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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 $K$













2












$begingroup$


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?










share|cite|improve this question









$endgroup$
















    2












    $begingroup$


    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?










    share|cite|improve this question









    $endgroup$














      2












      2








      2


      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 $operatornameSpec B$ 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?










      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 $operatornameSpec B$ 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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      zxcvzxcv

      1609




      1609




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$


          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









          $endgroup$




















            1












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$












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              active

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              2












              $begingroup$


              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









              $endgroup$

















                2












                $begingroup$


                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









                $endgroup$















                  2












                  2








                  2





                  $begingroup$


                  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 30 mins ago









                  darij grinbergdarij grinberg

                  11.2k33167




                  11.2k33167





















                      1












                      $begingroup$

                      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.






                      share|cite|improve this answer









                      $endgroup$

















                        1












                        $begingroup$

                        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.






                        share|cite|improve this answer









                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          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.






                          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



                          $d displaystyle sum_1^deg p p_j d^j - 1 = 0, tag10$



                          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.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 27 mins ago









                          Robert LewisRobert Lewis

                          48.3k23167




                          48.3k23167



























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