Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
$\langle x^2, y \rangle$ 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. – Ted Feb 2 '13 at 20:38
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?? – delfin Feb 2 '13 at 20:40
  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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.