Determining the structure of the quotient ring $\mathbb{Z}[x]/(x^2+3,p)$

Question

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 $p \in \{3,5\}$.

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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.

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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 $(\mathbb{Z}/5 \mathbb{Z})^2$.

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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!

Answer 1

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 $\mathbb F_5^2$.