**3. Connectivity of** *TI***(Γ(***R***))**

In this section, we study the connectivity of *TI*(Γ(R)).

**Theorem 7.** *Let I be a nonzero ideal of* R*. Then, TI*(Γ(R)) *has no cut-vertex.*

**Proof.** Assume that the vertex *u* of *TI*(Γ(R)) is a cut-vertex. Then, there exist *x*, *y* ∈ *TI*(Γ(R)) such that *u* lies on every path from *x* to *y*. Thus, we have the following cases.

Case (*i*): If *x* is adjacent to *y*, then there is a path from *x* to *y* in *TI*(Γ(R)) such that *u* does not lie on it. Hence, we obtain a contradiction.

Case (*ii*): If *x* is not adjacent to *y*, then *x*, *y* ∈/ *I*. Since *I* is nonzero ideal, *I* has at least two elements, and we have the following subcases.

Subcase(*a*): If *x*, *y* ∈ *X*, then *x* is not adjacent to *y*, and there exist *w*1, *w*<sup>2</sup> ∈ *I* such that *x* is adjacent to *w*1, *w*2, and similarly, *y* is adjacent to *w*1, *w*2. Therefore, if *u* is equal to *w*<sup>1</sup> or *w*2, then there is at least one path from *x* to *y*, and *u* does not lie on it, which is a contradiction. Moreover, we obtain the same contradiction when *u* is not equal to *w*<sup>1</sup> and *w*2.

Subcase(*b*): If *x* ∈ *X* and *y* ∈ *Y*, then *x* is not adjacent to *y*, and by the same arguments as used in the above subcase, we obtain the contradiction.

Subcase(*c*): If *x*, *y* ∈ *Y*, then *x* may or may not adjacent to *y*. Thus in both the cases and by the same arguments as used in the subcase(*a*), we obtain a contradiction.

**Corollary 9.** *Let I be an ideal of* R*. Then, TI*(Γ(R)) *has a cut-vertex if and only if I is a zero ideal.*

**Remark 3.** *If TI*(Γ(R)) *has a cut-vertex, then* 0 *is the cut-vertex of TI*(Γ(R))*.*

**Theorem 8.** *k*(*TI*(Γ(R))) = |*I*|.

**Proof.** By Remark 2, *δ*(*TI*(Γ(R))) = |*I*|. Moreover, for any graph *G*, *k*(*G*) *λ*(*G*) *δ*(*G*). Therefore, *k*(*TI*(Γ(R))) |*I*|. Now, if *u* ∈ *I*, then *u* is adjacent to each vertex *v* ∈ *TI*(Γ(R)). Thus, the minimum vertex-cut is the set of all those vertices in *I*. Therefore, *k*(*TI*(Γ(R))) ≥ |*I*|, and hence *k*(*TI*(Γ(R))) = |*I*|.

**Remark 4.** *For any commutative ring* R *with* 1 = 0*, the elements of I with some elements of Y form a vertex-cut of TI*(Γ(R))*. However, only the elements of I is the minimum vertex-cut of TI*(Γ(R))*.*

**Theorem 9.** *TI*(Γ(R)) *has a bridge if and only if either TI*(Γ(R)) *is a graph with two vertices (i.e., TI*(Γ(R)) ∼= *TI*(Γ(Z2))*), or I is the zero ideal of* R*.*

**Proof.** Suppose that *TI*(Γ(R)) has a bridge. Now, we have the following cases.

Case(*i*): If |R| = 2, then it is clear that *TI*(Γ(R)) ∼= *<sup>K</sup>*2, which has a bridge. Hence, *TI*(Γ(R)) ∼= *TI*(Γ(Z2)) .

Case(*ii*): If |R| - 3, then either *V*(*TI*(Γ(R))) ⊆ *I*, *V*(*TI*(Γ(R))) ⊆ *X*, or *V*(*TI*(Γ(R))) ⊆ *Y*. Let *e* = *uv* be the bridge of *TI*(Γ(R)). Since there is no edge neither between any two elements of *X* nor between any element of *X* with element of *Y*, we have the following subcases.

Subcase(*a*): If *u*, *v* ∈ *I*, and |R| - 3, then there exists *w* ∈ R such that *u* and *v* are adjacent to *w*. We note that *u* − *v* − *w* − *u* is a cycle, and there is no bridge between them; we obtain a contradiction.

Subcase(*b*): If *u*, *v* ∈ *Y*, and |R| - 3, then there exists *w* ∈ *I* such that *u* and *v* are adjacent to *w*. We note that *u* − *v* − *w* − *u* is a cycle, and there is no bridge between them; we obtain a contradiction.

Subcase(*c*): If *u* ∈ *I*, *v* ∈ *X*, and |R| -3, then there are two possibilities.

If |*I*| = 1 (i.e., *I* is a zero ideal of R), then *v* is adjacent to *u* = 0 only, and *uv* is a bridge of *TI*(Γ(R)). Moreover, each vertex of *X* with *u* forms a bridge of *TI*(Γ(R)) (i.e., we have |*X*| bridges).

If |*I*| ≥ 2 (i.e., *I* is not zero ideal of R), then there exists *u* = *w* ∈ *I* such that *u* and *v* are adjacent to *w*. We note that *u* − *v* − *w* − *u* is a cycle, and there is no bridge between them; we obtain a contradiction.

Subcase(*d*): If *u* ∈ *I*, *v* ∈ *Y*, and |R| -3, then there are three possibilities.

If |*I*| = 1 (i.e., *I* is a zero ideal of R) and |*Y*| = 1, then *v* is adjacent to *u* = 0 only, and *uv* is a bridge of *TI*(Γ(R)).

If |*I*| = 1 (i.e., *I* is a zero ideal of R) and |*Y*| ≥ 2, then there exists at least one element *v* = *w* ∈ *Y* such that *v* and *w* are connected vertices by a path *P*. Since each elements of *Y* are adjacent to elements of *I*, *v* − *P* − *w* − 0 − *v* is a cycle and there is no bridge between them. This is a contradiction.

If |*I*| ≥ 2 (i.e., *I* is not zero ideal of R), then there exists *u* = *w* ∈ *I* such that *v* is adjacent to *w*. Since each element of *I* is adjacent, *u* − *v* − *w* − *u* is a cycle, and there is no bridge between them; we obtain a contradiction.

Conversely, suppose that *TI*(Γ(R)) is a graph with two vertices. Then, it is clear that *TI*(Γ(R)) has a bridge. Let us suppose that *I* is the zero ideal of R (i.e., *I* = {0}) and |R| ≥ 3. Then, we have at least one element *u* in *X*. Hence, 0*u* is a bridge in *TI*(Γ(R)).

**Remark 5.** *If the ring* R ∼= Z<sup>2</sup> *or <sup>I</sup> is the zero ideal of* R*, then TI*(Γ(R)) *has a bridge and vice versa, i.e., if TI*(Γ(R)) *has a bridge, then the ring* R ∼= Z<sup>2</sup> *or I is the zero ideal of* R*.*
