[proofplan] We reduce the problem from irreducibility over $\mathbb{Q}[t]$ to irreducibility over $\mathbb{Z}[t]$ using Gauss's Lemma. We then argue by contradiction: assuming $f = gh$ is a nontrivial factorisation in $\mathbb{Z}[t]$, we reduce modulo $p$ and exploit the unique factorisation in $\mathbb{F}_p[t]$ to show that $p$ divides the constant terms of both factors. This forces $p^2 \mid a_0$, contradicting hypothesis (3). [/proofplan] [step:Reduce to irreducibility over $\mathbb{Z}[t]$ via Gauss's Lemma] Write $f = d \cdot f_0$ where $d = \gcd(a_0, a_1, \ldots, a_n)$ is the content of $f$ and $f_0 \in \mathbb{Z}[t]$ is primitive. Since $p \nmid a_n$ by hypothesis (1), the prime $p$ does not divide $d$. Consequently, $f_0$ inherits all three Eisenstein conditions with the same prime $p$: its leading coefficient is not divisible by $p$, its lower coefficients are divisible by $p$, and its constant term is not divisible by $p^2$. Since $f_0$ is primitive, [Gauss's Lemma](/theorems/???) applies: $f_0$ is irreducible in $\mathbb{Q}[t]$ if and only if $f_0$ is irreducible in $\mathbb{Z}[t]$. Since $f = d \cdot f_0$ with $d \in \mathbb{Z} \setminus \{0\}$, irreducibility of $f$ in $\mathbb{Q}[t]$ is equivalent to irreducibility of $f_0$ in $\mathbb{Q}[t]$. Replacing $f$ by $f_0$, we may assume without loss of generality that $f$ is primitive, and it suffices to show that $f$ admits no nontrivial factorisation in $\mathbb{Z}[t]$. [guided] The theorem asks us to prove irreducibility in $\mathbb{Q}[t]$, but the hypotheses — divisibility conditions on the integer coefficients by a prime $p$ — are statements about $\mathbb{Z}[t]$. The bridge between these two rings is Gauss's Lemma, which asserts that a primitive polynomial in $\mathbb{Z}[t]$ is irreducible in $\mathbb{Q}[t]$ if and only if it is irreducible in $\mathbb{Z}[t]$. We must verify that $f$ is primitive, i.e., that $\gcd(a_0, a_1, \ldots, a_n) = 1$. Suppose for contradiction that some prime $q$ divides every coefficient $a_i$. In particular, $q \mid a_n$. But hypothesis (1) states $p \nmid a_n$, so $q \neq p$. On the other hand, hypothesis (2) states $p \mid a_i$ for all $0 \leq i \leq n-1$, which places no constraint on primes other than $p$. There is no reason a prime $q \neq p$ cannot divide $a_n$. However, the content $\gcd(a_0, \ldots, a_n)$ need not equal $1$ for Gauss's Lemma to be useful: we can always factor out the content and apply the criterion to the primitive part. More directly: even if $f$ is not primitive, we can write $f = d \cdot f_0$ where $d = \gcd(a_0, \ldots, a_n)$ and $f_0$ is primitive. Since $p \nmid a_n$, the prime $p$ does not divide $d$, so $f_0$ still satisfies all three Eisenstein conditions with the same prime $p$. It therefore suffices to prove $f_0$ is irreducible in $\mathbb{Z}[t]$, which by Gauss's Lemma is equivalent to irreducibility of $f_0$ in $\mathbb{Q}[t]$. Since $f = d \cdot f_0$ with $d \in \mathbb{Z}$ a unit-free constant, $f$ is irreducible in $\mathbb{Q}[t]$ if and only if $f_0$ is. Without loss of generality, we may therefore assume $f$ is primitive, and it suffices to show $f$ is irreducible in $\mathbb{Z}[t]$. [/guided] [/step] [step:Assume a nontrivial factorisation $f = gh$ in $\mathbb{Z}[t]$ and set up notation] Suppose for contradiction that $f$ is reducible in $\mathbb{Z}[t]$. Then there exist polynomials \begin{align*} g &= b_r t^r + b_{r-1} t^{r-1} + \cdots + b_0 \in \mathbb{Z}[t], \\ h &= c_s t^s + c_{s-1} t^{s-1} + \cdots + c_0 \in \mathbb{Z}[t], \end{align*} with $r, s \geq 1$, $b_r \neq 0$, $c_s \neq 0$, and $r + s = n$, such that $f = gh$. Comparing leading coefficients yields $a_n = b_r c_s$, and comparing constant terms yields $a_0 = b_0 c_0$. [guided] We aim to derive a contradiction, so we assume $f$ factors nontrivially. Note that the factorisation must have both factors of degree $\geq 1$, since if one factor were a constant $d \in \mathbb{Z}$, then the factorisation $f = d \cdot (f/d)$ would require $d \mid a_i$ for all $i$, making $d$ a divisor of the content of $f$. Since $f$ is primitive, $d = \pm 1$, which is a unit in $\mathbb{Z}$, contradicting the assumption that the factorisation is nontrivial. The coefficient identities $a_n = b_r c_s$ and $a_0 = b_0 c_0$ follow from expanding the product $gh$ and matching the coefficient of $t^n$ (the highest degree term) and the constant term (the coefficient of $t^0$), respectively. [/guided] [/step] [step:Reduce modulo $p$ and use unique factorisation in $\mathbb{F}_p[t]$ to determine the forms of $\bar{g}$ and $\bar{h}$] Let $\pi \colon \mathbb{Z}[t] \to \mathbb{F}_p[t]$ denote the canonical reduction-mod-$p$ ring homomorphism, and write $\bar{f} = \pi(f)$, $\bar{g} = \pi(g)$, $\bar{h} = \pi(h)$. By hypothesis (2), $p \mid a_i$ for $0 \leq i \leq n-1$, so these coefficients vanish in $\mathbb{F}_p$. By hypothesis (1), $p \nmid a_n$, so $\bar{a}_n \neq 0$ in $\mathbb{F}_p$. Therefore \begin{align*} \bar{f}(t) = \bar{a}_n \, t^n \in \mathbb{F}_p[t]. \end{align*} Since $\pi$ is a ring homomorphism, $\bar{f} = \bar{g} \cdot \bar{h}$ in $\mathbb{F}_p[t]$. Now $\mathbb{F}_p[t]$ is a polynomial ring over a field, hence a unique factorisation domain. The element $t \in \mathbb{F}_p[t]$ is irreducible (it is a degree-one polynomial over a field). Since $t^n = \bar{a}_n^{-1} \bar{f} = \bar{a}_n^{-1} \bar{g} \cdot \bar{h}$, uniqueness of factorisation implies that each irreducible factor of $\bar{g}$ and $\bar{h}$ is an associate of $t$. Since $t$ has zero constant term, the only monic polynomials whose irreducible factors are all associates of $t$ are powers of $t$. Therefore there exist integers $r', s' \geq 0$ and elements $\bar{b}, \bar{c} \in \mathbb{F}_p \setminus \{0\}$ such that \begin{align*} \bar{g}(t) &= \bar{b} \, t^{r'}, \\ \bar{h}(t) &= \bar{c} \, t^{s'}, \end{align*} with $r' + s' = n$ and $\bar{b} \, \bar{c} = \bar{a}_n$. [guided] The key idea is to transfer the factorisation problem from $\mathbb{Z}[t]$ — where unique factorisation is delicate — to the simpler ring $\mathbb{F}_p[t]$, where unique factorisation holds and the polynomial $\bar{f}$ takes a particularly simple form. The reduction homomorphism $\pi \colon \mathbb{Z}[t] \to \mathbb{F}_p[t]$ acts coefficient-by-coefficient: it sends each $a_i \in \mathbb{Z}$ to its residue class $\bar{a}_i \in \mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$. Because $\pi$ is a ring homomorphism, it respects multiplication: $\pi(gh) = \pi(g)\pi(h)$. Hypotheses (1) and (2) together force $\bar{f}(t) = \bar{a}_n t^n$: every coefficient except the leading one is killed by reduction mod $p$, while the leading coefficient survives. So $\bar{f}$ is, up to a unit scalar, a perfect $n$-th power of the irreducible element $t$. Why does unique factorisation in $\mathbb{F}_p[t]$ force $\bar{g}$ and $\bar{h}$ to be monomials? The factorisation $\bar{g} \cdot \bar{h} = \bar{a}_n t^n$ in the UFD $\mathbb{F}_p[t]$ means that the multiset of irreducible factors of $\bar{g} \cdot \bar{h}$ consists of exactly $n$ copies of $t$ (together with the unit $\bar{a}_n$). By uniqueness, the irreducible factors of $\bar{g}$ and $\bar{h}$ must each be associates of $t$, so $\bar{g} = \bar{b} \, t^{r'}$ and $\bar{h} = \bar{c} \, t^{s'}$ for some $r' + s' = n$ and units $\bar{b}, \bar{c} \in \mathbb{F}_p^\times$. Note that $r' \leq r$ and $s' \leq s$ (the degree of $\bar{g}$ cannot exceed the degree of $g$, since reduction mod $p$ can only kill leading terms, not create new ones). Moreover, since $r, s \geq 1$, we know $r' \geq 0$ and $s' \geq 0$, but the crucial point is what happens to the constant terms. [/guided] [/step] [step:Extract $p \mid b_0$ and $p \mid c_0$ from the monomial forms] Since $\bar{g}(t) = \bar{b} \, t^{r'}$ and $r' + s' = n$ with $r, s \geq 1$ and $r + s = n$, the degree constraints force $r' \geq 1$ or $s' \geq 1$. In fact, since $\bar{g}$ has degree at most $r$ and $\bar{h}$ has degree at most $s$, and $r' + s' = n = r + s$, we must have $r' = r$ and $s' = s$. In particular, since $r \geq 1$ and $s \geq 1$, both $\bar{g}$ and $\bar{h}$ have zero constant term: \begin{align*} \bar{b}_0 &= 0 \in \mathbb{F}_p, \\ \bar{c}_0 &= 0 \in \mathbb{F}_p. \end{align*} Lifting back to $\mathbb{Z}$, this means $p \mid b_0$ and $p \mid c_0$. [guided] We need to pin down the exponents $r'$ and $s'$. We know that $\deg \bar{g} \leq \deg g = r$ and $\deg \bar{h} \leq \deg h = s$ (reduction mod $p$ cannot increase degree). Since $r' + s' = n = r + s$ and $r' \leq r$ and $s' \leq s$, the only possibility is $r' = r$ and $s' = s$. To see this: from $r' \leq r$ and $s' \leq s$, we get $r' + s' \leq r + s = n$. But we also have $r' + s' = n$. So both inequalities $r' \leq r$ and $s' \leq s$ are in fact equalities. This equality $r' = r \geq 1$ means that $\bar{g}(t) = \bar{b} \, t^r$ has no terms of degree less than $r$, and in particular its constant term $\bar{b}_0$ vanishes. Similarly, $s' = s \geq 1$ gives $\bar{c}_0 = 0$. Since the constant term of a polynomial $g = b_r t^r + \cdots + b_0$ reduces to $\bar{b}_0 = b_0 \bmod p$, the vanishing $\bar{b}_0 = 0$ means exactly $p \mid b_0$. The same reasoning gives $p \mid c_0$. [/guided] [/step] [step:Derive $p^2 \mid a_0$ and reach a contradiction with hypothesis (3)] Since $a_0 = b_0 c_0$ and $p \mid b_0$ and $p \mid c_0$, we conclude \begin{align*} p^2 \mid b_0 c_0 = a_0. \end{align*} This contradicts hypothesis (3), which asserts $p^2 \nmid a_0$. The assumption that $f$ admits a nontrivial factorisation in $\mathbb{Z}[t]$ is therefore false, so $f$ is irreducible in $\mathbb{Z}[t]$. By the reduction in the first step (via Gauss's Lemma), $f$ is irreducible in $\mathbb{Q}[t]$. [guided] We have established two divisibility facts: $p \mid b_0$ and $p \mid c_0$. Since $a_0 = b_0 c_0$ (matching constant terms in the identity $f = gh$), and since $b_0 = p \beta$ and $c_0 = p \gamma$ for some $\beta, \gamma \in \mathbb{Z}$, we obtain \begin{align*} a_0 = b_0 c_0 = (p\beta)(p\gamma) = p^2 \beta \gamma, \end{align*} which gives $p^2 \mid a_0$. But hypothesis (3) explicitly forbids this: $p^2 \nmid a_0$. This is a genuine contradiction, so our assumption — that $f = gh$ for some $g, h \in \mathbb{Z}[t]$ with $\deg g \geq 1$ and $\deg h \geq 1$ — must be false. Hence $f$ is irreducible in $\mathbb{Z}[t]$. The first step of the proof reduced the problem to $\mathbb{Z}[t]$-irreducibility via Gauss's Lemma, so we conclude that $f$ is irreducible in $\mathbb{Q}[t]$, as desired. It is worth noting where each of the three hypotheses was consumed: - Hypothesis (1), $p \nmid a_n$: used to ensure that $\bar{f}$ has degree exactly $n$ (and also to verify primitivity for Gauss's Lemma). - Hypothesis (2), $p \mid a_i$ for $0 \leq i \leq n-1$: used to force $\bar{f}(t) = \bar{a}_n t^n$, which via unique factorisation in $\mathbb{F}_p[t]$ constrains the factors $\bar{g}$ and $\bar{h}$ to be monomials. - Hypothesis (3), $p^2 \nmid a_0$: used to obtain the final contradiction once $p \mid b_0$ and $p \mid c_0$ have been established. All three hypotheses are necessary. Dropping hypothesis (1) allows $\bar{f}$ to be the zero polynomial, and the argument breaks down entirely. Dropping hypothesis (2) permits non-monomial factors in $\mathbb{F}_p[t]$, so the constant terms of $\bar{g}$ and $\bar{h}$ need not vanish. Dropping hypothesis (3) removes the source of the contradiction — for instance, $f(t) = t^2 - p^2$ satisfies (1) and (2) but not (3), and indeed $f = (t - p)(t + p)$ is reducible. [/guided] [/step]