To show that the linear map $T : P_{n-1}(\mathbb{C}) \times P_{m-1}(\mathbb{C}) \to P_{m+n-1}(\mathbb{C})$, defined by $T(r, s) = rp + sq$, is injective and surjective, we proceed as follows:

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1. Definitions and Notations

• $P_k(\mathbb{C})$: the space of polynomials with complex coefficients of degree at most $k$.
• $\deg(p) = m$, $\deg(q) = n$, with $m, n \geq 1$.
• $p$ and $q$ are nonconstant polynomials with no common roots.

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Injectivity

We need to show that $T(r, s) = 0$ implies $r = 0$ and $s = 0$.

Proof:

1. Suppose $T(r, s) = rp + sq = 0$.
2. Since $rp$ and $sq$ are multiples of $p$ and $q$, respectively, the roots of $p$ and $q$ impose constraints:
   • $rp$ has all roots of $p$, and $sq$ has all roots of $q$.
   • Because $p$ and $q$ have no common roots, $rp$ and $sq$ have disjoint sets of roots.
3. The equality $rp + sq = 0$ implies that $rp$ and $sq$ must both be zero since their roots are disjoint.
   • For $rp = 0$, since $p$ is nonconstant, $r = 0$.
   • For $sq = 0$, since $q$ is nonconstant, $s = 0$.
4. Hence, $r = 0$ and $s = 0$.

Thus, $T$ is injective.

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Surjectivity

We need to show that for any polynomial $h \in P_{m+n-1}(\mathbb{C})$, there exist $r \in P_{n-1}(\mathbb{C})$ and $s \in P_{m-1}(\mathbb{C})$ such that $T(r, s) = rp + sq = h$.

Proof:

1. Let $h \in P_{m+n-1}(\mathbb{C})$. We aim to solve $rp + sq = h$ for $r \in P_{n-1}(\mathbb{C})$ and $s \in P_{m-1}(\mathbb{C})$.
2. The degrees of $rp$ and $sq$ are as follows:
   • $\deg(rp) \leq n-1 + m = m+n-1$.
   • $\deg(sq) \leq m-1 + n = m+n-1$.
3. Therefore, $rp + sq$ has degree at most $m+n-1$, matching the degree of $h$.
4. Construct a basis for $P_{m+n-1}(\mathbb{C})$ using the bases of $P_{n-1}(\mathbb{C})$ and $P_{m-1}(\mathbb{C})$:
   • Let $\{1, x, \dots, x^{n-1}\}$ be the basis for $P_{n-1}(\mathbb{C})$.
   • Let $\{1, x, \dots, x^{m-1}\}$ be the basis for $P_{m-1}(\mathbb{C})$.
5. By choosing appropriate coefficients $r$ and $s$, any polynomial $h$ can be written as a linear combination of $rp$ and $sq$. This follows from the fact that $p$ and $q$ have no common roots, ensuring that their contributions are linearly independent.

Thus, $T$ is surjective.

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Conclusion

Since $T$ is both injective and surjective, it is a linear isomorphism. This establishes the result.

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Would you like further clarifications or examples? Here are some related questions to explore:

1. What are the implications of $p$ and $q$ having no common roots in this problem?
2. How does the degree constraint on $r$ and $s$ affect the proof of surjectivity?
3. Can the result extend to polynomials over other fields, like $\mathbb{R}$ or $\mathbb{Q}$?
4. What happens if $p$ and $q$ do share a common root? How does it affect $T$?
5. How does the structure of $T$ depend on the choice of basis for the polynomial spaces?

Tip: When proving injectivity, always ensure the disjoint nature of constraints imposed by linearly independent terms!