A space for which both terms in the Universal Coefficient Theorem are non-zero The Universal Coefficient Theorem gives us a powerful tool to calculate cohomology with coefficients, if one already knows the homology of the space. For the sake of completeness, here it is:$$0\rightarrow Ext_R^1(H_{n-1}(X;R),N)\rightarrow H^n(X;N)\rightarrow Hom_R(H_n(X;R),N)\rightarrow 0.$$
However, in many of the standard examples (torus, klein bottle, projective spaces, complex projective spaces), there is always a zero term in the UCT. That is, either $Ext_R^1(H_{n-1}(X;R),N)$ is zero or $Hom_R(H_n(X;R),N)$ is zero. 
Can anyone give me an example of a space $X$ and a ring $R$ in which really both terms contribute to the cohomology of $X$?
Thanks! 
Edit: As mentioned in the comments, the space $X=\mathbb{R}P^2\vee S^2$ indeed works. I would still be interested in hearing an example which is not necessarily a wedge of spaces. 
 A: I will flesh out E. KOW's comment into an answer.
Consider the space $X = \mathbb{RP}^2\vee S^2$, and take $N=\mathbb{Z}$ (unless otherwise specified, all my (co)homology groups will implicitly have integer coefficients, and $Ext$ and $Hom$ will also be over $\mathbb{Z}$). Then $H_0(X) \cong \mathbb{Z}$, $H_1(X)\cong \mathbb{Z}/2$, and $H_2(X) \cong \mathbb{Z}$.
Now if we try to compute $H^2(X)$ with UTC we get the terms $$Hom(H_2(X), \mathbb{Z}) \cong Hom(\mathbb{Z}, \mathbb{Z})\cong \mathbb{Z}$$ and $$Ext^1(H_1(X), \mathbb{Z})\cong Ext(\mathbb{Z}/2, \mathbb{Z})\cong \mathbb{Z}/2. $$

Edit: There is an example over $\mathbb{Z}$ where the space is not "decomposable" into a wedge sum/connected sum/cartesian product (or at least such a decomposition is far from obvious): $H^4(BO(n);\mathbb{Z})$ for $n\geq 2$ (my proof is non-trivial, there may be a simpler example).
Observe two things: for any abelian group $G$, 1) $Hom_{\mathbb{Z}}(G, \mathbb{Z})$ cannot contain torsion and 2) $Ext^1_{\mathbb{Z}}(G, \mathbb{Z})$ is entirely torsion (this second statement follows from formal properties of ext, namely additivity, $Ext^1(\mathbb{Z}, G) = 0$, and $Ext^1(\mathbb{Z}/n, \mathbb{Z})\cong \mathbb{Z}/n$). Therefore if $X$ is a space such that $H^n(X;\mathbb{Z})$ contains non-trivial free and torsion parts then by UCT both $Ext^1(H_{n-1}(X), \mathbb{Z})$ and $Hom(H_n(X), \mathbb{Z})$ must be non-trivial.
The integral cohomology ring of $BO(n)$ is known to have an additive decomposition into two parts: a free polynomial ring generated by the Pontryagin classes $\{p_1,\dots,p_{\lfloor \frac{n}{2} \rfloor}\}$, and a torsion part given by $im(\beta)$, the image of the Bockstein homomorphism coming from the short exact sequence $\mathbb{Z}\to\mathbb{Z} \to \mathbb{Z}/2$ (see for example Appendix B of Lawson and Michelsohn's "Spin Geometry"). The first Pontryagin class $p_1\in H^4(BO(n))$ generates a torsion-free subgroup, and in my answer to this other question I show that $\beta(w_i)\neq 0$ for all $i > 0$, so in particular there is a non-trivial order $2$ element $\beta(w_3) \in H^4(BO(n);\mathbb{Z})$. Therefore by the above general paragraph about the UCT it follows that both $Ext^1(H_3(BO(n)), \mathbb{Z})$ and $Hom(H_4(BO(n)), \mathbb{Z})$ must be non-trivial.
As for whether it's indecomposable, I'm pretty sure but I don't have proofs yet. I believe the ring structure rules out writing it non-trivially as a wedge product, but I'm not sure how to argue $BO(n)$ isn't a cartesian product.
