Find the $Ker(\theta)$ of the given ring homomorphism $\theta :$ $K[x,y]$$\to$ $K[t]$ a ring homomorphism such that
$x$$\to$$t^a$  and
$y$$\to$$t^b$ given, $gcd(a,b)=1$
Prove that $Ker(\theta)=\langle y^a-x^b\rangle$ .
 A: Any element $p(x,y)$ $\epsilon$ $K[x,y]$ such that $p(x,y)$ $\epsilon$ $\langle y^a-x^b\rangle$ implies $p(x,y)=(y^a-x^b)q(x,y)$, where $q(x,y)$ $\epsilon$ $K[x,y]$, then we can easily say $\theta(p(x,y))=0$ ,
Hence $p(x,y)$ $\epsilon$ $Ker(\theta)$, so, $\langle y^a-x^b\rangle$ $\subseteq$ $Ker(\theta)$
Now conversely, say any element $p(x,y)$ $\epsilon$ $Ker(\theta)$.  So, $\theta(p(x,y))=0$
Say,  $p(x,y)=(y^a-x^b)q(x,y)+\sum_{i=1}^a P_i (x) y^{a-i}$ , where $P_i (x)$ $\epsilon$ $K[x]$. Since $\theta(p(x,y))=0$ , so, $\theta(\sum_{i=1}^a P_i (x) y^{a-i})=0$
$\to$ $\sum_{i=1}^a P_i (t^a) (t^b)^{a-i}=0$ .
Now suppose, $P_i(x)=a_{0_{i}}+a_{1_{i}}x+....+a_{n_{i_{i}}}x^{n_{i}}$, where $a_{n_{i}}\neq0$.
So, now $\theta(P_i (x) y^{a-i})=$ $P_i (t^a) (t^b)^{a-i}=[a_{0_{i}}+a_{1_{i}}(t^a)+....+a_{n_{i_{i}}}(t^a)^{n_{i}}]\times(t^b)^{a-i}$
$=a_{0_{i}}t^{(ab-ib)}+a_{1_{i}}t^{(a+ab-ib)}+....+a_{n_{i_{i}}}t^{(a{n_{i}+ab-ib)}}$
$=\sum_{j=0}^{n_i}a_{j_i}t^{(ja+ab-ib)}$, since we have $\sum_{i=1}^a P_i (t^a) (t^b)^{a-i}=0$
$\to$ $\sum_{i=0}^a\sum_{j=0}^{n_i}a_{j_i}t^{(ja+ab-ib)}=0$, now all these $\sum_{j=0}^{n_i}a_{j_i}t^{(ja+ab-ib)}$ are polynomials of $t$ having degree $(a{n_{i}+ab-ib})$ i.e  of different degree. So the sum of these polynomials is equals to $0$ implies all the polynomials are individually equals to $0$. So, $\sum_{j=0}^{n_i}a_{j_i}t^{(ja+ab-ib)}=0$ where, $1\leq i \leq a$ . This implies that $\sum_{j=0}^{n_i}a_{j_i}t^{ja}=0$, since $t^b=y$ is a variable then $t^{b(a-i)}$$\neq0$ for non zero $t$. So,$\sum_{j=0}^{n_i}a_{j_i}(t^{a})^j=0$
$\to$ $\sum_{j=0}^{n_i}a_{j_i}(x)^j=0$ , [here $t^a=x$ so replacing]
$\to$ $a_{0_{i}}+a_{1_{i}}x+....+a_{n_{i_{i}}}x^{n_{i}}=0$
$\to$ $P_i(x)=0$ , where $1\leq i \leq a$. So, we get, $p(x,y)=(y^a-x^b)q(x,y)$, hence $p(x,y)$ $\epsilon$ $\langle y^a-x^b\rangle$ and
So $Ker(\theta)$ $\subseteq$ $\langle y^a-x^b\rangle$
So, $Ker(\theta)=\langle y^a-x^b\rangle$ (Hence proved).
