Q:

If $P^h$ is a prime ideal can we say that $P$ is also prime ideal?

Suppose $P$ is a proper ideal in a graded ring $A$ and $P^h$ the largest homogeneous ideal of $A$ contained in $P$. If $P^h$ be a prime ideal can we show that $P$ is also prime ideal in $A$ ?

I tried to prove this but I couldn't. If it is true then how can i prove it $?$
My effort: Taking $a$ and $b$ from $A-P$ want to show that $ab \in A-P$. Now writing $a= a_i+a_{i+1}+\cdots+a_{i+m}$ and $b=b_j+b_{j+1}+\cdots+b_{j+n},$ where $a_k$'s and $b_l$'s are homogeneous elements of degree respectively $k$ and $l$ in $A.$  WLOG one can assume that $a_i\notin P$ and $b_j\notin P.$ I cannot proceed further. Any help will be appreciated.
Thanks.

A:

It is false. Take $R = k[t]$ and let $I$ be any nonzero non graded proper ideal of R such as $I = (t-1)^2$. Then $I^h = 0$ which is prime. 

