Let \[ R_1=\mathbb F_p[x]/(x^2-2), \qquad R_2=\mathbb F_p[x]/(x^2-3). \] Determine whether \(R_1\) and \(R_2\) are isomorphic in each of the cases \[ p=2,\qquad p=5,\qquad p=11. \]
Solution
Write \[ \alpha\coloneqq[x]_{R_1}, \qquad \beta\coloneqq[x]_{R_2}. \]
1. For \(p=2\), the rings \(R_1\) and \(R_2\) are isomorphic.
In \(R_1\), \[ \alpha^2=2=0. \] In \(R_2\), \[ \beta^2=3=1 \] in \(\FF_2\), so \[ (\beta+1)^2 = \beta^2+2\beta+1 = 1+0+1 =0. \] Hence the assignment \[ \alpha\longmapsto\beta+1 \] defines a ring homomorphism \[ R_1\to R_2. \] It is surjective because \[ \beta=(\beta+1)+1 \] lies in its image. Both quotients are two-dimensional vector spaces over \(\FF_2\), hence both have four elements. A surjective map between these finite sets is bijective, so it is an isomorphism.
2. For \(p=5\), the rings \(R_1\) and \(R_2\) are isomorphic.
In \(R_1\), \[ \alpha^2=2. \] Therefore \[ (2\alpha)^2 = 4\alpha^2 = 8 = 3 \] in \(\FF_5\). Thus the assignment \[ \beta\longmapsto2\alpha \] respects the defining relation \(\beta^2=3\) and induces a ring homomorphism \[ R_2\to R_1. \] It is surjective because \(2\) is invertible modulo \(5\) and \[ \alpha=3(2\alpha). \] Again both rings have \(5^2=25\) elements, so the surjective homomorphism is an isomorphism.
3. For \(p=11\), the polynomial \(x^2-2\) is irreducible over\(\FF_{11}\).
The quadratic residues modulo \(11\) are \[ 0,\ 1,\ 3,\ 4,\ 5,\ 9. \] Thus \(2\) is not a square in \(\FF_{11}\), so \(x^2-2\) has no root there. A quadratic polynomial over a field is irreducible exactly when it has no root.
4. For \(p=11\), the polynomial \(x^2-3\) splits into distinct linearfactors:\[ x^2-3=(x-5)(x+5). \]
Since \[ 5^2=25=3 \] in \(\FF_{11}\), both \(5\) and \(-5\) are roots. They are distinct because \(5\neq-5\) modulo \(11\).
5. For \(p=11\), the rings \(R_1\) and \(R_2\) are not isomorphic.
By step 3, the ideal \((x^2-2)\) is maximal in \(\FF_{11}[x]\), so \(R_1\) is a field.
By step 4, in \(R_2\) the nonzero classes \[ [x-5] \qquad\text{and}\qquad [x+5] \] have product zero. They are nonzero because neither degree-one polynomial is divisible by \(x^2-3\). Thus \(R_2\) has zero divisors and is not a field. A field cannot be isomorphic to a ring with nonzero zero divisors.
6. Therefore\[ \boxed{ \begin{array}{c|ccc} p&2&5&11\\ \hline R_1\cong R_2&\text{yes}&\text{yes}&\text{no} \end{array} }. \]
Steps 1, 2, and 5 settle the three cases respectively.
7. Q.E.D.
Step 6 gives the complete comparison.