= Solution
Assume the forward field is sufficiently differentiable in maturity to differentiate its deterministic and <stochastic integrals>. Moving along the diagonal adds the maturity derivative to the fixed-maturity dynamics:
$$
\boxed{dr_t=\left[\left.\partial_Tf(t,T)\right|_{T=t}+\alpha_Q(t,t)\right]dt+\sigma(t,t)dW_t^Q.}
$$
The <Heath-Jarrow-Morton model> no-arbitrage drift is $\alpha_Q(t,T)=\sigma(t,T)\int_t^T\sigma(t,u)du$, so $\alpha_Q(t,t)=0$. Equivalently, from the integrated forward equation,
$$
\left.\partial_Tf(t,T)\right|_{T=t}
=f_0'(t)+\int_0^t\partial_T\alpha_Q(s,t)ds+\int_0^t\partial_T\sigma(s,t)dW_s^Q.
$$
That drift can depend on the entire previous forward curve or driving path, rather than on the current scalar rate.
\b[The <short rate> need not be <Markov>.] For a concrete <One Brownian factor does not imply a Markov short rate> example, take the deterministic volatility $\sigma(t,T)=T-t$. Its admissible risk-neutral drift is $(T-t)^3/2$, and
$$
r_t=f_0(t)+\frac{t^4}{8}+\int_0^t(t-s)dW_s
=f_0(t)+\frac{t^4}{8}+\int_0^tW_sds.
$$
Subtracting the deterministic part leaves $I_t=\int_0^tW_sds$. The past of $I$ reveals its derivative $W_t$, and $\mathbb E[I_{t+h}-I_t\mid\mathcal F_t^I]=hW_t$. Yet $W_t$ is not determined by $I_t$: the <Gaussian> covariance calculation gives
$$
\operatorname{Var}I_t=t^3/3,\qquad \operatorname{Cov}(W_t,I_t)=t^2/2,\qquad
\operatorname{Var}(W_t\mid I_t)=t/4>0.
$$
Consequently the conditional law of the future depends on more than $I_t$, proving failure of the <Markov property> for the scalar <short rate>. The enlarged pair $(I_t,W_t)$ is <Markov>. Special maturity-volatility structures, such as those producing a Vasicek-type short-rate diffusion, can instead admit a scalar <Markov> realization.
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