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Suppose $R=k[x,y,z]$ and $S=k[t]$. Consider the map $f:R\to S$ s.t. $f(x)=t$, $f(y)=t^2$ and $f(z)=t^3$. I suspect the kernel of this map is the ideal $(y-x^2,z-x^3)R$.

It's clearly contained in the kernel, but I am not sure how to prove the reverse inclusion.

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    Small comment - I presume you mean $f(z)=x^3$?2011-10-20
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    @Juan: Thanks, fixed.2011-10-20

2 Answers 2

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Since $\mathfrak I = (y-x^2,z-x^3)$ is contained in the kernel, your map corresponds to a map $R/\mathfrak I\to S$. Show that this map is injective.

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    Thanks, Henning. If I substitute $x^2$ for $y$ and $x^3$ for $z$ in a polynomial $g(x,y,z)$, I get a polynomial in the class of $x$ in $R$. Then this maps to zero in $S$ under $f$ if and only if it is zero to begin with since $f$ maps the class of $x$ in $R$ to $x$ in $S$ under the induced map. Is that OK?2011-10-20
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    Yes, that's about it.2011-10-20
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Hint: If $I = (y-x^2, z-x^3)$, then since $I$ is in the kernel of $f$, we have an induced map $f : R/I \to S$ which is clearly surjective, so we just need to show that it's injective. Now use the fact that every element of the coset $R/I$ contains an element of $R$ which is purely a polynomial in $x$.

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    Thanks, Ted. I think Henning made a similar suggestion.2011-10-20
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    Why is $f:R/I \rightarrow S$ surjective? Is it because $f:R \rightarrow R/I$ and $f:R \rightarrow S$ are both surjective? Also, would another way of showing injectivity be to use the First Isomorphism Theorem to say that $R/I \cong S$ meaning $f$ is injective? Thanks.2016-11-24