Not finitely generated ideal but only contained in finitely generated ideals implies ideal is prime I would like to prove the following.

Let $\mathfrak{a}$ be an ideal in a commutative ring, $R$, with the property that $\mathfrak{a}$ is not finitely generated but every ideal properly containing $\mathfrak{a}$ is finitely generated then in fact $\mathfrak{a}$ is a prime ideal.

My idea is to try contradiction. 
So suppose there are $a,b\in R$ with $ab\in \mathfrak{a}$ but $a, b\notin \mathfrak{a}$. Then $(a)+\mathfrak{a}$ and $(b)+\mathfrak{a}$ are ideals that properly contain $\mathfrak{a}$ and so they are finitely generated. However, at this point I'm not really sure what I should do next.
Thanks for any help!
 A: First, here is a general lemma:

Lemma: Let $I,J\subseteq R$ be ideals and suppose $I\cap J$ and $I+J$ are finitely generated.  Then $I$ and $J$ are finitely generated. 

Proof: Let $x_1,\dots, x_n$ generate $I\cap J$, and let $y_1+z_1,\dots,y_m+z_m$ generate $I+J$, for $y_i\in I$, $z_i\in J$.  If $w\in I$, then $w\in I+J$, so we can write $w=\sum c_i(y_i+z_i)$.  Note that $\sum c_iz_i=w-\sum c_i y_i\in I\cap J$, so we can write $\sum c_iz_i=\sum d_j x_j$.  We have now written $w=\sum c_iy_i+\sum d_j x_j$.  Since the $y_i$ and $x_j$ are all in $I$ and $w\in I$ was arbitrary, this means $\{x_1,\dots,x_n,y_1,\dots,y_m\}$ is a finite set of generators for $I$.  By swapping the roles of $I$ and $J$, we see $J$ is also finitely generated.
To solve your problem, we will apply the Lemma to $I=\mathfrak{a}$ and $J=(a)$.  You have already noted that $\mathfrak{a}+(a)$ is finitely generated.  To show that $\mathfrak{a}\cap (a)$ is finitely generated, first consider the ideal $K=\{c\in R:ac\in\mathfrak{a}\}$.  Clearly $\mathfrak{a}\subseteq K$, and by hypothesis $b\in K$ as well, so $K$ is strictly larger than $\mathfrak{a}$ and must be finitely generated.  But $aK=\mathfrak{a}\cap(a)$, so this implies $\mathfrak{a}\cap(a)$ is also finitely generated.  By the Lemma, we get that $\mathfrak{a}$ is finitely generated, which is a contradiction.
