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This is part of a larger homework problem. I am trying to prove that a quotient ring is not a field by showing that $\langle x^3+x+1\rangle$ is not maximal in the ring of polynomials in the integers mod $3$. I've tried to factor it. I think I can work with the simpler condition that it is prime. I have zero experience with this situation. Apparently this same ideal is maximal in the polynomial ring of the integers mod $2$, but I can't yet prove that either. Any help is appreciated and that includes general advice about algebra.

Milo Brandt
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OLP
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  • A challenge here for me is working in Zsub3. Even factoring simple polynomials seems difficult in this new terrain. – OLP Jan 27 '15 at 05:06

3 Answers3

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Hint: The polynomial is not irreducible.

André Nicolas
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  • Thanks for the hint. How does one recognize such a fact in general? – OLP Jan 27 '15 at 05:03
  • For degree $2$ or $3$, the polynomial is irreducible iff it has no root. Here we have the obvious root $x=1$, so $x-1$, or if you prefer $x+2$, is a factor. For higher degree polynomials, showing reducibility/irreducibility can take much more effort. – André Nicolas Jan 27 '15 at 05:13
  • Is that all I need to show? Or must I factor this expression in order to show that the ideal is not prime? I'm working ahead on this problem, so I've been taught next to nothing so far and my book is somewhat obscure on this issue, focusing as it does on R[x] and Z[x]. Thanks for the help. – OLP Jan 27 '15 at 05:16
  • We don't have a field because we don't even have an integral domain. The polynomial factors as $(x+2)(x^2+x+2)$ so $[x+2][x^2+x+2]=[0]$. In terms of ideals, the ideal generated by $x+2$ is strictly bigger. – André Nicolas Jan 27 '15 at 05:19
  • You don't need to factor explicitly, as I did above. It is enough to know that a factorization exists. – André Nicolas Jan 27 '15 at 05:23
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A $3$ degree polynomial is irreducible over $\mathbb Z_p$ $\iff$ it has no roots in it.Check that $1$ is a root here

Learnmore
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  • I see that, thanks. I'm trying to factor this expression. – OLP Jan 27 '15 at 05:10
  • How does irreducible imply non maximal? Does it factor into elements that are not in the generated ideal? Since the degrees of its factors are less than ? – OLP Jan 27 '15 at 05:13
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$$x^3+x+1=(x+2)(x^2+x+2)\pmod3\implies$$

$$\Bbb Z_3[x]/\langle x^3+x+1\rangle\cong\Bbb Z_3[x]/\langle x+2\rangle\times\Bbb Z_3[x]/\langle x^2+x+2\rangle$$

and you can check at once the direct product above isn't a field (not even an integer domain), as for example $\;\overline{(x+2)}\cdot\overline{(x^2+x+2)}=\overline0\;$ in the quotient ring, and none of these elements is zero there.

Timbuc
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