Question

Considere the ideal I=<x²-4x+3,x³+3x²-x-3> of the ring Q[x]. Find a polynomial p such that I=<p>.Is Q[x]/I a field? Give reasons for your answer.

Answer 1

Since $\mathbb Q[x]$ is a Principal ideal domain, $p$ is the gratest common divisor of the polinomials $x^2-4x+3$ and $x^3+3x^2-x-3$. Taking into account that $x^2-4x+3=x^2-x-3x+3=x(x-1)-3(x-1)=(x-3)(x-1)$ and

$x^3+3x^2-x-3=x(x^2-1)+3(x^2-1)=(x+3)(x^2-1)=(x+3)(x+1)(x-1)$, we conclude that $p=x-1$.

Consider the ring $\mathbb Q[x]/I=\mathbb Q[x]/\langle x-1\rangle$.

Polynomial remainder theorem implies that each element of $\mathbb Q[x]/\langle x-1\rangle$ is of the form $[f(x)]=f(x)+\langle x-1\rangle=f(x)+(x-1)\mathbb Q[x]$, where $\deg f(x)<\deg(x-1)=1$, and therefore, $f(x)\in\mathbb Q$. Consequently, $\mathbb Q[x]/\langle x-1\rangle=\{[a]\ : \ a\in\mathbb Q\}$.

Consider the map $\psi: \mathbb Q\to \mathbb Q[x]/\langle x-1\rangle, \psi(a)=[a].$ Taking into account that $\psi(ab)=[ab]=[a][b]=\psi(a)\psi(b)$ and $\psi(a+b)=[a+b]=[a]+[b]=\psi(a)+\psi(b)$, we conclude that $\psi$ is a homomorphism. If $a\ne b$, then $\psi(a)=[a]\ne [b]=\psi(b)$, and consequently $\psi$ is injective. Since $\psi(a)=[a]$ for any $[a]\in \mathbb Q[x]/\langle x-1\rangle$, $\psi$ is surjective. Therefore, $\psi$ is a isomorphism. Thus, $\mathbb Q$ and $\mathbb Q[x]/\langle x-1\rangle$ are isomorphic.

Since $\mathbb Q$ is a field, we conclude that $\mathbb Q[x]/I$ is a field.