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Problem 1
(+) Let $G=\mathbb R[x]$ denote the group of all polynomials with real coefficients under addition.
(a) For any $f\in\mathbb R[x]$,define $\phi:\mathbb R[x]\rightarrow\mathbb R[x]$ by $\phi(f)=f'$,the derivative of $f$. Show that $\phi$ defines a homomorphism.
(b) Let $\int f$ denote the antiderivative of $f$ through the point (0,0). Show that the mapping $f\rightarrow\int f$ from $G$ to $G$ defines a homomorphism. What is the kernel of the this mapping?
(c) Is the map in (b) still a homomorphism if $\int f$ denotes the antiderivative of $f$ through the point (0,1)?
Solution
(a) $\phi:\mathbb R[x]\rightarrow\mathbb R[x]$ by $\phi(f)=f'$ for any $f\in\mathbb R[x]$.
Because $f$ is polynomial with real coefficients,$\forall f_1,f_2\in\mathbb R[x]$,let $f_1$ and $f_2$ be of the form $f_1=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$,
$f_2=b_nx^n+b_{n-1}x^{n-1}+\dots+b_1x+b_0$.
Then $f_1'=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\dots+a_1$,
$f_2'=nb_nx^{n-1}+(n-1)b_{n-1}x^{n-2}+\dots+b_1$.
$\phi(f_1+f_2)=(f_1+f_2)'=[(a_n+b_n)x^n+(a_{n-1}+b_{n-1}x^{n-1}+\dots+(a_1+b_1)x+(a_0+b_0)]'$ $=n(a_n+b_n)x^{n-1}+\dots+(a_1+b_1)=f_1'+f_2'=\phi(f_1)+\phi(f_2)$.So $\phi$ is a homomorphism.
(b) Let $\psi:G\rightarrow G$ by $\psi(f)=\int f$. Use $f_1,f_2$ declared in (a).
$\int f_1=\frac{a_n}{n+1}x^{n+1}+\frac{a_{n-1}}{n}x^n+\dots+\frac{a_1}{2}x^2+a_0x$,
$\int f_2=\frac{b_n}{n+1}x^{n+1}+\frac{b_{n-1}}{n}x^n+\dots+\frac{b_1}{2}x^2+b_0x$,
$\int (f_1+f_2)=\int (f_1+f_2)=\frac{a_n+b_n}{n+1}x^{n+1}+\frac{a_{n-1}+b_{n-1}}{n}x^n+\dots+\frac{a_1+b_1}{2}x^2+(a_0+b_0)x=\int f_1+\int f_2=\psi(f_1)+\psi(f_2)$
So this map defines a homomorphism.
The kernel is $ker\psi={0}$.
(c)Because $\int f$ goes through (0,1)
So $f_1=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$,
$f_2=b_nx^n+b_{n-1}x^{n-1}+\dots+b_1x+b_0$.
$\int f_1=\frac{a_n}{n+1}x^{n+1}+\frac{a_{n-1}}{n}x^n+\dots+\frac{a_1}{2}x^2+a_0x+1$,
$\int f_2=\frac{b_n}{n+1}x^{n+1}+\frac{b_{n-1}}{n}x^n+\dots+\frac{b_1}{2}x^2+b_0x+1$,
$\int (f_1+f_2)=\int (f_1+f_2)=\frac{a_n+b_n}{n+1}x^{n+1}+\frac{a_{n-1}+b_{n-1}}{n}x^n+\dots+\frac{a_1+b_1}{2}x^2+(a_0+b_0)x+1\neq\int(f_1)+\int(f_2)$
So it is not a homomorphism.