Question

Let K be the ring that contains ${\mathrm{\mathbb{Z}}}_6$ as a subring. Show that $\mathbf{p}\left(x\right)=\mathbf{3}{x}^{\mathbf{2}}\mathbf{+}\mathbf{1}\in {\mathrm{\mathbb{Z}}}_{\mathbf{6}}\left[x\right]$has no roots in K. Thus, Corollary 5.12 may be false if F is not a field. [Hint: If u were a root, then $\mathbf{0}\mathbf{=}\mathbf{2}·\mathbf{3}$, and $\mathbf{3}{\mathbf{u}}^{\mathbf{2}}\mathbf{+}\mathbf{1}=0$. Derive a contradiction.]

Short answer

It is proved that$p\left(x\right)=3x^2+1\in {\mathrm{\mathbb{Z}}}_6\left[x\right]$ has no roots in K.

Step-by-step solution

Statement of Exercise 3.2.17

Exercise 3.2.17states that if $u$isa unit in a ring$R$ with an identity, then $u$willnot be a zero divisor.

Show that px=3x2+1∈ℤ6x has no roots in K

Assume that$u$is a root of $p\left(x\right)=3x^2+1$; therefore, $3u^2+1=0$. There is $1=3\left(-u^2\right)$, andas a result, 3 is a unit in $K$.

Moreover, $3·2=0$; therefore, 3willbe a zero divisor in $K$, and $\raisebox{1ex}{$\mathrm{\mathbb{Q}}\left[x\right]$}\!\left/ \!\raisebox{-1ex}{$\left(x^2-3\right)$}\right.\cong \mathrm{\mathbb{Q}}\left(\sqrt{3}\right)$.

It is also observed that$x^2-3$contains a root in$\mathrm{\mathbb{Q}}\left(\sqrt{3}\right)$(according toTheorem5.11).

This contradictsthe result that units in a ring with unity are not zero divisors.

Hence, it is proved$p\left(x\right)=3x^2+1\in {\mathrm{\mathbb{Z}}}_6\left[x\right]$ has no roots in K.