Question

 Find all the units of equation. equation  Check whether or not equation is a field. equation 

10 Jan 2026
Answer :
Word Count : 418
The ring (\mathbb{Z}[\sqrt{-7}]) consists of elements of the form (a + b\sqrt{-7}), where (a, b \in \mathbb{Z}). An element (u = a + b\sqrt{-7}) is a unit if it has a multiplicative inverse in the ring. This is equivalent to saying there exists (v = c + d\sqrt{-7} \in \mathbb{Z}[\sqrt{-7}]) such that (uv = 1). Compute the norm (N(a + b\sqrt{-7}) = a^2 + 7b^2). The key property: if (uv = 1), then (N(u)N(v) = N(uv) = N(1) = 1). So the norm must satisfy (N(a + b\sqrt{-7}) = a^2 + 7b^2 = 1). Now solve (a^2 + 7b^2 = 1) in integers: * If (b = 0), then (a^2 = 1 \implies a = \pm 1). * If (b \neq 0), then (7b^2 \ge 7 > 1), impossible. Thus the only solutions are (a = \pm 1, b = 0). Units in (\mathbb{Z}[\sqrt{-7}]): [ \boxed{\pm 1}. ] --- Next, consider (\mathbb{Q}[x]/\langle 8x^3 + 6x^2 - 9x + 24 \rangle). A quotient ring (\mathbb{Q}[x]/\langle f(x) \rangle) is
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