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Question

Prove that if F is a subfield of a field E, and cE,c \in E, then θ:F[x]E\theta: F[x] \rightarrow E defined by θ(f(x))=f(c)\theta(f(x))= f(c) is a homomorphism.

Solution

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F\mathbb{F} is a subfield of E\mathbb{E} and cEc \in \mathbb{E}. θ:F[x]E\theta: \mathbb{F}[x] \rightarrow \mathbb{E} is defined by θ(f(x))=f(c)\theta(f(x))=f(c). This map makes sense because θ\theta is the evaluation map and f(c)Ef(c) \in \mathbb{E}. For f(x),g(x)F[x]f(x),g(x) \in \mathbb{F}[x] we also have f(x),g(x)E[x]f(x),g(x) \in \mathbb{E}[x] and hence by the definitions of addition and multiplication in E[x]\mathbb{E}[x]

θ((f+g)(x))=(f+g)(c)=f(c)+g(c)=θ(f(x))+θ(g(x))\begin{align*} \theta((f+g)(x)) &= (f+g)(c) \\ &=f(c)+g(c) \\ &=\theta(f(x))+\theta(g(x)) \end{align*}

and also

θ((fg)(x))=(fg)(c)=f(c)g(c)=θ(f(x))θ(g(x))\begin{align*} \theta((fg)(x)) &= (fg)(c) \\ &=f(c)g(c) \\ &=\theta(f(x))\theta(g(x)) \end{align*}

Therefore, θ\theta is a homomorphism.

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