What does it mean that field $mathbb{F}_{p^n}$ “contains” the prime field $mathbb{Z}_p$?












4












$begingroup$


I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?



What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?



For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?



This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?



Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?






number-theory field-theory finite-fields extension-field computational-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
    $endgroup$
    – ancientmathematician
    Jan 6 at 7:51












  • $begingroup$
    @ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
    $endgroup$
    – Jus12
    Jan 6 at 8:09
















4












$begingroup$


I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?



What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?



For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?



This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?



Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?






number-theory field-theory finite-fields extension-field computational-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
    $endgroup$
    – ancientmathematician
    Jan 6 at 7:51












  • $begingroup$
    @ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
    $endgroup$
    – Jus12
    Jan 6 at 8:09














4












4








4





$begingroup$


I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?



What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?



For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?



This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?



Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?






number-theory field-theory finite-fields extension-field computational-algebra







share|cite|improve this question











$endgroup$




I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?



What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?



For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?



This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?



Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?






number-theory field-theory finite-fields extension-field computational-algebra




number-theory field-theory finite-fields extension-field computational-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 8:08







Jus12

















asked Jan 6 at 7:30









Jus12Jus12

1808




1808








  • 1




    $begingroup$
    Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
    $endgroup$
    – ancientmathematician
    Jan 6 at 7:51












  • $begingroup$
    @ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
    $endgroup$
    – Jus12
    Jan 6 at 8:09














  • 1




    $begingroup$
    Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
    $endgroup$
    – ancientmathematician
    Jan 6 at 7:51












  • $begingroup$
    @ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
    $endgroup$
    – Jus12
    Jan 6 at 8:09








1




1




$begingroup$
Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
$endgroup$
– ancientmathematician
Jan 6 at 7:51






$begingroup$
Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
$endgroup$
– ancientmathematician
Jan 6 at 7:51














$begingroup$
@ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
$endgroup$
– Jus12
Jan 6 at 8:09




$begingroup$
@ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
$endgroup$
– Jus12
Jan 6 at 8:09










2 Answers
2






active

oldest

votes


















6












$begingroup$

Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.



begin{align*}
Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
& ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
&={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
end{align*}



Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.




With your edited version:



$Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).



So consider the set of the vectors
$$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
      $endgroup$
      – Jus12
      Jan 6 at 8:28












    • $begingroup$
      Indeed, ''double and repeat'' can be used.
      $endgroup$
      – Wuestenfux
      Jan 6 at 8:28











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063576%2fwhat-does-it-mean-that-field-mathbbf-pn-contains-the-prime-field-mat%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.



    begin{align*}
    Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
    & ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
    &={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
    end{align*}



    Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.




    With your edited version:



    $Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).



    So consider the set of the vectors
    $$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
    This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.



      begin{align*}
      Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
      & ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
      &={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
      end{align*}



      Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.




      With your edited version:



      $Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).



      So consider the set of the vectors
      $$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
      This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.



        begin{align*}
        Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
        & ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
        &={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
        end{align*}



        Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.




        With your edited version:



        $Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).



        So consider the set of the vectors
        $$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
        This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.






        share|cite|improve this answer











        $endgroup$



        Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.



        begin{align*}
        Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
        & ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
        &={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
        end{align*}



        Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.




        With your edited version:



        $Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).



        So consider the set of the vectors
        $$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
        This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 6 at 7:50

























        answered Jan 6 at 7:42









        Anurag AAnurag A

        25.8k12249




        25.8k12249























            2












            $begingroup$

            The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
              $endgroup$
              – Jus12
              Jan 6 at 8:28












            • $begingroup$
              Indeed, ''double and repeat'' can be used.
              $endgroup$
              – Wuestenfux
              Jan 6 at 8:28
















            2












            $begingroup$

            The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
              $endgroup$
              – Jus12
              Jan 6 at 8:28












            • $begingroup$
              Indeed, ''double and repeat'' can be used.
              $endgroup$
              – Wuestenfux
              Jan 6 at 8:28














            2












            2








            2





            $begingroup$

            The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.






            share|cite|improve this answer









            $endgroup$



            The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 8:10









            WuestenfuxWuestenfux

            4,0051411




            4,0051411












            • $begingroup$
              I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
              $endgroup$
              – Jus12
              Jan 6 at 8:28












            • $begingroup$
              Indeed, ''double and repeat'' can be used.
              $endgroup$
              – Wuestenfux
              Jan 6 at 8:28


















            • $begingroup$
              I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
              $endgroup$
              – Jus12
              Jan 6 at 8:28












            • $begingroup$
              Indeed, ''double and repeat'' can be used.
              $endgroup$
              – Wuestenfux
              Jan 6 at 8:28
















            $begingroup$
            I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
            $endgroup$
            – Jus12
            Jan 6 at 8:28






            $begingroup$
            I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
            $endgroup$
            – Jus12
            Jan 6 at 8:28














            $begingroup$
            Indeed, ''double and repeat'' can be used.
            $endgroup$
            – Wuestenfux
            Jan 6 at 8:28




            $begingroup$
            Indeed, ''double and repeat'' can be used.
            $endgroup$
            – Wuestenfux
            Jan 6 at 8:28


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063576%2fwhat-does-it-mean-that-field-mathbbf-pn-contains-the-prime-field-mat%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            Display a custom attribute below product name in the front-end Magento 1.9.3.8

            Image
            0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
            Read more