Quotient rings of polynomial rings

Question

I have come across a quite difficult question while I am studying for a test:

Let $F=\Bbb Z[x]/(7,x^2-3)$. Let $u$ denote the image of $x$ under the canonical epimorphism from $\Bbb Z[x]$ to $F$.
A. Show that $F$ is a field. How many elements does it have?
B. Find all the elements of the group $\{z\in F\mid z^2+uz+1=0\}$ and express them by using $u$.

Please help me with an answer.

Answer 1

Question A has been adequatelly handled by others, so I will concentrate on Question B. The formula for the solutions of a quadratic equation works over any field of characteristic $\neq2$. Thus $$ z=\frac{-u\pm\sqrt{u^2-4}}2. $$ Because $u$ is a zero of $x^2-3$, we know that $u^2=3$. Therefore (remember that we are working modulo $7$) $$u^2-4=-1=27=9\cdot 3=9u^2=(3u)^2.$$ So the solutions of that equation are $$ z_1=\frac{-u+3u}2=u\qquad\text{and}\qquad z_2=\frac{-u-3u}2=-2u. $$

Answer 2

Hints: understand and justify/prove each of the following steps (hint of hint:isomorphism theorems!):

$$\Bbb Z[x]/\langle\,7\,,\,x^2-3\,\rangle\stackrel\bullet\cong\left(\Bbb Z[x]/\langle 7\rangle\right)/\langle\,x^2-3\,\rangle\cong\Bbb Z_7[x]/\langle\,x^2-3\,\rangle$$

• On the RHS it must be understood the ideal $\;\langle\; (x^2-3)+\overbrace{7\Bbb Z[x]}^{\langle7\rangle\;\text{in}\;\Bbb Z[x]/\langle7\rangle}\;\rangle\;$ , and similarly (but

  not identically!) in the last expression

Answer 3

First, we remark that we are basically looking at $(\mathbb{Z}/(7))[x]/(x^2-3)$.

A. Since $f(x)=x^2-3$ has no zeroes in $\mathbb{Z}/(7)$, it is irreducible, so $F$ is a field. Moreover, $F$ is a quadratic field extension of $\mathbb{Z}/(7)$, so $F$ contains $7^2=49$ elements.

B. Does the $|$ mean 'such that' and do you mean 'field' instead of 'group'? Assuming this: Each element in $F$ can be written in the form $z=a+bu$, where $a,b\in\mathbb{Z}/(7)$. Expanding the given expression gives $$z^2+uz+1 = (a+bu)^2+u(a+bu)+1 = 1+ a^2 + u(2ab+a) + u^2(b^2+b). $$ Since we have modded out $f(x)$, we know that $x^2-3=0\in F$, or equivalently that $x^2\equiv 3$, hence $u^2=3$. Using this, we find that $$z^2+uz+1 = 1+ a^2 + 3b(b+1) + u(2ab+a).$$ If this ever wants to equal $1$, we need $$\left\{ \begin{array}{l}a^2+3b(b+1)=0\\a(2b+1)=0.\end{array}\right.$$

Does this help?