Question

Consider the following statements:
(I) The kernel of the ring homomorphism \(f:\mathbb{Z}[x]\to\mathbb{Z}\) given by \(f(p(x))=p(1)\) is \(\{(x-1)q(x): q(x)\in\mathbb{Z}[x]\}\).
(II) The ring \(\mathbb{Z}_4\) is isomorphic to \(\mathbb{Z}_2\times\mathbb{Z}_2\).
Choose the correct answer:

• A. Only (I) is true
• B. Only (II) is true
• C. Both (I) and (II) are true
• D. Both (I) and (II) are false

Hint:

Use the division algorithm at \(x=1\) for (I); compare element orders in \(\mathbb{Z}_4\) versus \(\mathbb{Z}_2\times\mathbb{Z}_2\) for (II).

Answer 1

The Correct Option is A

Step 1: The kernel of \(f\) is \(\ker f=\{p(x)\in\mathbb{Z}[x] : p(1)=0\}\).

Step 2: Since \(x-1\) is monic, the division algorithm applies in \(\mathbb{Z}[x]\): for any \(p(x)\in\mathbb{Z}[x]\), write \(p(x)=(x-1)q(x)+r\) with \(q(x)\in\mathbb{Z}[x]\) and \(r\in\mathbb{Z}\) constant.

Step 3: Evaluating at \(x=1\) gives \(p(1)=(1-1)q(1)+r=r\). So \(p(1)=0 \iff r=0 \iff p(x)=(x-1)q(x)\) for some \(q(x)\in\mathbb{Z}[x]\).

Step 4: Hence \(\ker f=\{(x-1)q(x):q(x)\in\mathbb{Z}[x]\}\), exactly as stated, so statement (I) is TRUE.

Step 5: In \(\mathbb{Z}_4=\{0,1,2,3\}\), the element 1 has additive order 4, since \(1,2,3,0\) are the successive sums, first returning to 0 after 4 steps.

Step 6: In \(\mathbb{Z}_2\times\mathbb{Z}_2\), every element \((a,b)\) satisfies \((a,b)+(a,b)=(0,0)\), so every element has additive order at most 2.

Step 7: Since \(\mathbb{Z}_4\) has an element of order 4 while \(\mathbb{Z}_2\times\mathbb{Z}_2\) does not, the two rings are not isomorphic, so statement (II) is FALSE.

\[\boxed{\text{Only (I) is true}}\]