I'm interested in the following problem from Artin's Algebra text:

Determine the structure of the ring $\mathbb Z[x]/(x^2 + 3,p)$, where (a) p = 3, (b) p = 5.

I know that by the isomorphism theorems for rings we can take the quotients successively, and so

$$\mathbb{Z}[x]/(p) \cong (\mathbb{Z}/p \mathbb{Z})[x] $$

as the map $\mathbb{Z}[x] \to (\mathbb{Z}/p \mathbb{Z})[x]$ defined by $\sum_{n} a_n x^n \mapsto \sum_{n} \overline{a_n} x^n$ is a surjective ring homomorphism with kernel $(p)$. Thus it remains to study the quotients

$$(\mathbb{Z}/p \mathbb{Z})[x]/(x^2+3) $$ for $p \in \{3,5\}$.

If $p=3$, $(x^2+3)=(x^2)$ in $(\mathbb{Z}/3 \mathbb{Z})[x]$, and by using polynomial division all distinct coset representatives can be reduced to the following list of 9 elements

$$\{0,1,2,x,1+x,2+x,2x,1+2x,2+2x\}. $$

Moreover, it can shown that the list above gives 9 distinct cosets, as no difference of two distinct elements of the list is a multiple of $x^2$. Since $1$ and $x$ generate two distinct additive groups of order $3$, the additive group of our quotient ring is not cyclic. Elementary group theory then shows

$$(\mathbb{Z}/3 \mathbb{Z})[x]/(x^2)^+ \cong (\mathbb{Z}/ 3 \mathbb{Z})^2 $$

as additive groups.

I was then about to conclude that the multiplication on the quotient is then compatible with the usual one in $(\mathbb{Z}/3\mathbb{Z})^2$, but this is wrong!

It can be seen that the quotient is not isomorphic to $(\mathbb{Z}/3\mathbb{Z})^2$ as a ring, because the former contains a nonzero element (represented by $x$) whose square is zero, while the latter contains no such elements.

If $p=5$, a full list of coset representatives is of length 25

$$\{0,1,2,3,4,x,1+x,2+x,3+x,4+x,2x,1+2x,2+2x,3+2x,4+2x,3x,1+3x,2+3x,3+3x,4+3x,4x,1+4x,2+4x,3+4x,4+4x \} .$$

And once again, one can see that these represent 25 distinct cosets. Similarly to the $p=5$ case, I've managed to prove that the additive group of this ring is isomorphic to $(\mathbb{Z}/5 \mathbb{Z})^2$.

My questions:

  • Have I made any mistakes in my argument?
  • What exactly am I supposed to do in this question? determine the number of elements? Write down the tables for addition and multiplication?

Any further information about these quotients will be appreciated, thanks!


I would guess the author just wants you to simplify the definition of the rings, $\mathbb Z[x]/(x^2+3,p)$, in the special cases that $p=3,5$.

If $p=3$, you've seen yourself that the ring is $ \mathbb F_3[x]/(x^2)$, where $\mathbb F_3$ denotes the field with three elements. This is a free abelian group with multiplication defined by $(a+bx)(c+dx)=ac+(ad+bc)x$.

For $p=5$, we find that the ring is isomorphic to $\mathbb F_5[x]/(x^2+3)$. Now you can check that $-3$ is not a square modulo $5$, so the polynomial is irreducible. Hence the ring is a quadratic extension of $\mathbb F_5$. But these are all isomorphic, so we have that the ring is isomorphic to $\mathbb F_{5^2}$ (in standard notation). Alternatively, it can be described as $\mathbb F_5[\sqrt{-3}]$, with multiplication defined by $(a+b\sqrt{-3})(c+d\sqrt{-3})=(ac-3bd)+(ad+bc)\sqrt{-3}$, with underlying abelian group isomorphic to $\mathbb F_5^2$.

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  • $\begingroup$ Thanks. So since this is an introductory exercise, would you say that just taking the first quotient is sufficient at this point? The book hasn't talked about free abelian groups nor $\mathbb{F}_5[\sqrt{-3}]$ yet. $\endgroup$ – user1337 Dec 23 '15 at 9:44
  • $\begingroup$ @user1337 It is hard to understand precisely what is ment by "structure" here, so I'm not completely sure what is "right" answer to this exercise. To me it seems that you have ample understanding of quotient rings, though. $\endgroup$ – Fredrik Meyer Dec 23 '15 at 11:32

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