Finding the integral closure of k[x] in k ( x ) ( f ), where f ( x ) = x 6 + t x 5 + t 2 x 3 + t ∈ k [ x ]Let F be a field of characteristic 2. Let k := F ( t ), the field of rational functions in the variable t. Let f ( x ) = x 6 + t x 5 + t 2 x 3 + t ∈ k [ x ].We want to show that the integral closure of k[x] in k ( x ) ( f ) is k [ x ] [ f ].So let α = m + n f ∈ k ( x ) ( f ), where m , n ∈ k ( x ). Since F(t) is a field, k[x] is a PID and therefore α is integral over k[x] if and only if its minimal polynomial in k(x) has coefficients in k[x]. Let's then find the minimal polynomial of α. p ( y ) = ( y − ( m + n f ) ) ( y − ( m − n f ) ) = y 2 − 2 m y + ( m 2 − n 2 f ) and since m ∈ k ( x ) = F ( t ) k ( x ) and the characteristic of F equals 2 we get that 2 m y = 0. Hence p ( y ) = y 2 + ( m 2 − n 2 f )Thus, α = m + n f is integral over k[x] iff m 2 − n 2 f ∈ k [ x ]. I' m pretty sure that I'm on the right course but I don' t know how to continue.

Question

Finding the integral closure of k[x] in   k  (  x  )  (      f    ), where   f  (  x  )  =      x    6    +  t      x    5    +      t    2        x    3    +  t  ∈  k  [  x  ]Let F be a field of characteristic 2. Let   k  :=  F  (  t  ), the field of rational functions in the variable t. Let   f  (  x  )  =      x    6    +  t      x    5    +      t    2        x    3    +  t  ∈  k  [  x  ].We want to show that the integral closure of k[x] in   k  (  x  )  (      f    ) is   k  [  x  ]  [      f    ].So let   α  =  m  +  n      f    ∈  k  (  x  )  (      f    ), where   m  ,  n  ∈  k  (  x  ). Since F(t) is a field, k[x] is a PID and therefore   α is integral over k[x] if and only if its minimal polynomial in k(x) has coefficients in k[x]. Let&#039;s then find the minimal polynomial of   α.                    p        (        y        )                            =        (        y        −        (        m        +        n                  f                )        )        (        y        −        (        m        −        n                  f                )        )                                          =                  y          2                −        2        m        y        +        (                  m          2                −                  n          2                f        )            and since   m  ∈  k  (  x  )  =  F  (  t  )  k  (  x  ) and the characteristic of F equals 2 we get that   2  m  y  =  0. Hence  p  (  y  )  =      y    2    +  (      m    2    −      n    2    f  )Thus,   α  =  m  +  n      f   is integral over k[x] iff       m    2    −      n    2    f  ∈  k  [  x  ]. I&#039; m pretty sure that I&#039;m on the right course but I don&#039; t know how to continue.

granatimahs8 · Accepted Answer

Step 1Denote as B the integral closure of k[x] in   k  (  x  )  (      f    ). The inclusion   k  [  x  ]  [      f    ]  ⊆  B follows immediately from the fact that the coefficients of the minimal polynomial of an arbitrary element of   k  [  x  ]  [      f    ] lay in k[x]. For the reverse inclusion you have to use exercise 1.14. First observe that f is squarefree in       k    ¯    [  x  ] and then that       f    ′    (  x  )  =  5  t      x    4    +  3      t    2        x    2    =      x    2    (  5  t      x    2    +  3      t    2    ). Now, let   l  =  a  +  b      f        /    c  ∈  B where   a  ,  b  ,  c  ∈  k  [  x  ] and   (  a  ,  b  ,  c  )  =  1.Step 2Note that since l is integral we have             (      a      +      b              f                    /            c      )        2    ∈  k  [  x  ] which implies that       c    2        |        a    2    +      b    2    f. From exercise 1.14 we have that       c    2   divides f′ and since   5  t      x    2    +  3      t    2   is squarefree(why?) we conclude that c divides       x    2   and hence   c  =  1 or   c  =  x. If   c  =  1 we are done. Suppose now for a contradiction that   c  =  x. Also note that when   a  ∈  k  [  x  ], since k is of characteristic 2, we have that       a    2   has only powers divisible by 2 and its constant term squared. Since   c  =  x we have that       x    2    h  =      a    2    +      b    2    f for some   h  ∈  k  [  x  ] thus, the constant term of       a    2    +      b    2    f i.e       a    0    2    +      b    0    2    t must be zero. Finally,      a    0    2    +      b    0    2    t  =  0  ⇒      a    0    2    =      b    0    2    t  ⇒      a    0    2        b    0          −      2        =  t  ⇒  (      a    0        b    0          −      1            )    2    =  twhich implies that       t    ∈  F  (  t  ) a contradiction and you get the desired result.

Finding the integral closure of k[x] in k(x)(sqrtf), where f(x)=x^6+tx^5+t^2x^3+t in k[x]

Answered question

Answer & Explanation

New Questions in Algebra II