Find all $c \in \Bbb Z_5$ such that $\Bbb Z_5[X]/ \langle x^2+cx+1 \rangle$ be a field.

Question

Attempts:

$\Bbb Z_5[X]/ \langle x^2+cx+1 \rangle$ be a field iff $p(x) = x^2 + cx +1$ is irreducible in $\Bbb Z_5$ iff $p(x)$ has no roots in $\Bbb Z_5$. I found that $2 \ne c \in \Bbb Z_5$ are the answer.

Am I true?

I don't see why it shouldn't mean $c\ne-2$, but yes, it means that $c\ne3$ is necessary. — , Nov 12 '20 at 07:21
As usual (see the linked dupe) you need only test if the discriminant $,c^2-4\equiv c^2+1,$ is a square $!\bmod 5,,$ i.e. is one of ${0,\pm1,\pm2}^2\equiv 0,1,4,,$ which is true $\iff c^2\not\equiv 1\ \ $ — Bill Dubuque, Nov 12 '20 at 09:29

score 1 · Answer 1 · answered Nov 12 '20 at 07:30

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Just make a little table of the values of $x^2+cx+1$, its values are $1,2+c,2c, -2c, 2-c$. Hence see that $c=\pm 2$ and $c=0$ give reducible quadratics.

answered Nov 12 '20 at 07:30

ancient mathematician

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$x^2+1$ has root ? – lap lapan Nov 12 '20 at 07:35
the answer is $c=1$ and $c=4$ ? – lap lapan Nov 12 '20 at 07:54

J. W. Tanner · Accepted Answer · 2020-11-12T08:26:06.773

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You are correct that $\Bbb Z_5[X]/ \langle x^2+cx+1 \rangle$ is a field iff $x^2+cx+1$ has no roots in $\Bbb Z_5$.

For $c=0$, $x^2+1\equiv x^2-4=(x+2)(x-2)$ is reducible.

For $c=1$, $x^2+x+1$ is $1,3,2,3,1$ for $x=0,1,2,3,4$, respectively, so there are no roots.

For $c=2$, $x^2+2x+1=(x+1)^2$ is reducible.

For $c=3$, $x^2+3x+1\equiv x^2-2x+1=(x-1)^2$ is reducible.

For $c=4$, $x^2+4x+1\equiv x^2-x+1$ is $1,1,3,2,3$ for $x=0,1,2,3,4$, respectively, so there are no roots.

edited Nov 12 '20 at 08:26

answered Nov 12 '20 at 08:24

J. W. Tanner

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nice same. thanks – lap lapan Nov 12 '20 at 08:25
i.e. it's solvable $\iff c^2\not\equiv 1,,$ as I explain in a comment on the question. – Bill Dubuque Nov 12 '20 at 09:30

Find all $c \in \Bbb Z_5$ such that $\Bbb Z_5[X]/ \langle x^2+cx+1 \rangle$ be a field.

2 Answers2