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The problem is to confirm that the ideal generated by $x^3+x+1$ in $\mathbb{Z}/2\mathbb{Z}[x]$ is prime and the ideal generated by $x^3-x-1$ is maximal in $\mathbb{Z}/3\mathbb{Z}[x]$.

I tried to take quotient ring and see if they are integral domain or field. But it was hard for me to tell if the resulting quotient rings are integral domain or field. Any help will be greatly appreciated.

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2 Answers 2

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The second problem: The coefficients have only 3 possibilities, 0,1 or 2. And the power of $x$ in the quotient ring by the ideal of $x^3-x-1$ has to be at most two and so the quotient ring will have $3\times3\times3=27$ elements. Considering you want to avoid any concept like irreducibility you can simply write the multiplication table for all the non-zero elements and realise 1 as a product making everything invertible, hence a field.

For example constants have inverses, and in the quotient ring $x^3-x-1=0$ which is the same as saying $x^3-x=1$ or $x$ and $x^2-1$ are inverses of each other.

And $(x^2-1)^2$ will be the inverse of $x^2$.

First problem is also easier: you have just 4 elements in the quotient ring and need to carry out all multiplications and see that product of non-zero elements yield again a non-zero element.

After doing all these and convincing yourself that those ideals are prime and maximal you will realize that alternatives such as Euclidean domain concepts may be worth learning in order to avoid massive calculations.

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  • $\begingroup$ thank you so much this is somewhat brute but very clear. addition questions if you let me do so: then do i have to multiply polynomials 26*26 times to find inverses of each polynomials in prob.2? and are there only 4 elements in prob1? not 8 since the highest possible order of an element in the quotient is 3? $\endgroup$
    – Mathcho
    Oct 3, 2015 at 18:03
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    $\begingroup$ It is not exactly $26\times 26$. Because $ab=ba$, only half of it. Also note that if $a,b$ have $a',\ b'$ as their inverses then $a'b'$ will be the inverse of $ab$ and $(a')^m$ will be the inverse of $a^m$. This will reduce the number of calculations from 676 to perhaps 200 or less! $\endgroup$ Oct 4, 2015 at 1:39
  • $\begingroup$ thank you so much. i solved all the problems i mentioned here, refering to your advice. using x(x^2+2)=1 and multiplying x, (x^2+2) to the equation several times, i got all the inverses fairly quickly. thank you again $\endgroup$
    – Mathcho
    Oct 4, 2015 at 3:38
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Since $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ are fields, their respective polynomial rings are Euclidean domains, hence PIDs. It therefore suffices to check that $x^{3}+x+1$ and $x^{3}-x-1$ are irreducible over each respective polynomial ring. Feel free to comment further if you need more than this.

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  • $\begingroup$ thank you for your nice answer alex. but i haven't learned about euclidean domain or irreducibility yet. i just learned about rings, ideals, iso thms, sum, product, direct product/sum of ideals and rings and some thms about maximal, prime ideals. if you can give me a hint or answer with only those materials, i will very appreciate it. $\endgroup$
    – Mathcho
    Oct 3, 2015 at 7:27
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    $\begingroup$ Dear @Mathcho: would a reference do for now? Dummit and Foote Chapter 13, section 1 has some very useful information on this. In particular, example (5) given on page 516 should be helpful. $\endgroup$ Oct 3, 2015 at 7:42
  • $\begingroup$ ohh field extension...... that frightens me though haha. but i will check those materials. thank you. $\endgroup$
    – Mathcho
    Oct 3, 2015 at 8:35

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