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Let $I = \{f(X) = \mathbb{C}[X] \mid f(0) = f(1) = f(−1) = 0\}$. Then $I$ is an ideal of $\mathbb{C}[X]$.

(iii) Show that all elements of $\mathbb{C}[X]/I$ can be written in the form $aX^2 + bX + c + I$, where $a,b,c \in \mathbb{C}$.

(iv) For $a_2,a_1,a_0,b_2,b_1,b_0 \in C$, show that $a_2X^2 +a_1X +a_0 +I = b_2X^2 +b_1X +b_0 +I$ if and only if $a_2 = b_2,\,\,a_1 = b_1$ and $a_0 = b_0$.

I am completely stuck on both. I do have solutions of these but I really need someone to explain how to do these or guide me to do these types of questions. Correct me if I am wrong but is $\mathbb{C}[X]/I=\{a(X)+I \, : \, a(X)\in\mathbb{C}[X]\}$?

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    $\begingroup$ why am I getting negged? $\endgroup$ – snowman Apr 13 '15 at 19:34
  • $\begingroup$ I have written the question correctly but is there a typo on f(X)=C[X]? $\endgroup$ – snowman Apr 13 '15 at 20:53
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Some hints.

Saying that a polynomial $f$ satisfies $f(a)=0$ is equivalent to saying it's divisible by $X-a$.

The polynomials is $I$ are all divisible by $X$, $X-1$ and $X+1$, hence by their product (why?). Conversely, a polynomial divisible by $X^3-X$ is in $I$ (why?).

Therefore $I$ is the ideal generated by $X^3-X$.

The rest is done by the division algorithm.


1. Characterize $I$

Polynomials in $I$ are divisible by $X$, $X-1$ and $X+1$, because they have $0$, $1$ and $-1$ as roots. Since these three polynomials are irreducible and coprime, any polynomial in $I$ is divisible by $f(X)=X(X-1)(X+1)=X^3-X$.

Conversely, any polynomial divisible by $f(X)$ has $0$, $1$ and $-1$ as roots.

Therefore $I=(f(X))$ is the set of polynomials divisible by $f(X)$.

2. A “canonical” form for the elements of the quotient ring

Let $a(X)\in\mathbb{C}[X]$. We want to write $a(X)+I$ in a better way. The division algorithm says we can find $q(X)$ and $r(X)$ such that $$ a(X)=q(X)f(X)+r(X) $$ and $r(X)=0$ or its degree is less than $3$. Since certainly $q(X)f(X)\in I$, we have that $$ a(X)+I=r(X)+I $$ because $a(X)-r(X)=q(X)f(X)\in I$.

Since the degree of $r(X)$ is less than $3$, we can write $r(X)=a_0+a_1X+a_2X^2$, so we have proved that any element in $\mathbb{C}[X]/I$ can be written in the form $$ a_0+a_1X+a_2X^2+I $$ for some $a_0,a_1,a_2\in\mathbb{C}$.

3. The representation is unique

Suppose $a_0+a_1X+a_2X^2+I=b_0+b_1X+b_2X^2+I$. By definition we must have $$ (a_0+a_1X+a_2X^2)-(b_0+b_1X+b_2X^2)= (a_0-b_0)+(a_1-b_1)X+(a_2-b_2)X^2\in I $$ which, by point 1, means $f(X)=X^3-X$ divides $(a_0-b_0)+(a_1-b_1)X+(a_2-b_2)X^2$. If this polynomial is non zero, it has degree less than $3$, so this can't happen; therefore the only possibility is that $$ (a_0-b_0)+(a_1-b_1)X+(a_2-b_2)X^2=0 $$ (the zero polynomial), that is, $$ a_0=b_0,\quad a_1=b_1,\quad a_2=b_2 $$

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    $\begingroup$ @snowman It could be worded better, but that's the idea. $\endgroup$ – egreg Apr 13 '15 at 20:08
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    $\begingroup$ Your $a(X)\in f(X)$ means nothing; the quotient ring is the set of objects of the form $a(X)+I$ where $a$ is any polynomial. So it need not satisfy $a(0)=a(1)=a(-1)=0$. $\endgroup$ – egreg Apr 13 '15 at 20:12
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    $\begingroup$ @snowman No. As I said, it means nothing. $\endgroup$ – egreg Apr 13 '15 at 20:38
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    $\begingroup$ @snowman If you want to do it complicately, yes. Easier: $a(x)-r(x)\in I$. $\endgroup$ – egreg Apr 13 '15 at 21:40
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    $\begingroup$ Since $f(X)\in I$, any element of the form $q(X)f(X)\in I$ (by definition of ideal). $\endgroup$ – egreg Apr 13 '15 at 21:48

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