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Q11E.

Expert-verifiedFound in: Page 139

Book edition
3rd

Author(s)
Thomas W Hungerford, David Leep

Pages
608 pages

ISBN
9781111569624

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 $\mathbf{0}\mathbf{=}\mathbf{2}\xb7\mathbf{3}$ , and $\mathbf{3}{\mathbf{u}}^{\mathbf{2}}\mathbf{+}\mathbf{1}=0$. Derive a contradiction.]

It is proved that $p\left(x\right)=3{x}^{2}+1\in {\mathrm{\mathbb{Z}}}_{6}\left[x\right]$ has no roots in *K*.

**Exercise 3.2.17 **states that if $u$ is a unit in a ring $R$ with an identity, then $u$ will not be a **zero divisor**.

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

Moreover, $3\xb72=0$; therefore, 3 will be 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 to Theorem 5.11).

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

Hence, it is proved $p\left(x\right)=3{x}^{2}+1\in {\mathrm{\mathbb{Z}}}_{6}\left[x\right]$ has no roots in *K*.

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