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This is problem 14 in Herstein's Topics in Algebra. I'm having trouble with the problem (working through the text independently).

For $F$ a field, define $V_n=\{p(x)\in F(x) : \deg p(x)<n, n\in \mathbb{N}\}$. Then $W=\{a_0+a_1x+\dots+a_{n-1}x^{n-1}\in F[x] : \sum_{i=0}^{n-1}a_i=0\}$ is a subspace of $V_n$ over $F$. Indeed, we can show for $\alpha, \beta \in F$ and $p(x),q(x)\in W$ that $\alpha p(x) + \beta q(x)\in W$.

Question: What is a basis for $W$?

My thoughts: I gave up trying to guess a basis for $W$. I thought instead to define the function $\phi : V_n\rightarrow F$ by $\phi (p(x)=a_0+a_1x+...+a_{n-1}x^{n-1})=\sum_{i=0}^{n-1} a_i$. Then $\phi$ is a surjective homomorphism. From this we obtain $V_n/ \ker \phi\cong F$. But $\dim (V_n/\ker \phi)=\dim (F) = 1$ and since $1=\dim(V_n/\ker \phi)=\dim (V_n)-\dim (\ker\phi)=n-\dim (\ker \phi)$ we have $\dim (\ker\phi)= n-1$. So I'm looking for a basis of $n-1$ elements. But that's as far as I've gotten.

I appreciate the help, thanks!

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The set $\{1-x^i\mid 1\leq i<n\}$ is a basis for this space. Linear independence is easy to see, and you have already computed the dimension of the space.

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  • $\begingroup$ Wow. Haha, Thanks! $\endgroup$ – Eoin Jan 27 '15 at 0:36
  • $\begingroup$ @Eoin sure thing. $\endgroup$ – Matt Samuel Jan 27 '15 at 0:36
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You can use the arithmetic of the euclidean ring $F[x]$: the defining condition means $p(1)=0$, which in turn means $p(x)$ is divisible by $x-1$. So $W=\{(x-1)q(x)\::\:q(x)\in F[x], \deg q < n-1\}$.

From this we see a basis for $W$ corresponds bijectively to a basis for polynomials of degree less than $n-1$. For instance, a basis could be $$\{x-1, x(x-1), x^2(x-1),\dots, x^{n-2}(x-1)\}.$$

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