Here is a proof for any graded (not necessary Notherian) 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 $$\mathrm{begin} (f) = \min\{i | f_i \neq 0\},$$ $$\mathrm{end} (f) = \max\{i | f_i \neq 0\},$$ $$c(f) = \mathrm{end}(f) - \mathrm{begin}(f).$$ Thus $f$ is a non-zero homogeneous iff $c(f) = 0$

Assume can 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 $b = \mathrm{g}$ 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 minimaltity 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.

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.

Here is a proof for any graded (not necessary Notherian) 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 $$\mathrm{begin} (f) = \min\{i | f_i \neq 0\},$$ $$\mathrm{end} (f) = \max\{i | f_i \neq 0\},$$ $$c(f) = \mathrm{end}(f) - \mathrm{begin}(f).$$ Thus $f$ is a non-zero homogeneous iff $c(f) = 0$

Assume can 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 $b = \mathrm{g}$ 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 minimaltity 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.

Here is a proof for any graded (not necessary Notherian) 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 $$\mathrm{begin} (f) = \min\{i | f_i \neq 0\},$$ $$\mathrm{end} (f) = \max\{i | f_i \neq 0\},$$ $$c(f) = \mathrm{end}(f) - \mathrm{begin}(f).$$ Thus $f$ is a non-zero homogeneous iff $c(f) = 0$

Assume can 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 $b = \mathrm{g}$ 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 minimaltity 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.