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It seems that you are asking about smoothability of singularities. Some singularities are smoothable some are not.

I don't think there is a general criterion that tells you how to decide whether a specific singularity is smoothable or not and it may not be that a singularity that seems bad is necessarily not smoothable while one that looks OK is.

Hypersurface singularities are smoothable. I bet you can see why.

The simplest example I know of an innocent looking non-smoothable singularity is a cone over an abelian variety of dimension at least $2$. The proof is rather involved so it may not really qualify as "simplest" non-smoothable singularity, but it certainly does not seem bad. Anyway, the fact that it this is not smoothable follows from 1.3 that it is a Du Bois singularity and hence by a result of this paper, but Schwede if it were smoothable, the total space would have rational singularities, which is possible to prove CM, and then all fibers are CM, but this other wayssingularity is not. The same argument shows that any projective variety (or variety with isolated singularities) with DB but not CM singularities give examples of what cannot be the limit of smooth projective varieties.

Another set of examples is provided by quotient singularities. They are rigid and hence non-smoothable in dimension at least $3$, but there are two-dimensional quotient singularities that are smoothable, for instance $\big(\mathbb A^2/(x,y)\sim (-x,-y)\big)\simeq (x^2=yz)\subset \mathbb A^3$ is a hypersurface singularity.

If you put more conditions on $f$ that obviously limits further the possible singularities you can get.

EDIT incorporated Karl's comment into the example given.

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It seems that you are asking about smoothability of singularities. Some singularities are smoothable some are not.

I don't think there is a general criterion that tells you how to decide whether a specific singularity is smoothable or not and it may not be that a singularity that seems bad is necessarily not smoothable while one that looks OK is.

Hypersurface singularities are smoothable. I bet you can see why.

The simplest example I know of an innocent looking non-smoothable singularity is a cone over an abelian variety of dimension at least $2$. The proof is rather involved so it may not really qualify as "simplest" non-smoothable singularity, but it certainly does not seem bad. Anyway, the fact that it is not smoothable follows from 1.3 of this paper, but it is possible to prove this other ways.

Another set of examples is provided by quotient singularities. They are rigid and hence non-smoothable in dimension at least $3$, but there are two-dimensional quotient singularities that are smoothable, for instance $\big(\mathbb A^2/(x,y)\sim (-x,-y)\big)\simeq (x^2=yz)\subset \mathbb A^3$ is a hypersurface singularity.

If you put more conditions on $f$ that obviously limits further the possible singularities you can get.

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It seems that you are asking about smoothability of singularities. Some singularities are smoothable some are not.

I don't think there is a general criterion that tells you how to decide whether a specific singularity is smoothable or not and it may not be that a singularity that seems bad is necessarily not smoothable while one that looks OK is.

Hypersurface singularities are smoothable. I bet you can see why.

The simplest example I know of an innocent looking non-smoothable singularity is a cone over an abelian variety of dimension at least $2$. The proof is rather involved so it may not really qualify as "simplest" non-smoothable singularity, but it certainly does not seem bad. Anyway, the fact that it is not smoothable follows from 1.3 of this paper.

If you put more conditions on $f$ that obviously limits further the possible singularities you can get.

As for your question: Can $X_y$ have non-constant regular functions? The answer is no. Since the family is flat, the Hilbert polynomial of the fibers is constant. $h^0(X_y,\mathscr O_{X_y})$ is the constant term of that Hilbert polynomial, so if any fiber is connected and reduced, then $h^0(X_y,\mathscr O_{X_y})=1$ for all $y\in Y$.

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