Proof verification: Is 3 a square is every field $Bbb{F}_{p^2}$, where $p>3$ is a prime?
$begingroup$
I tried to prove it in this way:
Let $p(x) = x^2 - 3 in K[x]$, where $K = Bbb{F}_{p^2}$. Because $Bbb{F}_{p}$ is isomorphic to a subfield of $K$, we can make a reasoning in the following way:
If $p(x)$ is reducible in $Bbb{F}_{p}$, then we can write $p(x) = g(x)h(x)$, with $deg{g(x)} = deg{h(x)} = 1$. Then we, both $g(x), h(x)$ have a root in $Bbb{F}_p$, which is also a root of $p(x)$ in $K$.
If $p(x)$ is irreducible in $Bbb{F}_p$, then it has a root in $L = Bbb{F}_p/langle p(x)rangle$. Because $L = {a + bbar{x}: a, b in Bbb{F}_p}$ has $p^2$ elements, it is isomorphic to $K$. Therefore, $p(x)$ (or the image of it under a isomorphis) has a root in $K$.
Does this seem right?
field-theory finite-fields extension-field
asked Nov 30 '18 at 17:34
user480840user480840
61
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add a comment |
$begingroup$
I tried to prove it in this way:
Let $p(x) = x^2 - 3 in K[x]$, where $K = Bbb{F}_{p^2}$. Because $Bbb{F}_{p}$ is isomorphic to a subfield of $K$, we can make a reasoning in the following way:
If $p(x)$ is reducible in $Bbb{F}_{p}$, then we can write $p(x) = g(x)h(x)$, with $deg{g(x)} = deg{h(x)} = 1$. Then we, both $g(x), h(x)$ have a root in $Bbb{F}_p$, which is also a root of $p(x)$ in $K$.
If $p(x)$ is irreducible in $Bbb{F}_p$, then it has a root in $L = Bbb{F}_p/langle p(x)rangle$. Because $L = {a + bbar{x}: a, b in Bbb{F}_p}$ has $p^2$ elements, it is isomorphic to $K$. Therefore, $p(x)$ (or the image of it under a isomorphis) has a root in $K$.
Does this seem right?
field-theory finite-fields extension-field
asked Nov 30 '18 at 17:34
user480840user480840
61
$endgroup$
$begingroup$
This looks good, provided you already know that there is, up to isomorphism, only one field of cardinality $p^2$. (If you didn't know this, your proof would show only that, for each prime $p$, $3$ is a square ins some field of cardinality $p^2$.)
$endgroup$
– Andreas Blass
Nov 30 '18 at 18:44
add a comment |
$begingroup$
I tried to prove it in this way:
Let $p(x) = x^2 - 3 in K[x]$, where $K = Bbb{F}_{p^2}$. Because $Bbb{F}_{p}$ is isomorphic to a subfield of $K$, we can make a reasoning in the following way:
If $p(x)$ is reducible in $Bbb{F}_{p}$, then we can write $p(x) = g(x)h(x)$, with $deg{g(x)} = deg{h(x)} = 1$. Then we, both $g(x), h(x)$ have a root in $Bbb{F}_p$, which is also a root of $p(x)$ in $K$.
If $p(x)$ is irreducible in $Bbb{F}_p$, then it has a root in $L = Bbb{F}_p/langle p(x)rangle$. Because $L = {a + bbar{x}: a, b in Bbb{F}_p}$ has $p^2$ elements, it is isomorphic to $K$. Therefore, $p(x)$ (or the image of it under a isomorphis) has a root in $K$.
Does this seem right?
field-theory finite-fields extension-field
asked Nov 30 '18 at 17:34
user480840user480840
61
$endgroup$
I tried to prove it in this way:
Let $p(x) = x^2 - 3 in K[x]$, where $K = Bbb{F}_{p^2}$. Because $Bbb{F}_{p}$ is isomorphic to a subfield of $K$, we can make a reasoning in the following way:
If $p(x)$ is reducible in $Bbb{F}_{p}$, then we can write $p(x) = g(x)h(x)$, with $deg{g(x)} = deg{h(x)} = 1$. Then we, both $g(x), h(x)$ have a root in $Bbb{F}_p$, which is also a root of $p(x)$ in $K$.
If $p(x)$ is irreducible in $Bbb{F}_p$, then it has a root in $L = Bbb{F}_p/langle p(x)rangle$. Because $L = {a + bbar{x}: a, b in Bbb{F}_p}$ has $p^2$ elements, it is isomorphic to $K$. Therefore, $p(x)$ (or the image of it under a isomorphis) has a root in $K$.
Does this seem right?
field-theory finite-fields extension-field
field-theory finite-fields extension-field
asked Nov 30 '18 at 17:34
user480840user480840
61
asked Nov 30 '18 at 17:34
user480840user480840
61
asked Nov 30 '18 at 17:34
user480840user480840
61
asked Nov 30 '18 at 17:34
user480840user480840
61
asked Nov 30 '18 at 17:34
user480840user480840
61
61
$begingroup$
This looks good, provided you already know that there is, up to isomorphism, only one field of cardinality $p^2$. (If you didn't know this, your proof would show only that, for each prime $p$, $3$ is a square ins some field of cardinality $p^2$.)
$endgroup$
– Andreas Blass
Nov 30 '18 at 18:44
add a comment |
$begingroup$
This looks good, provided you already know that there is, up to isomorphism, only one field of cardinality $p^2$. (If you didn't know this, your proof would show only that, for each prime $p$, $3$ is a square ins some field of cardinality $p^2$.)
$endgroup$
– Andreas Blass
Nov 30 '18 at 18:44
$begingroup$
This looks good, provided you already know that there is, up to isomorphism, only one field of cardinality $p^2$. (If you didn't know this, your proof would show only that, for each prime $p$, $3$ is a square ins some field of cardinality $p^2$.)
$endgroup$
– Andreas Blass
Nov 30 '18 at 18:44
$begingroup$
This looks good, provided you already know that there is, up to isomorphism, only one field of cardinality $p^2$. (If you didn't know this, your proof would show only that, for each prime $p$, $3$ is a square ins some field of cardinality $p^2$.)
$endgroup$
– Andreas Blass
Nov 30 '18 at 18:44
add a comment |
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$begingroup$
This looks good, provided you already know that there is, up to isomorphism, only one field of cardinality $p^2$. (If you didn't know this, your proof would show only that, for each prime $p$, $3$ is a square ins some field of cardinality $p^2$.)
$endgroup$
– Andreas Blass
Nov 30 '18 at 18:44