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Expert-verified Found in: Page 139 ### Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624 # Let K be the ring that contains ${\mathrm{ℤ}}_{6}$ as a subring. Show that $\mathbf{p}\left(x\right)=\mathbf{3}{x}^{\mathbf{2}}\mathbf{+}\mathbf{1}\in {\mathrm{ℤ}}_{\mathbf{6}}\left[x\right]$ 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.]

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

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## Statement of Exercise 3.2.17

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.

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

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, $3·2=0$; therefore, 3 will be a zero divisor in $K$, and $\mathrm{ℚ}\left[x\right]}{\left({x}^{2}-3\right)}\cong \mathrm{ℚ}\left(\sqrt{3}\right)$.

It is also observed that ${x}^{2}-3$ contains a root in $\mathrm{ℚ}\left(\sqrt{3}\right)$ (according to Theorem 5.11).

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

Hence, it is proved $p\left(x\right)=3{x}^{2}+1\in {\mathrm{ℤ}}_{6}\left[x\right]$ has no roots in K. ### Want to see more solutions like these? 