Let be $(A,M)$ a local ring (if it needs noetherian ring), $P_1$ and $P_2$ two prime ideals with $P_1\subseteq P_2$, $\ a_1\in M\setminus P_2$. I don't know if it suffices to conclude this sentence:
If $P_1+(a)=P_2+(a)$ so $P_1=P_2$.
Someone can help me? Maybe in a more general setting it is possible to say that if $P+I=Q+I$ so $P=Q$.
Thank you.
On
Here is an example to show that this need not be true if $A$ is not Noetherian. Let $A$ be a valuation ring with value group $\mathbb{Z}^2$ with the lexicographic order. Let $P_1=0$ and let $P_2$ be the prime ideal consisting of elements whose valuation has positive first coordinate. Let $a\in A$ be any element with valuation $(0,1)$. Then $P_1\subset P_2\subset (a)$ so $P_1+(a)=P_2+(a)=(a)$ but $P_1\neq P_2$.
I assume $A$ is commutative since you have commutative algebra as a tag, and I do use a hypothesis that $A$ is noetherian. I am not sure whether the result holds in more general settings.
Lemma: if $R$ is a noetherian integral domain, and $x\in R$ is a non-unit, then the only prime ideal strictly contained in $xR$ is $\{0\}$.
Proof: Let $P\subset xR$ a prime ideal strictly contained in $xR$. Certainly $(xR)P=xP=P$: indeed, given any $p\in P$, since $p\in xR$ there is $r\in R$ such that $p=xr$, whence – because $P$ is prime and $x\notin P$ – we must have $r\in P$. Now, $R$ is noetherian, so $P$ is a finitely generated ideal – ie a finitely generated $R$-module – and thus we may apply Nakayama's lemma to obtain an element $xy\in xR$ such $(1-xy)P=\{0\}$. Now, if $P$ is non-zero – say $p\in P\setminus\{0\}$ – then, because $(1-xy)p=0$ and $R$ is an integral domain, we have that $xy=1$ and so $x$ is a unit, a contradiction. Thus $P$ must indeed be the zero ideal.
Proof of main result: We consider the quotient ring $R:=A/P_1$. Note, since $P_1$ is prime, $R$ is an integral domain. Also, since $(a)+P_1\subseteq M$ is a proper ideal of $A$, the principal ideal $(a+P_1)R$ is certainly a proper ideal of $R$.
Now note, $P_2/P_1\subseteq (P_2+(a))/P_1=(P_1+(a))/P_1=(a+P_1)R$. We claim this inclusion is proper; indeed, otherwise we have $a+P_1\in P_2/P_1$, whence there are $p_2\in P_2$ and $p_1\in P_1$ such that $a=p_2+p_1\in P_2$, a contradiction.
So $P_2/P_1$ is a strict subset of the proper principal ideal $(a+P_1)R$. It is also a prime ideal, since $P_2$ is, and hence we may apply our lemma above to obtain $P_2/P_1=0$. But this means precisely that $P_2=P_1$, as desired.