Does a field extension L of F have the same characteristic as F?

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Is the following proof sound? Is it true for all field extension finite and infinite?



Suppose $F$ has characteristic $p$. Since $1_l=1_f$ this implies that $p*1_f=p*1_l=0$. $textchar(L)=p_1$ cannot be less than $p$ as this implies that $p_1*1_f=0$. Hence, $textchar(L)=p$. Similarly if $F$ has characteristic $0$ this implies that $L$ cant have finite characteristic (same reasoning as above.)






abstract-algebra proof-verification galois-extensions







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    up vote
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    down vote

    favorite












    Is the following proof sound? Is it true for all field extension finite and infinite?



    Suppose $F$ has characteristic $p$. Since $1_l=1_f$ this implies that $p*1_f=p*1_l=0$. $textchar(L)=p_1$ cannot be less than $p$ as this implies that $p_1*1_f=0$. Hence, $textchar(L)=p$. Similarly if $F$ has characteristic $0$ this implies that $L$ cant have finite characteristic (same reasoning as above.)






    abstract-algebra proof-verification galois-extensions







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      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Is the following proof sound? Is it true for all field extension finite and infinite?



      Suppose $F$ has characteristic $p$. Since $1_l=1_f$ this implies that $p*1_f=p*1_l=0$. $textchar(L)=p_1$ cannot be less than $p$ as this implies that $p_1*1_f=0$. Hence, $textchar(L)=p$. Similarly if $F$ has characteristic $0$ this implies that $L$ cant have finite characteristic (same reasoning as above.)






      abstract-algebra proof-verification galois-extensions







      share|cite|improve this question















      Is the following proof sound? Is it true for all field extension finite and infinite?



      Suppose $F$ has characteristic $p$. Since $1_l=1_f$ this implies that $p*1_f=p*1_l=0$. $textchar(L)=p_1$ cannot be less than $p$ as this implies that $p_1*1_f=0$. Hence, $textchar(L)=p$. Similarly if $F$ has characteristic $0$ this implies that $L$ cant have finite characteristic (same reasoning as above.)






      abstract-algebra proof-verification galois-extensions




      abstract-algebra proof-verification galois-extensions






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      edited 2 hours ago









      Babelfish

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      1,099315










      asked 2 hours ago









      Jhon Doe

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          3 Answers
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          up vote
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          accepted










          If $L$ is a field extension of $K$, then $K$ is additively a subgroup of the additive group of $L$. This inherits the characteristic.






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            up vote
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            This is correct. To be a bit nit-picky, you should really think of $p * 1$ as $underbrace1+1+1+dots+1_ptext times$ and not as a multiplication in the field.






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            • yeah i did. Didn't want to write out the 1s. Thanks.
              – Jhon Doe
              2 hours ago

















            up vote
            1
            down vote













            As an alternative way to see this, a field $k$ has characteristic $p$ precisely if it contains $mathbb F_p$ (and characteristic $0$ if and only if it contains $mathbb Q$). This is proved by considering the unique ring map $mathbb Zto k$, which has the property that the unique nonnegative generator of its kernel is the characteristic of $k$. Then one can factor this map uniquely through $mathbb F_p$ (or $mathbb Q$).






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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              2
              down vote



              accepted










              If $L$ is a field extension of $K$, then $K$ is additively a subgroup of the additive group of $L$. This inherits the characteristic.






              share|cite|improve this answer
























                up vote
                2
                down vote



                accepted










                If $L$ is a field extension of $K$, then $K$ is additively a subgroup of the additive group of $L$. This inherits the characteristic.






                share|cite|improve this answer






















                  up vote
                  2
                  down vote



                  accepted







                  up vote
                  2
                  down vote



                  accepted






                  If $L$ is a field extension of $K$, then $K$ is additively a subgroup of the additive group of $L$. This inherits the characteristic.






                  share|cite|improve this answer












                  If $L$ is a field extension of $K$, then $K$ is additively a subgroup of the additive group of $L$. This inherits the characteristic.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Wuestenfux

                  1,211128




                  1,211128




















                      up vote
                      1
                      down vote













                      This is correct. To be a bit nit-picky, you should really think of $p * 1$ as $underbrace1+1+1+dots+1_ptext times$ and not as a multiplication in the field.






                      share|cite|improve this answer




















                      • yeah i did. Didn't want to write out the 1s. Thanks.
                        – Jhon Doe
                        2 hours ago














                      up vote
                      1
                      down vote













                      This is correct. To be a bit nit-picky, you should really think of $p * 1$ as $underbrace1+1+1+dots+1_ptext times$ and not as a multiplication in the field.






                      share|cite|improve this answer




















                      • yeah i did. Didn't want to write out the 1s. Thanks.
                        – Jhon Doe
                        2 hours ago












                      up vote
                      1
                      down vote










                      up vote
                      1
                      down vote









                      This is correct. To be a bit nit-picky, you should really think of $p * 1$ as $underbrace1+1+1+dots+1_ptext times$ and not as a multiplication in the field.






                      share|cite|improve this answer












                      This is correct. To be a bit nit-picky, you should really think of $p * 1$ as $underbrace1+1+1+dots+1_ptext times$ and not as a multiplication in the field.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 hours ago









                      Babelfish

                      1,099315




                      1,099315











                      • yeah i did. Didn't want to write out the 1s. Thanks.
                        – Jhon Doe
                        2 hours ago
















                      • yeah i did. Didn't want to write out the 1s. Thanks.
                        – Jhon Doe
                        2 hours ago















                      yeah i did. Didn't want to write out the 1s. Thanks.
                      – Jhon Doe
                      2 hours ago




                      yeah i did. Didn't want to write out the 1s. Thanks.
                      – Jhon Doe
                      2 hours ago










                      up vote
                      1
                      down vote













                      As an alternative way to see this, a field $k$ has characteristic $p$ precisely if it contains $mathbb F_p$ (and characteristic $0$ if and only if it contains $mathbb Q$). This is proved by considering the unique ring map $mathbb Zto k$, which has the property that the unique nonnegative generator of its kernel is the characteristic of $k$. Then one can factor this map uniquely through $mathbb F_p$ (or $mathbb Q$).






                      share|cite|improve this answer
























                        up vote
                        1
                        down vote













                        As an alternative way to see this, a field $k$ has characteristic $p$ precisely if it contains $mathbb F_p$ (and characteristic $0$ if and only if it contains $mathbb Q$). This is proved by considering the unique ring map $mathbb Zto k$, which has the property that the unique nonnegative generator of its kernel is the characteristic of $k$. Then one can factor this map uniquely through $mathbb F_p$ (or $mathbb Q$).






                        share|cite|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          As an alternative way to see this, a field $k$ has characteristic $p$ precisely if it contains $mathbb F_p$ (and characteristic $0$ if and only if it contains $mathbb Q$). This is proved by considering the unique ring map $mathbb Zto k$, which has the property that the unique nonnegative generator of its kernel is the characteristic of $k$. Then one can factor this map uniquely through $mathbb F_p$ (or $mathbb Q$).






                          share|cite|improve this answer












                          As an alternative way to see this, a field $k$ has characteristic $p$ precisely if it contains $mathbb F_p$ (and characteristic $0$ if and only if it contains $mathbb Q$). This is proved by considering the unique ring map $mathbb Zto k$, which has the property that the unique nonnegative generator of its kernel is the characteristic of $k$. Then one can factor this map uniquely through $mathbb F_p$ (or $mathbb Q$).







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 53 mins ago









                          asdq

                          1,5791417




                          1,5791417



























                               

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