Too many times I've seen the following type of isomorphism:
Let $K[x,y]$ where $K$ is a field, and if we consider the quotient of $K[x,y]/(y^3 - x^5)$ for example, then we have $K[x,y]/(y^3 - x^5) \cong K[t^3,t^5]$ where the isomorphism is $x \mapsto t^3$ and $y \mapsto t^5$.
Clearly, $x \mapsto t^3$ and $y \mapsto t^5$ defines a $K$-algebra homomorphism from $K[x,y]$ to $K[t^3,t^5]$, and it is also clear why $(y^3 - x^5)$ is included in the kernel. However, I do not understand why the kernel is included in $(y^3 - x^5)$? Please if you can help me with this, I'll really appreciate it. Thanks!
We certainly have a homomorphism from $K[x,y]/(y^3-x^5)$ to $K[t^3,t^5]$. To show this is an isomorphism we must show there's an inverse homomorphism. For any $n\in\Bbb Z$ such that $n=3a+5b$ has a nonnegative solution just pick one $(a_n,b_n)$. Define $\phi$ on $K[t^3,t^5]$ by $\phi:t^n\mapsto x^{a_n}y^{b_n}+(y^3-x^5)$. All we need to show this defines a homomorphism is that $x^{a_m+a_n}y^{a_m+b_n}\equiv x^{a_{m+n}}y^{b_{m+n}}\in(y^3-x^5)$ for all $m$, $n$. This amounts to $x^ay^b-x^{a'}y^{b'}\in(y^3-x^5)$ whenever $3a+5b=3a'+5b'$ which is straightforward. It's then clear this is the inverse homomorphism we need.