Bourbaki

Exercise 8, § 2

Algebra, IV, p. 88

Let BB be the additive group Z⊕(Z/2Z)\mathbf{Z} \oplus (\mathbf{Z}/2\mathbf{Z}), with the commutative ring structure obtained by putting (a,b)⋅(a′,b′)=(aa′,ab′+ba′)(a, b) \cdot (a', b') = (aa', ab' + ba') for (a,b)∈B(a, b) \in B and (a′,b′)∈B(a', b') \in B. Let ε=(0,1)∈B\varepsilon = (0, 1) \in B, and f(X)=εX(X−1)∈B[X]f(X) = \varepsilon X(X-1) \in B[X]. Then f(α)=0f(\alpha) = 0 for all α∈B\alpha \in B.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.