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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 $\mathbb{Q[x]}/{(x^2-1)}$?

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  • $\begingroup$ Write out $(f(x) + (x^2-1))(g(x) + (x^2-1)) = 0 + (x^2-1)$ and start using what you know about ideals... $\endgroup$
    – Randall
    Commented Jun 3, 2023 at 20:07
  • $\begingroup$ 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.) $\endgroup$
    – user26857
    Commented Jun 5, 2023 at 19:52
  • $\begingroup$ Remember that elements of this ring are not polynomials. They are equivalence classes. $\endgroup$ Commented Jun 5, 2023 at 20:32

1 Answer 1

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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 $1=x^2=2x^2-1=\cdots$.

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  • $\begingroup$ 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? $\endgroup$
    – jontao
    Commented Jun 3, 2023 at 20:51
  • $\begingroup$ @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 $g$...?) $\endgroup$
    – A.M.
    Commented Jun 5, 2023 at 21:01

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