Question

I know that not every local seminormal ring is Cohen-Macaulay. But are 1-dimensional local seminormal rings Cohen-Macaulay?

Answer 1

Your two sentences present some discrepancy.

Not all 1-dimensional local rings are semi-normal, so the first sentence has nothing to do with the second.

If all 1-dimensional local rings were Cohen-Macaulay, then all local rings would be Cohen-Macaulay by virtue of the definition of the Cohen-Macaulay property.

Semi-normalization can make things worse. Three lines meeting in a point that are not contained in a plane (say the 3 coordinate axes in 3-space) is semi-normal and it is not Gorenstein. On the other hand, 3 lines meeting in a point and contained in a plane is Gorenstein, but it is not semi-normal. Its semi-normalization is exactly the above union of 3 lines meeting in a point that are not contained in a plane.

Answer 2

I agree with Sándor's, I'm not quite sure what you mean. And JSpecter provides a counter-example.

However, 1 dimensional seminormal rings are Cohen-Macaulay. Indeed, 1 dimensional reduced rings are Cohen-Macaulay (you just need a single non-zero-divisor).