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Use $v={\sigma }^2$ as the variance parameter, with prior density relative to $d\mu dv$. Let $\bar{x}=n^{-1}\sum x_i$ and $S=\sum (x_i-\bar{x}{)}^2$. Multiplying the prior by the likelihood function gives the unnormalized Bayesian posteriorThe printed request for an “improper posterior” should not be read as a claim that this posterior is always improper. Although the prior is improper, integrating over $\mu$ gives a marginal kernel $v^{-(n+1)/2}{e}^{-S/(2v)}$, which is integrable exactly when $n>1$ and $S>0$. Under these usual nondegenerate-data conditions it is a proper posterior. If those conditions fail, the displayed kernel is improper and there is no posterior probability distribution for a sampler to target.

Completing the square and using the inverse-gamma distribution give the full conditionalsThe inverse-gamma distribution convention is density proportional to $v^{-\alpha -1}{e}^{-\beta /v}$. Equivalently the precision satisfiesusing shape and rate, as in the question.

A Gibbs sampler starts from $v^{(0)}>0$ and alternatesThis is the normal mean-variance posterior with a log-uniform variance prior. For a marginal check, $v\mid x\sim \mathrm{IG}((n-1)/2,S/2)$: integrating out $\mu$ reduces the shape by $1/2$. This marginal shape is different from the full-conditional shape $n/2$.