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Let $S=\{f \in P_n(F) : f(1)=0\}$. Clearly, the polynomial $f(x)=0 \in S$ because $f(c)=0$ for any choice of $c\in F$. To demonstrate closure under addition and multiplication consider the fact that $cf(1)+g(1)=c\cdot 0+0=(cf+g)(1)=0$ for $f,g\in S$. To find dimension of $P_n(F)$, consider the linear transformation $T:P_n \rightarrow F$ defined by $T(f(x))=f(1)$, namely evaluating a polynomial at $1$ is a linear transformation from $P_n$ to $F$: \begin{eqnarray} T(f(x)+cg(x))=T(f(x))+cT(g(x))=f(1)+cg(1) \end{eqnarray} Thus, $S$ is the null space of $T$, and from Lemma 1 and Lemma 2 (don't worry about these) we see that \begin{eqnarray} \dim R(T) + \dim N(T) = \dim P_n(F) \leadsto 1 + \dim N(T) = n+1 \leadsto \dim N(T) = n \end{eqnarray}

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We all are guessing that $\,P_n(F)\,$ is the linear (vector) space of all the polynomials of degree not more than $\,n\,$ over some field $\,F\,$ ...Yet I find it odd that in two cases to show linearity you choose to show $\,cf+g\;,\;\;c\in F\;,\;f,g\in P_n(F)\;$ , instead of $\,cf+dg\,$...why? –  DonAntonio Apr 18 '13 at 2:23
    
BTW, basically your proof is correct. –  DonAntonio Apr 18 '13 at 2:23
    
Wait, what is the problem? –  Trancot Apr 18 '13 at 3:43
    
@DonAntonio I was just using the general idea expounded on page 35 of Hoffmam's "Linear Algebra," namely Theorem 1 on that page. –  Trancot Apr 18 '13 at 3:45
    
Indeed so, @Trancot. Never saw it that way and it looks odd to me. –  DonAntonio Apr 18 '13 at 3:56
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Yes, your proof is correct. But the first part is superfluous. Define $T: P_n(F)\to F$ by $f\mapsto f(1)$. In your question you checked that $T$ is linear. Then as you noted, $S$ is the null space (sometimes also called kernel) of $T$. As the null space of a linear map it is thus a subspace of $P_n(F)$.

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Yes, your proof is correct. Alternatively, note that $\{1 - x, 1 - x^2, \dots , 1 - x^n\} \subseteq S$ is linearly independent. Since it's clear that $S \ne P_n(F)$, it follows that $\dim S = n$.

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