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Given the polynomials $g(t) = t$ and $h(t) = (t-3)^2 \in \mathbb{C}[t]$, I want to find the smallest (in terms of degree) polynomial $f(t) \in \mathbb{C}$ satisfying $f \equiv 0$ mod $g$ and $f \equiv 3$ mod $h$.

However, I've been struggling so far finding any (not just the smallest) polynomial that satisfies these two properties. I guess this can be done via the Chinese Remainder Theorem?

What I thought of so far: $g$ and $h$ are relatively prime (their largest common divisor is $1$), which can easily be seen when writing $h(t) = t^2 - 6 t + 9$. Therefore, one can apply the Chinese Remainder Theorem, which (in the form I've seen it) gives us that

$$\mathbb{C}[t]/(g h) \cong \mathbb{C}[t]/(g) \oplus \mathbb{C}[t]/(h)$$

I don't really know though how this helps me. How would I proceed?

Alternatively, I've noticed that $g(0) = 0$ and $h(3) = 0$, and $0$ and $3$ are exactly the values that $f$ must be equal to, mod $g$ and mod $h$. Could I somehow use this anyhow to solve it easier, without the Chinese Remainder Theorem potentially?

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Chinese remainder theorem guarantees a unique solution modulo $gh$ (I think... I'm not an expert on this subject). We can solve for $f$ by solving the system of linear congruences. First, we have \begin{align} f &\equiv 0 \pmod{t} \\ f &\equiv 3 \pmod{(t-3)^2} \end{align} We can convert this to an equation relating polynomials. \begin{align} f &= 0 + pt \\ f &= 3 + q(t-3)^2 \end{align} where $p, q\in\mathbb{C}[t]$. Thus, $$ pt = 3 + q(t - 3)^2.$$ Take modulo $t$ both sides to get $$ 0 \equiv 3 + q(-3)^2 \pmod{t} \implies q \equiv -\frac{1}{3} \pmod{t}.$$ This gives $q = -\frac 13 + rt$ for some $r \in \mathbb{C}[t]$, and substituting this into our original expression gives $$ f = 3 + \left(-\frac{1}{3}+rt\right)(t-3)^2 $$ $$ f = 3 - \frac{1}{3}(t-3)^2 + rt(t-3)^2 $$ $$ f \equiv 3 - \frac{1}{3}(t-3)^2 \pmod{t(t-3)}.$$

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