Consider the polynomial ring \(\mathbb{Z}[x]\). Let \(I = (2, x^2 + 3x - 1)\) be the ideal of \(\mathbb{Z}[x]\) generated by 2 and \(x^2 + 3x - 1\). Describe the structure of the quotient ring \(\mathbb{Z}[x]/I\) in as much detail as you can. Do the same for the quotient ring \(\mathbb{Z}[x]/J\) where \(J = (2, x^3 + 3x^2 - x + 4)\).
Solution
The two rings are respectively \(\mathbb F_4\) and \(\mathbb F_2\times\mathbb F_4\).
1. Reduction modulo \(2\) gives\[ \mathbb Z[x]/I\cong\mathbb F_2[x]/(q),\qquad \mathbb Z[x]/J\cong\mathbb F_2[x]/(xq), \qquad q=x^2+x+1. \]
Coefficient reduction \(\mathbb Z[x]\to\mathbb F_2[x]\) is surjective and has kernel \((2)\). For any \(f\in\mathbb Z[x]\), the inverse image of \((\overline f)\) is \((2,f)\): if \(\overline h=\overline f\,\overline a\), lift \(\overline a\) to an integer polynomial \(a\); then \(h-fa\in(2)\). The reverse inclusion follows immediately by reduction. Thus the induced quotient maps are isomorphisms. The polynomials in the problem reduce to \(q\) and \(x^3+x^2+x=xq\), respectively.
2. The first ring is the field\[ E=\mathbb F_2[\alpha],\qquad \alpha^2=\alpha+1, \qquad E=\{0,1,\alpha,1+\alpha\}. \]
The polynomial \(q\) has neither \(0\) nor \(1\) as a root. A reducible quadratic over a field has a linear factor and hence a root, so \(q\) is irreducible. Polynomial division gives unique representatives \(a+bx\), with \(a,b\in\mathbb F_2\), in \(\mathbb F_2[x]/(q)\). Let \(\alpha\) be the class of \(x\). Its defining relation is \(\alpha^2=\alpha+1\), and \(\alpha^3=\alpha(\alpha+1)=1\). The three nonzero elements are \(1,\alpha,\alpha^2\) and are all invertible, proving directly that \(E\) is a field of four elements.
It has characteristic \(2\), additive group \(C_2\times C_2\), and multiplicative group \(E^\times=\langle\alpha\rangle\cong C_3\). Its only ideals are \((0)\) and \(E\), since a nonzero ideal contains a unit. In particular \(I\) is maximal and prime. The only idempotents of \(E\) are \(0,1\), as \(e^2=e\) implies \(e(e-1)=0\) in a field; it has no nonzero zero divisors or nilpotents.
3. The second ring \(B=\mathbb F_2[x]/(xq)\) is explicitlyisomorphic to \(\mathbb F_2\times E\).
Define \[ \Theta:B\longrightarrow\mathbb F_2\times E, \qquad [h(x)]\longmapsto(h(0),h(\alpha)). \] Both evaluations are ring homomorphisms, and \(xq\) vanishes at \(0\) and \(\alpha\), so the map is well defined. If \(\Theta([h])=0\), then \(x\mid h\) and \(q\mid h\). The identity \(q-x(x+1)=1\) shows that \(x\) and \(q\) are coprime. More explicitly, write \(h=xk\). Modulo \(q\), the class of \(x\) is the unit \(\alpha\), so \(q\mid k\). Thus \(xq\mid h\), proving injectivity.
Let \[ e_0=[q],\qquad e_1=[x^2+x]. \] Then \(\Theta(e_0)=(1,0)\) and \(\Theta(e_1)=(0,1)\). For \(a,b,c\in\mathbb F_2\), the element \[ a e_0+(b+c[x])e_1 \] maps to \((a,b+c\alpha)\). This proves surjectivity and gives the inverse on every element of the product. Consequently \(B\cong\mathbb F_2\times E\) as rings.
4. The ring \(B\) has eight elements, unit group \(C_3\), four nonzero zero divisors, exactly the ideals \((0),(e_0),(e_1),B\), and no nonzero nilpotents.
The ring \(B\) has eight elements, characteristic \(2\), and additive group \(C_2^3\). Equivalently, polynomial division gives the basis \(1,[x],[x^2]\) over \(\mathbb F_2\). A pair \((a,b)\in\mathbb F_2\times E\) is a unit exactly when \(a=1\) and \(b\ne0\), so \(B^\times\cong E^\times\cong C_3\). A nonzero pair is a zero divisor exactly when at least one component is zero: multiply by a nonzero pair supported in that component to obtain zero; when both components are nonzero the pair is a unit. Thus there are four nonzero zero divisors, namely \((1,0)\) and the three \((0,b)\) with \(b\ne0\).
Every ideal \(L\) in the product decomposes into its two coordinate ideals: for \((a,b)\in L\), multiplication by \((1,0)\) and \((0,1)\) puts \((a,0)\) and \((0,b)\) in \(L\). Each coordinate ideal in a field is zero or the whole field. Therefore the complete ideal list in \(B\) is \[ (0),\quad (e_0),\quad(e_1),\quad B. \] The middle two are maximal, with quotients \(E\) and \(\mathbb F_2\), respectively, and are the only prime ideals: the zero ideal is not prime because \(e_0e_1=0\) with both factors nonzero. Hence \(B\) is neither a field nor a domain nor a local ring, and \(J\) is neither prime nor maximal.
The only idempotents are \(0,e_0,e_1,1\), because each coordinate of an idempotent in a field is \(0\) or \(1\). Finally, a nilpotent pair has nilpotent coordinates, which must both be zero. Thus \(B\) is reduced despite having nonzero zero divisors.