Element of given order in a finite field












2












$begingroup$


Find an element of multiplicative order 4 and an element of order 5 in $F_{121}$ defined by $x^{2} +x +7$ ∈ $Z_{11}$.



The most obvious way to go about this seems to find a generator and raise it to a quarter the order of the field, thus producing an element that is equal to 1 when raised to the power of 4, according to an analog of Fermat's Little theorem. But since the polynomial's coefficients are over $Z_{11}$, I can't seem to find an obvious generator, and the reduction mod the quadratic seems cumbersome. Is there a more efficient elegant way to gleam elements of a desired order from this finite field?






abstract-algebra field-theory finite-fields







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Find an element of multiplicative order 4 and an element of order 5 in $F_{121}$ defined by $x^{2} +x +7$ ∈ $Z_{11}$.



    The most obvious way to go about this seems to find a generator and raise it to a quarter the order of the field, thus producing an element that is equal to 1 when raised to the power of 4, according to an analog of Fermat's Little theorem. But since the polynomial's coefficients are over $Z_{11}$, I can't seem to find an obvious generator, and the reduction mod the quadratic seems cumbersome. Is there a more efficient elegant way to gleam elements of a desired order from this finite field?






    abstract-algebra field-theory finite-fields







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Find an element of multiplicative order 4 and an element of order 5 in $F_{121}$ defined by $x^{2} +x +7$ ∈ $Z_{11}$.



      The most obvious way to go about this seems to find a generator and raise it to a quarter the order of the field, thus producing an element that is equal to 1 when raised to the power of 4, according to an analog of Fermat's Little theorem. But since the polynomial's coefficients are over $Z_{11}$, I can't seem to find an obvious generator, and the reduction mod the quadratic seems cumbersome. Is there a more efficient elegant way to gleam elements of a desired order from this finite field?






      abstract-algebra field-theory finite-fields







      share|cite|improve this question









      $endgroup$




      Find an element of multiplicative order 4 and an element of order 5 in $F_{121}$ defined by $x^{2} +x +7$ ∈ $Z_{11}$.



      The most obvious way to go about this seems to find a generator and raise it to a quarter the order of the field, thus producing an element that is equal to 1 when raised to the power of 4, according to an analog of Fermat's Little theorem. But since the polynomial's coefficients are over $Z_{11}$, I can't seem to find an obvious generator, and the reduction mod the quadratic seems cumbersome. Is there a more efficient elegant way to gleam elements of a desired order from this finite field?






      abstract-algebra field-theory finite-fields




      abstract-algebra field-theory finite-fields






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      user613048user613048

      302




      302






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          I don't think there's a magic shortcut to this kind of question in general, but the specific cases we have here are simple enough that there are some tricks. For an element of order $5$, we can note that $5$ divides $11-1$, so there is actually an element of order $5$ in $mathbb{F}_{11}$, which is easy to find by trial and error. A bit more systematically, we can see that the subgroup of squares in $mathbb{F}_{11}^*$ is cyclic of order $5$ so any square besides $0$ and $1$ must have order $5$.



          For an element of order $4$, we can save some work by noticing that an element of order $4$ is just a square root of $-1$. So, we just need to solve for $a,binmathbb{F}_{11}$ such that $(a+bx)^2=-1$. Expanding out $(a+bx)^2$ using $x^2+x+7=0$ we get $$(a^2-7b^2)+(2ab-b^2)x=-1$$ and so we need $$a^2-7b^2=-1$$ and $$2ab-b^2=0.$$ The second equation gives $b=0$ or $2a=b$. The first case does not work (since there is no square root of $-1$ in $mathbb{F}_7$); in the second case the first equation simplifies to $$5a^2=1$$ which we can easily solve in $mathbb{F}_{11}$ to get $a=pm 3$. So, the elements of order $4$ are $pm(3+6x)$.






          share|cite|improve this answer









          $endgroup$













            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%2f3084188%2felement-of-given-order-in-a-finite-field%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            I don't think there's a magic shortcut to this kind of question in general, but the specific cases we have here are simple enough that there are some tricks. For an element of order $5$, we can note that $5$ divides $11-1$, so there is actually an element of order $5$ in $mathbb{F}_{11}$, which is easy to find by trial and error. A bit more systematically, we can see that the subgroup of squares in $mathbb{F}_{11}^*$ is cyclic of order $5$ so any square besides $0$ and $1$ must have order $5$.



            For an element of order $4$, we can save some work by noticing that an element of order $4$ is just a square root of $-1$. So, we just need to solve for $a,binmathbb{F}_{11}$ such that $(a+bx)^2=-1$. Expanding out $(a+bx)^2$ using $x^2+x+7=0$ we get $$(a^2-7b^2)+(2ab-b^2)x=-1$$ and so we need $$a^2-7b^2=-1$$ and $$2ab-b^2=0.$$ The second equation gives $b=0$ or $2a=b$. The first case does not work (since there is no square root of $-1$ in $mathbb{F}_7$); in the second case the first equation simplifies to $$5a^2=1$$ which we can easily solve in $mathbb{F}_{11}$ to get $a=pm 3$. So, the elements of order $4$ are $pm(3+6x)$.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              I don't think there's a magic shortcut to this kind of question in general, but the specific cases we have here are simple enough that there are some tricks. For an element of order $5$, we can note that $5$ divides $11-1$, so there is actually an element of order $5$ in $mathbb{F}_{11}$, which is easy to find by trial and error. A bit more systematically, we can see that the subgroup of squares in $mathbb{F}_{11}^*$ is cyclic of order $5$ so any square besides $0$ and $1$ must have order $5$.



              For an element of order $4$, we can save some work by noticing that an element of order $4$ is just a square root of $-1$. So, we just need to solve for $a,binmathbb{F}_{11}$ such that $(a+bx)^2=-1$. Expanding out $(a+bx)^2$ using $x^2+x+7=0$ we get $$(a^2-7b^2)+(2ab-b^2)x=-1$$ and so we need $$a^2-7b^2=-1$$ and $$2ab-b^2=0.$$ The second equation gives $b=0$ or $2a=b$. The first case does not work (since there is no square root of $-1$ in $mathbb{F}_7$); in the second case the first equation simplifies to $$5a^2=1$$ which we can easily solve in $mathbb{F}_{11}$ to get $a=pm 3$. So, the elements of order $4$ are $pm(3+6x)$.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                I don't think there's a magic shortcut to this kind of question in general, but the specific cases we have here are simple enough that there are some tricks. For an element of order $5$, we can note that $5$ divides $11-1$, so there is actually an element of order $5$ in $mathbb{F}_{11}$, which is easy to find by trial and error. A bit more systematically, we can see that the subgroup of squares in $mathbb{F}_{11}^*$ is cyclic of order $5$ so any square besides $0$ and $1$ must have order $5$.



                For an element of order $4$, we can save some work by noticing that an element of order $4$ is just a square root of $-1$. So, we just need to solve for $a,binmathbb{F}_{11}$ such that $(a+bx)^2=-1$. Expanding out $(a+bx)^2$ using $x^2+x+7=0$ we get $$(a^2-7b^2)+(2ab-b^2)x=-1$$ and so we need $$a^2-7b^2=-1$$ and $$2ab-b^2=0.$$ The second equation gives $b=0$ or $2a=b$. The first case does not work (since there is no square root of $-1$ in $mathbb{F}_7$); in the second case the first equation simplifies to $$5a^2=1$$ which we can easily solve in $mathbb{F}_{11}$ to get $a=pm 3$. So, the elements of order $4$ are $pm(3+6x)$.






                share|cite|improve this answer









                $endgroup$



                I don't think there's a magic shortcut to this kind of question in general, but the specific cases we have here are simple enough that there are some tricks. For an element of order $5$, we can note that $5$ divides $11-1$, so there is actually an element of order $5$ in $mathbb{F}_{11}$, which is easy to find by trial and error. A bit more systematically, we can see that the subgroup of squares in $mathbb{F}_{11}^*$ is cyclic of order $5$ so any square besides $0$ and $1$ must have order $5$.



                For an element of order $4$, we can save some work by noticing that an element of order $4$ is just a square root of $-1$. So, we just need to solve for $a,binmathbb{F}_{11}$ such that $(a+bx)^2=-1$. Expanding out $(a+bx)^2$ using $x^2+x+7=0$ we get $$(a^2-7b^2)+(2ab-b^2)x=-1$$ and so we need $$a^2-7b^2=-1$$ and $$2ab-b^2=0.$$ The second equation gives $b=0$ or $2a=b$. The first case does not work (since there is no square root of $-1$ in $mathbb{F}_7$); in the second case the first equation simplifies to $$5a^2=1$$ which we can easily solve in $mathbb{F}_{11}$ to get $a=pm 3$. So, the elements of order $4$ are $pm(3+6x)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                Eric WofseyEric Wofsey

                182k13209337




                182k13209337






























                    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%2f3084188%2felement-of-given-order-in-a-finite-field%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

                    How to label and detect the document text images

                    Image
                    1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C
                    Read more

                    Vallis Paradisi

                    Image
                    Coordinata: .mw-parser-output .geo-default,.mw-parser-output .geo-dms,.mw-parser-output .geo-dec{display:inline}.mw-parser-output .geo-nondefault,.mw-parser-output .geo-multi-punct{display:none}.mw-parser-output .longitude,.mw-parser-output .latitude{white-space:nowrap} 33°03′S 71°37′W  /  33.05°S 71.617°W  / -33.05; -71.617 Vallis Paradisi Res apud Vicidata repertae: Civitas : Chilia Numerus incolarum : 296 655 Zona horaria : UTC-4 Situs interretialis Nomen officiale : Valparaíso Gubernium Consilium : municipal council of Valparaíso Geographia Superficies : 401 600 000 Metrum quadratum Coniunctiones urbium Urbes gemellae : Ovetum , Badalona , Santa Fe , Ora longa , Sciamhaevum , Quantunia , Callao , Bat Yam , Melaka , Novorossiysk , Busan , Corduba , Odessa , Rosario , Malacca , Barcino , Soteropolis , Franciscopolis (California) , Guanajuato, Silao Tabula aut despectus Despectus in portum sinumque et urbem Vallis Paradisi Vallis Para
                    Read more

                    Tabula Rosettana

                    Image
                    Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a
                    Read more