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I have a ring $R$ unitary and commutative with four elements and characteristic $2$. I have $$I=\{f \in R[X,Y]; f(t,t^2)=0\ \forall t \in R\}.$$ I have to find a finite number of generators for this ideal. I have thought to use $\langle x^2,y \rangle$, but I can't show all the $f$ that are in $I$ are generated by $\langle x^2,y \rangle$, in fact I'm not sure if this example works. I thought it could because of the characteristic.

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  • $\begingroup$ $\langle x^2, y \rangle$ can't be right because $x^2$ and $y$ are not in the ideal. Also note that there are 2 different rings $R$ that satisfy your conditions, and the answer will be different for the 2 rings. $\endgroup$ – Ted Feb 2 '13 at 20:38
  • $\begingroup$ yes, you are rigth..I know there are differnet rings with this property8i thougt 3 not 2) but I dont know how could I start..there is not an usual procediment?? $\endgroup$ – delfin Feb 2 '13 at 20:40
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  1. Start by finding one nonzero polynomial $f$ satisfying $f(t, t^2) = 0$ for all $t \in R$.
  2. Taking the quotient $R/I$ where $I$ by the ideal generated by the $f$ you found in step 1.
  3. Repeat steps 1 and 2, with $R$ replaced by $R/I$, until you can't find any more nonzero polynomials in step 1. You will have to prove that there aren't any, of course.

At some point, you will need to break the problem into 2 cases, depending on what $R$ is. There are 2 commutative unital rings of characteristic 2.

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