Lorena Beard

Answered

2022-06-29

We can assume that $p\ne 2,3$ because 0 is a root of $f(x)$ in ${\mathbb{F}}_{2}$ and ${\mathbb{F}}_{3}$. Let Q be the "squares subgroup" (?) of index 2 in ${\mathbb{F}}_{p}^{\ast}$. If 2 and 3 are not in Q (this means that $({x}^{2}-2)({x}^{2}-3)$ doesn't have roots in ${\mathbb{F}}_{p}$), then $2\cdot 3\in Q$ (?), and then ${x}^{2}-6$ has a root in ${\mathbb{F}}_{p}$

Answer & Explanation

Karla Hull

Expert

2022-06-30Added 20 answers

By the squares subgroup, they mean the subgroup

$Q=\{a\in {\mathbb{F}}_{p}^{\ast}\mid a={b}^{2}\text{for some}b\in {\mathbb{F}}_{p}^{\ast}\}.$

In other words, Q is the image of the homomorphism $\phi :{\mathbb{F}}_{p}^{\ast}\to {\mathbb{F}}_{p}^{\ast}$ given as $x\mapsto {x}^{2}$

In the below, I am also going to assume that $p\notin \{2,3\}$

Since $\mathrm{ker}\phi =\{1,-1\}$ and $-1\ne 1$ (why?), we see that G has index 2 in ${\mathbb{F}}_{p}^{\ast}$

Now, if $2,3,6\notin Q$, then this means that the cosets $2Q,3Q,6Q$ are all not equal to Q. Since Q has index 2 in ${\mathbb{F}}_{p}$, this means that $2Q=3Q=6Q$

Thus, one of $2,3,6$ must be in Q. In turn, the given polynomial has a root in ${\mathbb{F}}_{p}$ (in fact, in ${\mathbb{F}}_{p}^{\ast}$, if $p\ne 2,3$)

$Q=\{a\in {\mathbb{F}}_{p}^{\ast}\mid a={b}^{2}\text{for some}b\in {\mathbb{F}}_{p}^{\ast}\}.$

In other words, Q is the image of the homomorphism $\phi :{\mathbb{F}}_{p}^{\ast}\to {\mathbb{F}}_{p}^{\ast}$ given as $x\mapsto {x}^{2}$

In the below, I am also going to assume that $p\notin \{2,3\}$

Since $\mathrm{ker}\phi =\{1,-1\}$ and $-1\ne 1$ (why?), we see that G has index 2 in ${\mathbb{F}}_{p}^{\ast}$

Now, if $2,3,6\notin Q$, then this means that the cosets $2Q,3Q,6Q$ are all not equal to Q. Since Q has index 2 in ${\mathbb{F}}_{p}$, this means that $2Q=3Q=6Q$

Thus, one of $2,3,6$ must be in Q. In turn, the given polynomial has a root in ${\mathbb{F}}_{p}$ (in fact, in ${\mathbb{F}}_{p}^{\ast}$, if $p\ne 2,3$)

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