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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!

user26857
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Robert
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  • You may find the technique I used in my answers to https://math.stackexchange.com/questions/1911270/every-element-of-mathbbrx-y-langle-x-x2y21-rangle-is-unit and https://math.stackexchange.com/questions/1854588/ker-phi-a-1-a-n-for-a-ring-homomorphism-phi-rx-1-x-n-t helpful. Basically, the trick is to reduce elements of the quotient ring to a nice form which allows you to show the induced map on the quotient ring is injective. – Eric Wofsey May 08 '17 at 05:40

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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.

Angina Seng
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