatorname{Spec} B$ means a maximal ideal $n$ of $B$. And $operatorname{Spec}(ϕ)(n) = ϕ^{−1}(n)$. 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






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asked 1 hour ago









zxcvzxcv

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  • $begingroup$
    This is close to being a duplicate of this old thread. But, it is not clear cut, so I won't use my dupehammer privilege to force my opinion.
    $endgroup$
    – Jyrki Lahtonen
    40 secs ago


















  • $begingroup$
    This is close to being a duplicate of this old thread. But, it is not clear cut, so I won't use my dupehammer privilege to force my opinion.
    $endgroup$
    – Jyrki Lahtonen
    40 secs ago
















$begingroup$
This is close to being a duplicate of this old thread. But, it is not clear cut, so I won't use my dupehammer privilege to force my opinion.
$endgroup$
– Jyrki Lahtonen
40 secs ago




$begingroup$
This is close to being a duplicate of this old thread. But, it is not clear cut, so I won't use my dupehammer privilege to force my opinion.
$endgroup$
– Jyrki Lahtonen
40 secs ago










2 Answers
2






active

oldest

votes


















4












$begingroup$


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 $g = psi$.






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$endgroup$





















    2












    $begingroup$

    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



    $d displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{10}$



    and via (4) this forces



    $displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{11}$



    since $D$ is an integral domain; but this asserts that $d$ satisfies the polynomial



    $displaystyle sum_1^{deg p} p_j x^{j - 1} in F[x] tag{12}$



    of degree $deg p - 1$, which contradicts the minimality of the degree of $p(x)$; therefore (8) binds and we may write



    $displaystyle sum_1^{deg p}p_j d^j = -p_0, tag{13}$



    or



    $d left( -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} right ) = 1, tag{14}$



    which shows that



    $d^{-1} = -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} in D; tag{15}$



    since every $0 ne d in D$ has in iverse in $D$ by (15), $D$ is indeed a field.






    share|cite|improve this answer











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      2 Answers
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      2 Answers
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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 $g = psi$.






      share|cite|improve this answer









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        4












        $begingroup$


        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 $g = psi$.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$


          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 $g = psi$.






          share|cite|improve this answer









          $endgroup$




          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 $g = psi$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          darij grinbergdarij grinberg

          11.3k33167




          11.3k33167























              2












              $begingroup$

              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



              $d displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{10}$



              and via (4) this forces



              $displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{11}$



              since $D$ is an integral domain; but this asserts that $d$ satisfies the polynomial



              $displaystyle sum_1^{deg p} p_j x^{j - 1} in F[x] tag{12}$



              of degree $deg p - 1$, which contradicts the minimality of the degree of $p(x)$; therefore (8) binds and we may write



              $displaystyle sum_1^{deg p}p_j d^j = -p_0, tag{13}$



              or



              $d left( -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} right ) = 1, tag{14}$



              which shows that



              $d^{-1} = -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} in D; tag{15}$



              since every $0 ne d in D$ has in iverse in $D$ by (15), $D$ is indeed a field.






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                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



                $d displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{10}$



                and via (4) this forces



                $displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{11}$



                since $D$ is an integral domain; but this asserts that $d$ satisfies the polynomial



                $displaystyle sum_1^{deg p} p_j x^{j - 1} in F[x] tag{12}$



                of degree $deg p - 1$, which contradicts the minimality of the degree of $p(x)$; therefore (8) binds and we may write



                $displaystyle sum_1^{deg p}p_j d^j = -p_0, tag{13}$



                or



                $d left( -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} right ) = 1, tag{14}$



                which shows that



                $d^{-1} = -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} in D; tag{15}$



                since every $0 ne d in D$ has in iverse in $D$ by (15), $D$ is indeed a field.






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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



                  $d displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{10}$



                  and via (4) this forces



                  $displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{11}$



                  since $D$ is an integral domain; but this asserts that $d$ satisfies the polynomial



                  $displaystyle sum_1^{deg p} p_j x^{j - 1} in F[x] tag{12}$



                  of degree $deg p - 1$, which contradicts the minimality of the degree of $p(x)$; therefore (8) binds and we may write



                  $displaystyle sum_1^{deg p}p_j d^j = -p_0, tag{13}$



                  or



                  $d left( -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} right ) = 1, tag{14}$



                  which shows that



                  $d^{-1} = -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} in D; tag{15}$



                  since every $0 ne d in D$ has in iverse in $D$ by (15), $D$ is indeed a field.






                  share|cite|improve this answer











                  $endgroup$



                  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



                  $d displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{10}$



                  and via (4) this forces



                  $displaystyle sum_1^{deg p} p_j d^{j - 1} = 0, tag{11}$



                  since $D$ is an integral domain; but this asserts that $d$ satisfies the polynomial



                  $displaystyle sum_1^{deg p} p_j x^{j - 1} in F[x] tag{12}$



                  of degree $deg p - 1$, which contradicts the minimality of the degree of $p(x)$; therefore (8) binds and we may write



                  $displaystyle sum_1^{deg p}p_j d^j = -p_0, tag{13}$



                  or



                  $d left( -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} right ) = 1, tag{14}$



                  which shows that



                  $d^{-1} = -p_0^{-1}displaystyle sum_1^{deg p} p_j d^{j- 1} in D; tag{15}$



                  since every $0 ne d in D$ has in iverse in $D$ by (15), $D$ is indeed a field.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 21 mins ago

























                  answered 1 hour ago









                  Robert LewisRobert Lewis

                  48.3k23167




                  48.3k23167






























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