Sorry to reply to a very old question, but I thought I should record one interesting class of examples of Krull dimension $1$, (completions of) numerical semigroup rings. Fix $T=\{t_1,...,t_n\}$ a set of natural numbers whose greatest common divisor is $1$. Let's also assume $t_i>1$ for each $i$.

Consider the ring $R=K[[x^{t_1},...,x^{t_n}]]$, viewed as a subring of $K[[x]]$. This is a local ring whose maximal ideal is generated by $(x^{t_1},...,x^{t_n})$, and it is an integral domain as it is a subring of $K[[x]]$ which is an integral domain.

I claim the normalization of $R$ is $K[[x]]$, a local ring. To see this, first note that since $\{t_1,\cdots,t_n\}$ has greatest common divisor $1$, there are integers $r_1,\cdots,r_n$ so that $r_1t_1+\cdots+r_nt_n=1$. Then, $(x^{t_i})^{r_i}$ is an element of the fraction field $F$ of $R$ for each $i$, so too is their product $x^{r_1t_1+\cdots+r_nt_n} =x$. Then, $x \in F$ satisfies the monic polynomial $X^{t_1}-x^{t_1}$ in $R[X]$, and thus $x$ is integral over $R$. Consequently, $K[[x]]$ is inside the normalization of $R$. Further, since $K[[x]]$ is itself normal, $K[[x]]$ must be normalization of $R$ as $\operatorname{Frac}(K[[x]])=F$.

Thus, $R$ is a local domain whose integral closure is local, but $R$ itself is not normal. This shows that many interesting examples of non-normal unibranched rings exist even in dimension $1$.

Sorry to reply to a very old question, but I thought I should record one interesting class of examples of Krull dimension $1$, (completions of) numerical semigroup rings. Fix $T=\{t_1,...,t_n\}$ a set of natural numbers whose greatest common divisor is $1$. Let's also assume $t_i>1$ for each $i$.

Consider the ring $R=K[x^{t_1},...,x^{t_n}]$, viewed as a subring of $K[x]$. Suppressing the process of localizing at the origin, this is a local ring whose maximal ideal is generated by $(x^{t_1},...,x^{t_n})$, and it is an integral domain as it is a subring of $K[x]$ which is an integral domain.

I claim the normalization of $R$ is $K[x]$, a local ring. To see this, first note that since $\{t_1,\cdots,t_n\}$ has greatest common divisor $1$, there are integers $r_1,\cdots,r_n$ so that $r_1t_1+\cdots+r_nt_n=1$. Then, $(x^{t_i})^{r_i}$ is an element of the fraction field $F$ of $R$ for each $i$, so too is their product $x^{r_1t_1+\cdots+r_nt_n} =x$. Then, $x \in F$ satisfies the monic polynomial $X^{t_1}-x^{t_1}$ in $R[X]$, and thus $x$ is integral over $R$. Consequently, $K[x]$ is inside the normalization of $R$. Further, since $K[x]$ is itself normal, $K[x]$ must be normalization of $R$ as $\operatorname{Frac}(K[x])=F$.

Thus, $R$ is a local domain whose integral closure is local, but $R$ itself is not normal. This shows that many interesting examples of non-normal unibranched rings exist even in dimension $1$.

Sorry to reply to a very old question, but I thought I should record one interesting class of examples of Krull dimension $1$, numerical semigroup rings. Fix $T=\{t_1,...,t_n\}$ a set of natural numbers whose greatest common divisor is $1$. Let's also assume $t_i>1$ for each $i$.

Consider the ring $R=K[[x^{t_1},...,x^{t_n}]]$, viewed as a subring of $K[[x]]$. This is a local ring whose maximal ideal is generated by $(x^{t_1},...,x^{t_n})$, and it is an integral domain as it is a subring of $K[[x]]$ which is an integral domain.

I claim the normalization of $R$ is $K[[x]]$, a local ring. To see this, first note that since $\{t_1,\cdots,t_n\}$ has greatest common divisor $1$, there are integers $r_1,\cdots,r_n$ so that $r_1t_1+\cdots+r_nt_n=1$. Then, $(x^{t_i})^{r_i}$ is an element of the fraction field $F$ of $R$ for each $i$, so too is their product $x^{r_1t_1+\cdots+r_nt_n} =x$. Then, $x \in F$ satisfies the monic polynomial $X^{t_1}-x^{t_1}$ in $R[X]$, and thus $x$ is integral over $R$. Consequently, $K[[x]]$ is inside the normalization of $R$. Further, since $K[[x]]$ is itself normal, $K[[x]]$ must be normalization of $R$ as $\operatorname{Frac}(K[[x]])=F$.

Thus, $R$ is a local domain whose integral closure is local, but $R$ itself is not normal. This shows that many interesting examples of non-normal unibranched rings exist even in dimension $1$.

Sorry to reply to a very old question, but I thought I should record one interesting class of examples of Krull dimension $1$, (completions of) numerical semigroup rings. Fix $T=\{t_1,...,t_n\}$ a set of natural numbers whose greatest common divisor is $1$. Let's also assume $t_i>1$ for each $i$.

Consider the ring $R=K[[x^{t_1},...,x^{t_n}]]$, viewed as a subring of $K[[x]]$. This is a local ring whose maximal ideal is generated by $(x^{t_1},...,x^{t_n})$, and it is an integral domain as it is a subring of $K[[x]]$ which is an integral domain.

I claim the normalization of $R$ is $K[[x]]$, a local ring. To see this, first note that since $\{t_1,\cdots,t_n\}$ has greatest common divisor $1$, there are integers $r_1,\cdots,r_n$ so that $r_1t_1+\cdots+r_nt_n=1$. Then, $(x^{t_i})^{r_i}$ is an element of the fraction field $F$ of $R$ for each $i$, so too is their product $x^{r_1t_1+\cdots+r_nt_n} =x$. Then, $x \in F$ satisfies the monic polynomial $X^{t_1}-x^{t_1}$ in $R[X]$, and thus $x$ is integral over $R$. Consequently, $K[[x]]$ is inside the normalization of $R$. Further, since $K[[x]]$ is itself normal, $K[[x]]$ must be normalization of $R$ as $\operatorname{Frac}(K[[x]])=F$.

Thus, $R$ is a local domain whose integral closure is local, but $R$ itself is not normal. This shows that many interesting examples of non-normal unibranched rings exist even in dimension $1$.