One knows that graded ideals in polynomial rings over a field are primary iff they are graded-primary. What about the irreducible ideals?
Let $I$ be a graded ideal in a polynomial ring over a field. Is it true that $I$ is irreducible if and only if it can't be written as an intersection of two graded ideals (properly containing $I$)?
If the answer is positive in this case, then what's going on if change the ring by an arbitrary $\Bbb N$ (or $\Bbb Z$)-graded ring?
Here is a proof for any graded (not necessary Noetherian) ring based on this proof. For simplify we assume $R = \oplus_{n \in \mathbb{N}} R_n$, although the proof hold for any graded ring.
Firstly, we give some definitions. For any $f = f_0 + \cdots + f_n$ we define $$\mathrm{begin} (f) = \min\{i \mid f_i \neq 0\},$$ $$\mathrm{end} (f) = \max\{i \mid f_i \neq 0\},$$ $$c(f) = \mathrm{end}(f) - \mathrm{begin}(f).$$ Thus $f$ is a non-zero homogeneous iff $c(f) = 0$.
Assume that $(0)$ is a graded-irreducible ideal of $R$ and prove that it is an irreducible ideal. Assume $(0)$ is the intersection of two non-zero ideals. We can choose two non-zero elements $f$ and $g$ such that $(0) = (f) \cap (g)$. Here we choose $f$ and $g$ such that $c(f) + c(g)$ is minimal. Since $(0)$ is graded irreducible we have $c(f) + c(g)>0$. Assume $c(f) \le c(g)$, thus $c(g)>0$ i.e., $g$ is non-homogeneous. Let $a = \mathrm{begin}(f)$ and $b = \mathrm{begin}(g)$ we have $(f_a) \cap (g_b) \neq 0$. Thus we have two homogeneous elements $d$ and $e$ such that $df_a = eg_b \neq 0$. Replacing $f$ and $g$ by $df$ and $eg$, respectively, we can assume henceforth that $\mathrm{begin}(f) = \mathrm{begin}(g):=b$ and $f_b = g_b$.
Similar to this proof we have
Claim 1: Let $r$ be a homogeneous element and $b = \mathrm{begin}(f)$ such that $rf_b = 0$. Then $rf = 0$ and $rg = 0$.
Claim 2: Let $h = h_0 + \cdots + h_t$ be an element such that $hf = 0$. Then $h_if = 0$ for all $i = 0, ..., t$.
Claim 3: $hf = 0$ iff $hg = 0$.
Let $g' = g-f$ we have $c(g')<c(g)$ since $f_b = g_b$. By the minimality we have $(f) \cap (g-f) \neq 0$. Thus there are $u, v$ and $w$ such that $$0\neq w = uf = v(g-f).$$ If $vg \neq 0$ then $vg = (u+v)f \in (f) \cap (g)$, a contradiction.
If $vg = 0$, then $vf = 0$ by Claim 3. So $w = 0$. This is also a contradiction.
The proof is complete.
There is a positive answer to this question now by Pham Hung Quy's post here and this paper, but one can also ask this question about other gradings. This is not an answer, but a long comment:
Descending further in the hierarchy of grading monoids, a commutative Noetherian monoids is any quotient of $\mathbb{N}^n$ by a congruence. An ideal is graded by such a monoid if and only if it is a binomial ideal (Proposition 1.11 in Binomial Ideals by Eisenbud and Sturmfels) and the theory of binomial ideals can be developed somewhat similarly to that of monomial ideals. However, an ideal graded by a fixed commutative Noetherian monoid $Q$ need not even have a primary decomposition into $Q$-graded ideals. For example, $(x^2-xy, xy-y^2) = (x,y^2) \cap (x-y)$. The ideal on the left hand side is (finely) $Q$-graded for some monoid $Q$ in which $x$ and $y$ have different degrees, but they must have the same degree in the unique minimal prime $(x-y)$.
Now one could ask if in general binomial ideals have binomial irreducible decompositions, that is, letting $Q$ vary in the decomposition. This was an open problem in Eisenbud/Sturmfels, but the answer is negative by Section 6 in Irreducible decomposition of binomial ideals by Kahle, Miller, O'Neill.
