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Can it happen that an irreducible degree $2$ polynomial becomes a perfect square in the algebraic closure?
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Question. Is it possible to find a field $K$ together with a degree-$2$ polynomial $P in K[x]$ such that $P$ is irreducible, but $P = Q^2$ for some $Q in overlineK[x]$?
This can't occur if $K = mathbbR$ due to the rational root theorem. I thought possibly that $x^2+x+1 in mathbbF_2[x]$ would be an example, since if $alpha$ is a root, then so too is $-alpha$. However, when I expand out $(x-alpha)^2$ I don't get $x^2+x+1$. Factorizing it carefully, we find that $$x^2+x+1 = (x-alpha)(x-(alpha+1)),$$ so this is not an example.
Ideas, anyone?
polynomials field-theory irreducible-polynomials
asked Jul 17 at 12:46
goblin
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up vote
2
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favorite
Question. Is it possible to find a field $K$ together with a degree-$2$ polynomial $P in K[x]$ such that $P$ is irreducible, but $P = Q^2$ for some $Q in overlineK[x]$?
This can't occur if $K = mathbbR$ due to the rational root theorem. I thought possibly that $x^2+x+1 in mathbbF_2[x]$ would be an example, since if $alpha$ is a root, then so too is $-alpha$. However, when I expand out $(x-alpha)^2$ I don't get $x^2+x+1$. Factorizing it carefully, we find that $$x^2+x+1 = (x-alpha)(x-(alpha+1)),$$ so this is not an example.
Ideas, anyone?
polynomials field-theory irreducible-polynomials
asked Jul 17 at 12:46
goblin
35.5k1153181
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Question. Is it possible to find a field $K$ together with a degree-$2$ polynomial $P in K[x]$ such that $P$ is irreducible, but $P = Q^2$ for some $Q in overlineK[x]$?
This can't occur if $K = mathbbR$ due to the rational root theorem. I thought possibly that $x^2+x+1 in mathbbF_2[x]$ would be an example, since if $alpha$ is a root, then so too is $-alpha$. However, when I expand out $(x-alpha)^2$ I don't get $x^2+x+1$. Factorizing it carefully, we find that $$x^2+x+1 = (x-alpha)(x-(alpha+1)),$$ so this is not an example.
Ideas, anyone?
polynomials field-theory irreducible-polynomials
asked Jul 17 at 12:46
goblin
35.5k1153181
Question. Is it possible to find a field $K$ together with a degree-$2$ polynomial $P in K[x]$ such that $P$ is irreducible, but $P = Q^2$ for some $Q in overlineK[x]$?
This can't occur if $K = mathbbR$ due to the rational root theorem. I thought possibly that $x^2+x+1 in mathbbF_2[x]$ would be an example, since if $alpha$ is a root, then so too is $-alpha$. However, when I expand out $(x-alpha)^2$ I don't get $x^2+x+1$. Factorizing it carefully, we find that $$x^2+x+1 = (x-alpha)(x-(alpha+1)),$$ so this is not an example.
Ideas, anyone?
polynomials field-theory irreducible-polynomials
asked Jul 17 at 12:46
goblin
35.5k1153181
asked Jul 17 at 12:46
goblin
35.5k1153181
asked Jul 17 at 12:46
goblin
35.5k1153181
asked Jul 17 at 12:46
goblin
35.5k1153181
35.5k1153181
add a comment |Â
add a comment |Â
2 Answers
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The standard example is $X^2 - T$ with $K = mathbbF_2(T)$.
It factors as $(X - sqrtT)^2$ in an extension containing $sqrtT$.
In order for this to work, you need the base field to not be perfect, which does not happen with finite extensions of $mathbbQ$, $mathbbR$ or $mathbbC$ (because they have characteristic $0$) or finite fields.
One of the simplest examples of a field which is not perfect is the function field in one variable $mathbbF_p(T)$, and we get similar examples of repeated roots for polynomials over these fields.
answered Jul 17 at 12:50
Tob Ernack
2,149318
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up vote
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This can happen in characteristic two. Let $F$ be a field of characteristic
two, and $K=F(X)$, the field of rational functions in $X$ over $F$. Then
take $P(t)=t^2-X$. This is irreducible, yet over the algebraic closure
it equals $(t-Y)^2$ where $Y$ is a square root of $X$.
This phenomenon is called inseparability.
answered Jul 17 at 12:51
Lord Shark the Unknown
85.5k951112
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The standard example is $X^2 - T$ with $K = mathbbF_2(T)$.
It factors as $(X - sqrtT)^2$ in an extension containing $sqrtT$.
In order for this to work, you need the base field to not be perfect, which does not happen with finite extensions of $mathbbQ$, $mathbbR$ or $mathbbC$ (because they have characteristic $0$) or finite fields.
One of the simplest examples of a field which is not perfect is the function field in one variable $mathbbF_p(T)$, and we get similar examples of repeated roots for polynomials over these fields.
answered Jul 17 at 12:50
Tob Ernack
2,149318
add a comment |Â
up vote
5
down vote
accepted
The standard example is $X^2 - T$ with $K = mathbbF_2(T)$.
It factors as $(X - sqrtT)^2$ in an extension containing $sqrtT$.
In order for this to work, you need the base field to not be perfect, which does not happen with finite extensions of $mathbbQ$, $mathbbR$ or $mathbbC$ (because they have characteristic $0$) or finite fields.
One of the simplest examples of a field which is not perfect is the function field in one variable $mathbbF_p(T)$, and we get similar examples of repeated roots for polynomials over these fields.
answered Jul 17 at 12:50
Tob Ernack
2,149318
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The standard example is $X^2 - T$ with $K = mathbbF_2(T)$.
It factors as $(X - sqrtT)^2$ in an extension containing $sqrtT$.
In order for this to work, you need the base field to not be perfect, which does not happen with finite extensions of $mathbbQ$, $mathbbR$ or $mathbbC$ (because they have characteristic $0$) or finite fields.
One of the simplest examples of a field which is not perfect is the function field in one variable $mathbbF_p(T)$, and we get similar examples of repeated roots for polynomials over these fields.
answered Jul 17 at 12:50
Tob Ernack
2,149318
The standard example is $X^2 - T$ with $K = mathbbF_2(T)$.
It factors as $(X - sqrtT)^2$ in an extension containing $sqrtT$.
In order for this to work, you need the base field to not be perfect, which does not happen with finite extensions of $mathbbQ$, $mathbbR$ or $mathbbC$ (because they have characteristic $0$) or finite fields.
One of the simplest examples of a field which is not perfect is the function field in one variable $mathbbF_p(T)$, and we get similar examples of repeated roots for polynomials over these fields.
answered Jul 17 at 12:50
Tob Ernack
2,149318
edited Jul 17 at 12:58
answered Jul 17 at 12:50
Tob Ernack
2,149318
answered Jul 17 at 12:50
Tob Ernack
2,149318
answered Jul 17 at 12:50
Tob Ernack
2,149318
2,149318
add a comment |Â
add a comment |Â
up vote
2
down vote
This can happen in characteristic two. Let $F$ be a field of characteristic
two, and $K=F(X)$, the field of rational functions in $X$ over $F$. Then
take $P(t)=t^2-X$. This is irreducible, yet over the algebraic closure
it equals $(t-Y)^2$ where $Y$ is a square root of $X$.
This phenomenon is called inseparability.
answered Jul 17 at 12:51
Lord Shark the Unknown
85.5k951112
add a comment |Â
up vote
2
down vote
This can happen in characteristic two. Let $F$ be a field of characteristic
two, and $K=F(X)$, the field of rational functions in $X$ over $F$. Then
take $P(t)=t^2-X$. This is irreducible, yet over the algebraic closure
it equals $(t-Y)^2$ where $Y$ is a square root of $X$.
This phenomenon is called inseparability.
answered Jul 17 at 12:51
Lord Shark the Unknown
85.5k951112
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This can happen in characteristic two. Let $F$ be a field of characteristic
two, and $K=F(X)$, the field of rational functions in $X$ over $F$. Then
take $P(t)=t^2-X$. This is irreducible, yet over the algebraic closure
it equals $(t-Y)^2$ where $Y$ is a square root of $X$.
This phenomenon is called inseparability.
answered Jul 17 at 12:51
Lord Shark the Unknown
85.5k951112
This can happen in characteristic two. Let $F$ be a field of characteristic
two, and $K=F(X)$, the field of rational functions in $X$ over $F$. Then
take $P(t)=t^2-X$. This is irreducible, yet over the algebraic closure
it equals $(t-Y)^2$ where $Y$ is a square root of $X$.
This phenomenon is called inseparability.
answered Jul 17 at 12:51
Lord Shark the Unknown
85.5k951112
answered Jul 17 at 12:51
Lord Shark the Unknown
85.5k951112
answered Jul 17 at 12:51
Lord Shark the Unknown
85.5k951112
answered Jul 17 at 12:51
Lord Shark the Unknown
85.5k951112
85.5k951112
add a comment |Â
add a comment |Â
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