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In $\Bbb Z_3[x]$ let $f(x)=x^3+x^2+x+1$ and let $A=\Bbb Z_3[x]/(f) $ be the quotient ring. I'm asked to find the cardinality of $A $ and whether it is a field.

For the cardinality, I should find that $A $ is in bijection with the set of polynomials with degree smaller than $3$ (and in $\Bbb Z_3[X]$), hence $|A|=3^3$.The thing is, I can't really visualise $A$. I know its elements are the classes of the remainders of Euclidean division by $f$, but how do I get these?

For the other question, the given answer is that $f(x)=(x^2+1)(x+1) $ is reducible in $\Bbb Z_3[X]$, thus $A$ is not a field. Why does this work? Is this a general property of quotient rings of polynomials, i.e. if $a\in A[X]$ is reducible then $A[X]/(a)$ is not a field?

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    $\begingroup$ $(a)$ is reducible $\implies (a)$ is not maximal $\implies A[X]/(a)$ is not a field. $\endgroup$ – Sahiba Arora Jul 10 '17 at 22:50
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    $\begingroup$ See this answer for some motivation (not quite "visualization") of normal forms in quotients of Euclidean domains. $\endgroup$ – Bill Dubuque Jul 11 '17 at 0:57
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$\frac {\mathbb Z_3[x]}{x^3+x^2+x+1}$ gives us a set of polynomials in $\mathbb Z_3[x]$ of degree 2 (or less).

This can be thought of as a 3 dimensional vector space. And there are $3^3$ elements.

Since $(x + 1)(x^2 +1) = 0$ in this ring, it is not a field. In a field $ab = 0$ if and only if $a = 0$ or $b = 0$

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  • $\begingroup$ Why does the quotient give us a set .... of degree 2? I don't know because I don't see how to treat the elements of it. Also, why is that product equal to zero? $\endgroup$ – Richard Jul 11 '17 at 0:13
  • $\begingroup$ If you take a polynomial and you divide it by another polynomial, the remainder will always be smaller than the polynomial that is doing the dividing. It is a consequence of the Euclidean algorithm. $\endgroup$ – Doug M Jul 11 '17 at 0:17
  • $\begingroup$ Ohhh, right, how could I not see it! Thank you! $\endgroup$ – Richard Jul 11 '17 at 0:38
  • $\begingroup$ Ah excuse me, why is $(x+1)(x^2+1)=0$? $\endgroup$ – Richard Jul 11 '17 at 7:58
  • $\begingroup$ $(x+1)(x^2+1) = x^3+x^2+x+1$ $\endgroup$ – Doug M Jul 11 '17 at 16:26
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I'm guessing by $\mathbb{Z}_3$ you mean $\mathbb{Z}/3\mathbb{Z}$ and not the $3$-adic integers.

To answer your second question, yes there is a general property of this sort. Namely, a quotient ring $R/I$ is a field if and only if $I$ is a maximal ideal. In your case, $I$ is the principal ideal generated by $f$, so $I$ is maximal iff $f$ is irreducible (prove this yourself!).

Your first question seems more conceptual. A way to "visualize" $A$ could be by using the chinese remainder theorem to write it as the product of a field with 9 elements and a field with 3 elements via the decomposition of $f$ you gave. Have a look at the related questions if you're still confused.

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