Prove that the elements $1, t-a, (t-a)^2, (t-a)^3,\dots, (t-a)^{n-1}$ form a $\mathbb{C}$-basis for the quotient ring $\mathbb{C}[t]/((t-a)^n)$.

$((t-a)^n)$ is the ideal generated by $(t-a)^n$.

There is a similar proposition but I am not sure if it is related. Let $R$ be a ring and $f(x)$ be a monic polynomial with coefficients in $R$. Let $R[a]$ denote the ring obtained by adjoining an element satisfying the relation $f(a) = 0$. The powers $1,a,a^2, \dots, a^{n-1}$ form a basis for $R[a]$ over $R$. Note that the elements of $R[a]$ are of the form of a linear combination $r_0 + r_1a + ...r_{n-1}a^{n-1}$ where each $r_i\in R$

I know that $\mathbb{C}$ is a ring.

  • $\begingroup$ You can always express a polynomial in terms of powers of $t-a$ by using derivatives. $\endgroup$
    – Pedro Tamaroff
    Oct 17 '13 at 1:45
  • $\begingroup$ @sarah $P(t)=P(a)+P'(a)(t-a)+P''(a)\dfrac{(t-a)^2}{2!}+\cdots$? $\endgroup$
    – Pedro Tamaroff
    Oct 17 '13 at 1:52
  • $\begingroup$ I still dont quite see how this proves it. I do see that there is a relation. Could you explain more in detail your proof? $\endgroup$
    – sarah
    Oct 17 '13 at 2:26

I'll give you some hints. In the comments you'll find a different hint.

Hint 1: The definition of a basis is being linearly independent and spanning.

Hint 2: For linearly independent assume that $$a_0\cdot 1+a_1\cdot (1-a)+\dots+a_{n-1}(t-a)^{n-1}=0$$ in the quotient ring. This means that $$a_0\cdot 1+a_1\cdot (1-a)+\dots+a_{n-1}(t-a)^{n-1}=(t-a)^n(p(t))$$ in $\mathbb{C}[t]$. Now compare coefficients.

Hint 3: Divide an arbitrary polynomial $p(t)$ by $(t-a)^n$. What will be the result? Alternatively you can apply the hint by Pedro Tamaroff in the comments.


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