Question about constructing fields












1












$begingroup$



Find all the monic irreducible polynomials in $F_5[x]$ of degree two (aside
from $x^2-2$ and $x^2-3$, there are eight of them) Adjoining a root u of
these polynomials to $F_5$, construct eight fields $F_5(u)$ of $25$ elements.
Prove that each of these fields is isomorphic to $F_5(sqrt 2)^times$




Apart from $x^2-2$ and $x^2-3$, I've found that there are $8$ monic irreducible polynomials in $F_5[x]$, which are:
$x^2+x+1,,, x^2+x+2,,, x^2+2x+3,,, x^2+2x+4,,, x^2+3x+3,,, x^2+3x+4,,, x^2+4x+1,,, x^2+4x+2$



For example for the polynomial $x^2+x+1$ we have
$F_5[u]=F_5[x]/x^2+x+1$ where we identify $u$ with the image of $x $ in $F_5[u]$



Now, how to construct such a field of $25$ elements and show that this field is isomorphic to $F_5(sqrt 2)^times$?






field-theory irreducible-polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 19:10


















1












$begingroup$



Find all the monic irreducible polynomials in $F_5[x]$ of degree two (aside
from $x^2-2$ and $x^2-3$, there are eight of them) Adjoining a root u of
these polynomials to $F_5$, construct eight fields $F_5(u)$ of $25$ elements.
Prove that each of these fields is isomorphic to $F_5(sqrt 2)^times$




Apart from $x^2-2$ and $x^2-3$, I've found that there are $8$ monic irreducible polynomials in $F_5[x]$, which are:
$x^2+x+1,,, x^2+x+2,,, x^2+2x+3,,, x^2+2x+4,,, x^2+3x+3,,, x^2+3x+4,,, x^2+4x+1,,, x^2+4x+2$



For example for the polynomial $x^2+x+1$ we have
$F_5[u]=F_5[x]/x^2+x+1$ where we identify $u$ with the image of $x $ in $F_5[u]$



Now, how to construct such a field of $25$ elements and show that this field is isomorphic to $F_5(sqrt 2)^times$?






field-theory irreducible-polynomials







share|cite|improve this question









$endgroup$












  • $begingroup$
    All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 19:10
















1












1








1





$begingroup$



Find all the monic irreducible polynomials in $F_5[x]$ of degree two (aside
from $x^2-2$ and $x^2-3$, there are eight of them) Adjoining a root u of
these polynomials to $F_5$, construct eight fields $F_5(u)$ of $25$ elements.
Prove that each of these fields is isomorphic to $F_5(sqrt 2)^times$




Apart from $x^2-2$ and $x^2-3$, I've found that there are $8$ monic irreducible polynomials in $F_5[x]$, which are:
$x^2+x+1,,, x^2+x+2,,, x^2+2x+3,,, x^2+2x+4,,, x^2+3x+3,,, x^2+3x+4,,, x^2+4x+1,,, x^2+4x+2$



For example for the polynomial $x^2+x+1$ we have
$F_5[u]=F_5[x]/x^2+x+1$ where we identify $u$ with the image of $x $ in $F_5[u]$



Now, how to construct such a field of $25$ elements and show that this field is isomorphic to $F_5(sqrt 2)^times$?






field-theory irreducible-polynomials







share|cite|improve this question









$endgroup$





Find all the monic irreducible polynomials in $F_5[x]$ of degree two (aside
from $x^2-2$ and $x^2-3$, there are eight of them) Adjoining a root u of
these polynomials to $F_5$, construct eight fields $F_5(u)$ of $25$ elements.
Prove that each of these fields is isomorphic to $F_5(sqrt 2)^times$




Apart from $x^2-2$ and $x^2-3$, I've found that there are $8$ monic irreducible polynomials in $F_5[x]$, which are:
$x^2+x+1,,, x^2+x+2,,, x^2+2x+3,,, x^2+2x+4,,, x^2+3x+3,,, x^2+3x+4,,, x^2+4x+1,,, x^2+4x+2$



For example for the polynomial $x^2+x+1$ we have
$F_5[u]=F_5[x]/x^2+x+1$ where we identify $u$ with the image of $x $ in $F_5[u]$



Now, how to construct such a field of $25$ elements and show that this field is isomorphic to $F_5(sqrt 2)^times$?






field-theory irreducible-polynomials




field-theory irreducible-polynomials






share|cite|improve this question













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asked Dec 10 '18 at 19:05









Leyla AlkanLeyla Alkan

1,5751724




1,5751724












  • $begingroup$
    All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 19:10




















  • $begingroup$
    All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 19:10


















$begingroup$
All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
$endgroup$
– Robert Israel
Dec 10 '18 at 19:10






$begingroup$
All fields of the same finite cardinality are isomorphic. Or are you trying to construct an explicit isomorphism?
$endgroup$
– Robert Israel
Dec 10 '18 at 19:10












1 Answer
1






active

oldest

votes


















0












$begingroup$

For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[sqrt{2}]$.
If $alpha^2 = 2$ and $beta = x alpha + y$ we have
$$beta^2 + beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[sqrt{2}]$. Since the cardinalities are equal it's also onto.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't know Herstein's version of the term "isomorphism" was widely used.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 20:04










  • $begingroup$
    I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 20:56











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1 Answer
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active

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$begingroup$

For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[sqrt{2}]$.
If $alpha^2 = 2$ and $beta = x alpha + y$ we have
$$beta^2 + beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[sqrt{2}]$. Since the cardinalities are equal it's also onto.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't know Herstein's version of the term "isomorphism" was widely used.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 20:04










  • $begingroup$
    I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 20:56
















0












$begingroup$

For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[sqrt{2}]$.
If $alpha^2 = 2$ and $beta = x alpha + y$ we have
$$beta^2 + beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[sqrt{2}]$. Since the cardinalities are equal it's also onto.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I didn't know Herstein's version of the term "isomorphism" was widely used.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 20:04










  • $begingroup$
    I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 20:56














0












0








0





$begingroup$

For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[sqrt{2}]$.
If $alpha^2 = 2$ and $beta = x alpha + y$ we have
$$beta^2 + beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[sqrt{2}]$. Since the cardinalities are equal it's also onto.






share|cite|improve this answer









$endgroup$



For example, let's see that there is a root of $x^2 + x + 1$ in $F_5[sqrt{2}]$.
If $alpha^2 = 2$ and $beta = x alpha + y$ we have
$$beta^2 + beta + 1 = (2 x^2 + y^2 + y + 1) + (2 x y + x)alpha = 0$$ if $x=y=2$. So we get an isomorphism from $F_5[x]/(x^2+x+1)$ into $F_5[sqrt{2}]$. Since the cardinalities are equal it's also onto.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 19:22









Robert IsraelRobert Israel

329k23217470




329k23217470












  • $begingroup$
    I didn't know Herstein's version of the term "isomorphism" was widely used.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 20:04










  • $begingroup$
    I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 20:56


















  • $begingroup$
    I didn't know Herstein's version of the term "isomorphism" was widely used.
    $endgroup$
    – Matt Samuel
    Dec 10 '18 at 20:04










  • $begingroup$
    I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
    $endgroup$
    – Robert Israel
    Dec 10 '18 at 20:56
















$begingroup$
I didn't know Herstein's version of the term "isomorphism" was widely used.
$endgroup$
– Matt Samuel
Dec 10 '18 at 20:04




$begingroup$
I didn't know Herstein's version of the term "isomorphism" was widely used.
$endgroup$
– Matt Samuel
Dec 10 '18 at 20:04












$begingroup$
I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
$endgroup$
– Robert Israel
Dec 10 '18 at 20:56




$begingroup$
I come by it honestly: Herstein taught my first abstract algebra course, back in 1969.
$endgroup$
– Robert Israel
Dec 10 '18 at 20:56


















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