# Find units and zero divisors in $\mathbb{Q}[x]/{(x^2-1)}$

I need to find units and zero divisors in $$\mathbb{Q}[x]/{(x^2-1)}$$.

I don't know how to start... I know that all elements in $$\mathbb{Q[x]}/{(x^2-1)}$$ are polynomials $$p(x)= ax+b$$ like $$0, 1, x ,x+1$$. Is there any way to find all elements in $$\mathbb{Q[x]}/{(x^2-1)}$$?

• Write out $(f(x) + (x^2-1))(g(x) + (x^2-1)) = 0 + (x^2-1)$ and start using what you know about ideals... Jun 3 at 20:07
• Zerodivisors: $a(x\pm1)$. Units: $ax+b$ with $b\ne\pm a$. (As one can see the two sets form a partition of the ring.) Jun 5 at 19:52
• Remember that elements of this ring are not polynomials. They are equivalence classes. Jun 5 at 20:32

You should also know how addition and multiplication work in $$\mathbb{Q[x]}/{(x^2-1)}$$: you do them as you would in $$\mathbb{Q[x]}$$ but remember that $$x^2-1=0$$.
Two polynomials $$f,g$$ are going to be zero divisors if $$f(x)g(x)=0$$. For instance, $$(x-1)(x+1)=x^2-1=0$$. Can you characterise all of the possible polynomials that equal 0?
Two polynomials $$f,g$$ are going to be units if $$f(x)g(x)=1$$. Think similarly to the above, and this time $$1=x^2=2x^2-1=\cdots$$.
• Two polynomials f,g are going to be zero divisors if $f(x)g(x)$ is a multiple of $x^2-1$. I have two cases: 1) $f(x) = (ax+b)(x+1) = ax^2+(a+b)x+b$ and $g(x) = (cx+d)(x-1)=cx^2+(d-c)x-d$ 2)$f(x)=(ax+b)(x^2-1)=ax^3-ax+bx^2$ and $g(x)$ is a polynomial $\in \mathbb{Q[x]}$...it's right? Jun 3 at 20:51
• @jontao in case (2), you just have $f(x)=0$ because $x^2-1=0$ and anything times 0 equals 0. Case (1) is true. If you want, you can rewrite $f$ and $g$ as linear functions using $x^2=1$, where you'll hopefully notice that $f(x)=r(x+1)$ for some $r\in \mathbb{Q}$ (and for $g$...?)