Question

Let \( \mathbb{Q}[x] \) be the ring of all polynomials with coefficients in \( \mathbb{Q} \) under the usual polynomial addition and multiplication. Let \( T: \mathbb{Q}[x] \to \mathbb{Q}[x] \) be defined by \[ T(p(x)) = p(x^2), \quad \text{for all } p(x) \in \mathbb{Q}[x]. \] Which of the following statements is/are TRUE?

• A. \(T\) is a ring homomorphism.
• B. \(T\) is one-to-one.
• C. \(T\) is onto.
• D. \(T\) is a ring isomorphism.

Hint:

Check T(p+q)=T(p)+T(q) and T(pq)=T(p)T(q) for the homomorphism property, then look at which polynomials (odd or even degree terms) can actually appear as p(x^2).

Answer 1

The Correct Option is A, B

Step 1: Check whether T preserves addition and multiplication.
A map \(T: R \to R\) between rings is a ring homomorphism if \(T(p+q) = T(p) + T(q)\) and \(T(pq) = T(p)T(q)\) for all \(p, q\), and it also sends the identity \(1\) to \(1\).
Here \(T\) is the substitution map that replaces \(x\) with \(x^2\) in a polynomial. For any two polynomials \(p(x), q(x) \in \mathbb{Q}[x]\):
\[ T(p(x) + q(x)) = (p+q)(x^2) = p(x^2) + q(x^2) = T(p(x)) + T(q(x)) \] \[ T(p(x)q(x)) = (pq)(x^2) = p(x^2)\,q(x^2) = T(p(x))\,T(q(x)) \] Also \(T(1) = 1(x^2) = 1\), so the identity is preserved too.
So \(T\) is a ring homomorphism, this is the general fact that substituting a fixed element (here \(x^2\)) for the variable in a polynomial ring always gives a ring homomorphism. Option (A) is TRUE.

Step 2: Check whether T is one-to-one.
A ring homomorphism is one-to-one exactly when its kernel is just the zero element.
Suppose \(T(p(x)) = 0\), that is \(p(x^2) = 0\) as a polynomial. If \(p(x)\) is a nonzero polynomial of degree \(n\) with leading coefficient \(a_n \neq 0\), then \(p(x^2)\) is a nonzero polynomial of degree \(2n\) with the same leading coefficient \(a_n\), so it cannot be the zero polynomial.
So \(p(x^2) = 0\) forces \(p(x) = 0\), which means the kernel of \(T\) is \(\{0\}\), so \(T\) is one-to-one. Option (B) is TRUE.

Step 3: Check whether T is onto.
Every polynomial in the image of \(T\) has the form \(p(x^2) = a_0 + a_1 x^2 + a_2 x^4 + \cdots\), which only has even powers of \(x\).
Take the polynomial \(q(x) = x\), which has an odd power with a nonzero coefficient. There is no \(p(x) \in \mathbb{Q}[x]\) with \(p(x^2) = x\), because substituting \(x^2\) for \(x\) can never produce an odd-degree term.
So \(x\) is not in the image of \(T\), which means \(T\) is not onto. Option (C) is FALSE.

Step 4: Decide about the isomorphism and give the final answer.
A ring isomorphism must be both one-to-one and onto. Since \(T\) is not onto, it cannot be a ring isomorphism, even though it is an injective ring homomorphism. Option (D) is FALSE.
So among the four statements, only (A) and (B) hold.
\[ \boxed{T \text{ is a ring homomorphism and one-to-one, but not onto and not an isomorphism}} \]