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?

Question

Let $A$ be a square matrix. If $A^2 = A$, then the matrix $A$ is called:

Answer 1

The map $T:\mathbb{Q}[x]\to\mathbb{Q}[x]$ sends a polynomial $p(x)$ to the polynomial obtained by replacing every $x$ with $x^2$. Replacing the variable of a polynomial ring by a fixed element of the same ring, here $x^2$, is called an evaluation or substitution map, and such maps are always ring homomorphisms. Let's check each option using this idea together with the degree of a polynomial.

T is a ring homomorphism: For any $p, q \in \mathbb{Q}[x]$, $(p+q)(x^2) = p(x^2)+q(x^2)$ and $(pq)(x^2) = p(x^2)q(x^2)$ hold because these are algebraic identities true for any value put in place of $x$, including $x^2$, and $T(1)=1$ too. So $T$ is a ring homomorphism. This statement is TRUE.
T is one-to-one: If $p(x)$ has degree $n$ and leading coefficient $a_n \neq 0$, then $p(x^2)$ has degree $2n$ with the same leading coefficient $a_n \neq 0$, so it is never the zero polynomial unless $p(x)$ itself is zero. So $T(p)=0$ forces $p=0$, meaning $T$ has a trivial kernel and is injective. This statement is TRUE.
T is onto: Every polynomial of the form $p(x^2)$ only has even powers of $x$, since each power of $x$ in $p(t)$ turns into a power of $x^2$, which is even. A polynomial like $x$, which has a nonzero odd-degree term, can never be written as $p(x^2)$ for any $p$. So $x \in \mathbb{Q}[x]$ is not in the image of $T$, hence $T$ is not onto. This statement is FALSE.
T is a ring isomorphism: An isomorphism needs $T$ to be both one-to-one and onto. Since $T$ is not onto, it fails to be an isomorphism, regardless of it being an injective homomorphism. This statement is FALSE.

So the statements that hold are that $T$ is a ring homomorphism and that $T$ is one-to-one.

Let's summarize:

Substituting $x^2$ for $x$ always gives a ring homomorphism, since sums and products of polynomials transform correctly under any substitution.
The map only reaches even-power polynomials, so it misses odd-degree ones like $x$, which is why it is not onto and therefore not an isomorphism.

Thus the correct choices are that $T$ is a ring homomorphism and that $T$ is one-to-one.

\[ \boxed{T \text{ is a homomorphism and injective, but not surjective or an isomorphism}} \]

Answer 2

The map $T:\mathbb{Q}[x]\to\mathbb{Q}[x]$ sends a polynomial $p(x)$ to the polynomial obtained by replacing every $x$ with $x^2$. Replacing the variable of a polynomial ring by a fixed element of the same ring, here $x^2$, is called an evaluation or substitution map, and such maps are always ring homomorphisms. Let's check each option using this idea together with the degree of a polynomial.

T is a ring homomorphism: For any $p, q \in \mathbb{Q}[x]$, $(p+q)(x^2) = p(x^2)+q(x^2)$ and $(pq)(x^2) = p(x^2)q(x^2)$ hold because these are algebraic identities true for any value put in place of $x$, including $x^2$, and $T(1)=1$ too. So $T$ is a ring homomorphism. This statement is TRUE.
T is one-to-one: If $p(x)$ has degree $n$ and leading coefficient $a_n \neq 0$, then $p(x^2)$ has degree $2n$ with the same leading coefficient $a_n \neq 0$, so it is never the zero polynomial unless $p(x)$ itself is zero. So $T(p)=0$ forces $p=0$, meaning $T$ has a trivial kernel and is injective. This statement is TRUE.
T is onto: Every polynomial of the form $p(x^2)$ only has even powers of $x$, since each power of $x$ in $p(t)$ turns into a power of $x^2$, which is even. A polynomial like $x$, which has a nonzero odd-degree term, can never be written as $p(x^2)$ for any $p$. So $x \in \mathbb{Q}[x]$ is not in the image of $T$, hence $T$ is not onto. This statement is FALSE.
T is a ring isomorphism: An isomorphism needs $T$ to be both one-to-one and onto. Since $T$ is not onto, it fails to be an isomorphism, regardless of it being an injective homomorphism. This statement is FALSE.

So the statements that hold are that $T$ is a ring homomorphism and that $T$ is one-to-one.

Let's summarize:

Substituting $x^2$ for $x$ always gives a ring homomorphism, since sums and products of polynomials transform correctly under any substitution.
The map only reaches even-power polynomials, so it misses odd-degree ones like $x$, which is why it is not onto and therefore not an isomorphism.

Thus the correct choices are that $T$ is a ring homomorphism and that $T$ is one-to-one.

\[ \boxed{T \text{ is a homomorphism and injective, but not surjective or an isomorphism}} \]

Show Hint

The Correct Option is A, B

Solution and Explanation

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