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Denote by $T_0,T_1$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(f)$. Denote by $X$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(C_0)$. Denote by $Y$ a global section of $\mathcal{O}_{\mathbb{F}_1}(C_0+f)$ that is linearly independent from $T_0X,T_1X$. Then the "total coordinate ring" or "Cox ring" of $\mathbb{F}_1$ is $S_*(\mathbb{F}_1) = k[T_0,T_1,X,Y]$. This is a $\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}$-graded ring where you can think of elements in $\mathbb{Z}\times \mathbb{Z}$ as linear combinations $af+bC_0$. Thus the degree of $T_0,T_1$ is $1f+0C_0$, the degree of $X$ is $0f+1C_0$, and the degree of $Y$ is $1f+1C_0$. The embedding of $\mathbb{F}_1 \hookrightarrow \mathbb{P}^{17}$ induces a graded algebra homomorphism $S_*(\mathbb{P}^{17}) \to S_*(\mathbb{F}_1)$, i.e., $$k[Z_{l_0,l_1,m,n}|(l_0,l_1,m,n) \in I] \to k[T_0,T_1,X,Y], \ Z_{l_0,l_1,m,n} \mapsto T_0^{l_0}T_1^{l_1}X^mY^n.$$ Here, by definition, $I$ is the set of $4$-tuples of nonegative integers such that $$(l_0+l_1)(1f+0C_0) + m(0f+1C_0) + n(1f+1C_0) = 5f+3C_0, $$ i.e., $l_0+l_1+n$ equals $5$ and $m+n$ equals $3$. In particular, this is a monomial homomorphism, or equivalently, a map of semigroup rings coming from the map of semigroups $\phi:\mathbb{Z}_{\geq 0}^{18} \mapsto \mathbb{Z}_{\geq 0}^4$ as indicated. Therefore the kernel is generated by the rank 4, quadratic binomials, $$ Z_{l_0,l_1,m,n}Z_{l_0',l_1',m',n'} - Z_{l_0'',l_1'',m'',n''}Z_{l_0''',l_1''',m''',n'''}, $$ $$ \phi(l_0+l_0',l_1+l_1',m+m',n+n') = \phi(l_0''+l_0''', l_1''+l_1''', m''+m''',n''+n'''). $$

$\textbf{Edit}.$ I realize now that I am not certain whether these binomials are sufficient to generate the kernel. Certainly the kernel is generated by binomials, and the kernel includes these binomials. I believe these binomials define the closed subscheme, i.e., the saturation of the ideal generated by these binomials equals the full kernel. However, I do not immediately see how to check that there are no further binomial generators. Eisenbud and Sturmfels do give algorithms to check whether a binomial ideal is prime; this might do the trick.

$\textbf{Second Edit.}$ Actually, there is a way to check this using Macaulay. Macaulay will tell you the Hilbert polynomial of the closed subscheme cut out by given generators. So you can input the generators above and compare the Hilbert polynomial to the Hilbert polynomial of the surface $\mathbb{F}_1$, namely $(7t+2)(3t+1)/2$.

$\textbf{Third Edit.}$ Hochster's theorem about Cohen-Macaulayness of toric ideals implies that the graded ideal of this projectively normal variety of dimension $2$ is generated by binomials of degree $\leq 2$. Thus the 115 quadratic binomials above generate the graded ideal of this toric surface.

Denote by $T_0,T_1$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(f)$. Denote by $X$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(C_0)$. Denote by $Y$ a global section of $\mathcal{O}_{\mathbb{F}_1}(C_0+f)$ that is linearly independent from $T_0X,T_1X$. Then the "total coordinate ring" or "Cox ring" of $\mathbb{F}_1$ is $S_*(\mathbb{F}_1) = k[T_0,T_1,X,Y]$. This is a $\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}$-graded ring where you can think of elements in $\mathbb{Z}\times \mathbb{Z}$ as linear combinations $af+bC_0$. Thus the degree of $T_0,T_1$ is $1f+0C_0$, the degree of $X$ is $0f+1C_0$, and the degree of $Y$ is $1f+1C_0$. The embedding of $\mathbb{F}_1 \hookrightarrow \mathbb{P}^{17}$ induces a graded algebra homomorphism $S_*(\mathbb{P}^{17}) \to S_*(\mathbb{F}_1)$, i.e., $$k[Z_{l_0,l_1,m,n}|(l_0,l_1,m,n) \in I] \to k[T_0,T_1,X,Y], \ Z_{l_0,l_1,m,n} \mapsto T_0^{l_0}T_1^{l_1}X^mY^n.$$ Here, by definition, $I$ is the set of $4$-tuples of nonegative integers such that $$(l_0+l_1)(1f+0C_0) + m(0f+1C_0) + n(1f+1C_0) = 5f+3C_0, $$ i.e., $l_0+l_1+n$ equals $5$ and $m+n$ equals $3$. In particular, this is a monomial homomorphism, or equivalently, a map of semigroup rings coming from the map of semigroups $\phi:\mathbb{Z}_{\geq 0}^{18} \mapsto \mathbb{Z}_{\geq 0}^4$ as indicated. Therefore the kernel is generated by the rank 4, quadratic binomials, $$ Z_{l_0,l_1,m,n}Z_{l_0',l_1',m',n'} - Z_{l_0'',l_1'',m'',n''}Z_{l_0''',l_1''',m''',n'''}, $$ $$ \phi(l_0+l_0',l_1+l_1',m+m',n+n') = \phi(l_0''+l_0''', l_1''+l_1''', m''+m''',n''+n'''). $$

$\textbf{Edit}.$ I realize now that I am not certain whether these binomials are sufficient to generate the kernel. Certainly the kernel is generated by binomials, and the kernel includes these binomials. I believe these binomials define the closed subscheme, i.e., the saturation of the ideal generated by these binomials equals the full kernel. However, I do not immediately see how to check that there are no further binomial generators. Eisenbud and Sturmfels do give algorithms to check whether a binomial ideal is prime; this might do the trick.

$\textbf{Second Edit.}$ Actually, there is a way to check this using Macaulay. Macaulay will tell you the Hilbert polynomial of the closed subscheme cut out by given generators. So you can input the generators above and compare the Hilbert polynomial to the Hilbert polynomial of the surface $\mathbb{F}_1$, namely $(7t+2)(3t+1)/2$.

$\textbf{Third Edit.}$ Hochster's theorem about Cohen-Macaulayness of toric ideals implies that the graded ideal of this projectively normal variety of dimension $2$ is generated by binomials of degree $\leq 2$. Thus the 115 quadratic binomials above generate the graded ideal of this toric surface.

$\textbf{Fourth Edit.}$ Although it is not so clear in the literature, Hochster's theorem actually says that the degrees of generating binomials are bounded by $r$ where $r$ is the dimension of the $\textit{affine}$ cone, not the projective toric variety. So Hochster's theorem implies, in this case, that the ideal is generated by binomials of degree at most $3$, not $2$. However, it is straightforward to check that every degree $3$ binomial is in the ideal generated by the 115 quadratic binomials.

Denote by $T_0,T_1$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(f)$. Denote by $X$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(C_0)$. Denote by $Y$ a global section of $\mathcal{O}_{\mathbb{F}_1}(C_0+f)$ that is linearly independent from $T_0X,T_1X$. Then the "total coordinate ring" or "Cox ring" of $\mathbb{F}_1$ is $S_*(\mathbb{F}_1) = k[T_0,T_1,X,Y]$. This is a $\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}$-graded ring where you can think of elements in $\mathbb{Z}\times \mathbb{Z}$ as linear combinations $af+bC_0$. Thus the degree of $T_0,T_1$ is $1f+0C_0$, the degree of $X$ is $0f+1C_0$, and the degree of $Y$ is $1f+1C_0$. The embedding of $\mathbb{F}_1 \hookrightarrow \mathbb{P}^{17}$ induces a graded algebra homomorphism $S_*(\mathbb{P}^{17}) \to S_*(\mathbb{F}_1)$, i.e., $$k[Z_{l_0,l_1,m,n}|(l_0,l_1,m,n) \in I] \to k[T_0,T_1,X,Y], \ Z_{l_0,l_1,m,n} \mapsto T_0^{l_0}T_1^{l_1}X^mY^n.$$ Here, by definition, $I$ is the set of $4$-tuples of nonegative integers such that $$(l_0+l_1)(1f+0C_0) + m(0f+1C_0) + n(1f+1C_0) = 5f+3C_0, $$ i.e., $l_0+l_1+n$ equals $5$ and $m+n$ equals $3$. In particular, this is a monomial homomorphism, or equivalently, a map of semigroup rings coming from the map of semigroups $\phi:\mathbb{Z}_{\geq 0}^{18} \mapsto \mathbb{Z}_{\geq 0}^4$ as indicated. Therefore the kernel is generated by the rank 4, quadratic binomials, $$ Z_{l_0,l_1,m,n}Z_{l_0',l_1',m',n'} - Z_{l_0'',l_1'',m'',n''}Z_{l_0''',l_1''',m''',n'''}, $$ $$ \phi(l_0+l_0',l_1+l_1',m+m',n+n') = \phi(l_0''+l_0''', l_1''+l_1''', m''+m''',n''+n'''). $$

$\textbf{Edit}.$ I realize now that I am not certain whether these binomials are sufficient to generate the kernel. Certainly the kernel is generated by binomials, and the kernel includes these binomials. I believe these binomials define the closed subscheme, i.e., the saturation of the ideal generated by these binomials equals the full kernel. However, I do not immediately see how to check that there are no further binomial generators. Eisenbud and Sturmfels do give algorithms to check whether a binomial ideal is prime; this might do the trick.

$\textbf{Second Edit.}$ Actually, there is a way to check this using Macaulay. Macaulay will tell you the Hilbert polynomial of the closed subscheme cut out by given generators. So you can input the generators above and compare the Hilbert polynomial to the Hilbert polynomial of the surface $\mathbb{F}_1$, namely $(7t+2)(3t+1)/2$.

Denote by $T_0,T_1$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(f)$. Denote by $X$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(C_0)$. Denote by $Y$ a global section of $\mathcal{O}_{\mathbb{F}_1}(C_0+f)$ that is linearly independent from $T_0X,T_1X$. Then the "total coordinate ring" or "Cox ring" of $\mathbb{F}_1$ is $S_*(\mathbb{F}_1) = k[T_0,T_1,X,Y]$. This is a $\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}$-graded ring where you can think of elements in $\mathbb{Z}\times \mathbb{Z}$ as linear combinations $af+bC_0$. Thus the degree of $T_0,T_1$ is $1f+0C_0$, the degree of $X$ is $0f+1C_0$, and the degree of $Y$ is $1f+1C_0$. The embedding of $\mathbb{F}_1 \hookrightarrow \mathbb{P}^{17}$ induces a graded algebra homomorphism $S_*(\mathbb{P}^{17}) \to S_*(\mathbb{F}_1)$, i.e., $$k[Z_{l_0,l_1,m,n}|(l_0,l_1,m,n) \in I] \to k[T_0,T_1,X,Y], \ Z_{l_0,l_1,m,n} \mapsto T_0^{l_0}T_1^{l_1}X^mY^n.$$ Here, by definition, $I$ is the set of $4$-tuples of nonegative integers such that $$(l_0+l_1)(1f+0C_0) + m(0f+1C_0) + n(1f+1C_0) = 5f+3C_0, $$ i.e., $l_0+l_1+n$ equals $5$ and $m+n$ equals $3$. In particular, this is a monomial homomorphism, or equivalently, a map of semigroup rings coming from the map of semigroups $\phi:\mathbb{Z}_{\geq 0}^{18} \mapsto \mathbb{Z}_{\geq 0}^4$ as indicated. Therefore the kernel is generated by the rank 4, quadratic binomials, $$ Z_{l_0,l_1,m,n}Z_{l_0',l_1',m',n'} - Z_{l_0'',l_1'',m'',n''}Z_{l_0''',l_1''',m''',n'''}, $$ $$ \phi(l_0+l_0',l_1+l_1',m+m',n+n') = \phi(l_0''+l_0''', l_1''+l_1''', m''+m''',n''+n'''). $$

$\textbf{Edit}.$ I realize now that I am not certain whether these binomials are sufficient to generate the kernel. Certainly the kernel is generated by binomials, and the kernel includes these binomials. I believe these binomials define the closed subscheme, i.e., the saturation of the ideal generated by these binomials equals the full kernel. However, I do not immediately see how to check that there are no further binomial generators. Eisenbud and Sturmfels do give algorithms to check whether a binomial ideal is prime; this might do the trick.

$\textbf{Second Edit.}$ Actually, there is a way to check this using Macaulay. Macaulay will tell you the Hilbert polynomial of the closed subscheme cut out by given generators. So you can input the generators above and compare the Hilbert polynomial to the Hilbert polynomial of the surface $\mathbb{F}_1$, namely $(7t+2)(3t+1)/2$.

$\textbf{Third Edit.}$ Hochster's theorem about Cohen-Macaulayness of toric ideals implies that the graded ideal of this projectively normal variety of dimension $2$ is generated by binomials of degree $\leq 2$. Thus the 115 quadratic binomials above generate the graded ideal of this toric surface.

Denote by $T_0,T_1$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(f)$. Denote by $X$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(C_0)$. Denote by $Y$ a global section of $\mathcal{O}_{\mathbb{F}_1}(C_0+f)$ that is linearly independent from $T_0X,T_1X$. Then the "total coordinate ring" or "Cox ring" of $\mathbb{F}_1$ is $S_*(\mathbb{F}_1) = k[T_0,T_1,X,Y]$. This is a $\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}$-graded ring where you can think of elements in $\mathbb{Z}\times \mathbb{Z}$ as linear combinations $af+bC_0$. Thus the degree of $T_0,T_1$ is $1f+0C_0$, the degree of $X$ is $0f+1C_0$, and the degree of $Y$ is $1f+1C_0$. The embedding of $\mathbb{F}_1 \hookrightarrow \mathbb{P}^{17}$ induces a graded algebra homomorphism $S_*(\mathbb{P}^{17}) \to S_*(\mathbb{F}_1)$, i.e., $$k[Z_{l_0,l_1,m,n}|(l_0,l_1,m,n) \in I] \to k[T_0,T_1,X,Y], \ Z_{l_0,l_1,m,n} \mapsto T_0^{l_0}T_1^{l_1}X^mY^n.$$ Here, by definition, $I$ is the set of $4$-tuples of nonegative integers such that $$(l_0+l_1)(1f+0C_0) + m(0f+1C_0) + n(1f+1C_0) = 5f+3C_0, $$ i.e., $l_0+l_1+n$ equals $5$ and $m+n$ equals $3$. In particular, this is a monomial homomorphism, or equivalently, a map of semigroup rings coming from the map of semigroups $\phi:\mathbb{Z}_{\geq 0}^{18} \mapsto \mathbb{Z}_{\geq 0}^4$ as indicated. Therefore the kernel is generated by the rank 4, quadratic binomials, $$ Z_{l_0,l_1,m,n}Z_{l_0',l_1',m',n'} - Z_{l_0'',l_1'',m'',n''}Z_{l_0''',l_1''',m''',n'''}, $$ $$ \phi(l_0+l_0',l_1+l_1',m+m',n+n') = \phi(l_0''+l_0''', l_1''+l_1''', m''+m''',n''+n'''). $$

$\textbf{Edit}.$ I realize now that I am not certain whether these binomials are sufficient to generate the kernel. Certainly the kernel is generated by binomials, and the kernel includes these binomials. I believe these binomials define the closed subscheme, i.e., the saturation of the ideal generated by these binomials equals the full kernel. However, I do not immediately see how to check that there are no further binomial generators. Eisenbud and Sturmfels do give algorithms to check whether a binomial ideal is prime; this might do the trick.

Denote by $T_0,T_1$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(f)$. Denote by $X$ a basis for the global sections of $\mathcal{O}_{\mathbb{F}_1}(C_0)$. Denote by $Y$ a global section of $\mathcal{O}_{\mathbb{F}_1}(C_0+f)$ that is linearly independent from $T_0X,T_1X$. Then the "total coordinate ring" or "Cox ring" of $\mathbb{F}_1$ is $S_*(\mathbb{F}_1) = k[T_0,T_1,X,Y]$. This is a $\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}$-graded ring where you can think of elements in $\mathbb{Z}\times \mathbb{Z}$ as linear combinations $af+bC_0$. Thus the degree of $T_0,T_1$ is $1f+0C_0$, the degree of $X$ is $0f+1C_0$, and the degree of $Y$ is $1f+1C_0$. The embedding of $\mathbb{F}_1 \hookrightarrow \mathbb{P}^{17}$ induces a graded algebra homomorphism $S_*(\mathbb{P}^{17}) \to S_*(\mathbb{F}_1)$, i.e., $$k[Z_{l_0,l_1,m,n}|(l_0,l_1,m,n) \in I] \to k[T_0,T_1,X,Y], \ Z_{l_0,l_1,m,n} \mapsto T_0^{l_0}T_1^{l_1}X^mY^n.$$ Here, by definition, $I$ is the set of $4$-tuples of nonegative integers such that $$(l_0+l_1)(1f+0C_0) + m(0f+1C_0) + n(1f+1C_0) = 5f+3C_0, $$ i.e., $l_0+l_1+n$ equals $5$ and $m+n$ equals $3$. In particular, this is a monomial homomorphism, or equivalently, a map of semigroup rings coming from the map of semigroups $\phi:\mathbb{Z}_{\geq 0}^{18} \mapsto \mathbb{Z}_{\geq 0}^4$ as indicated. Therefore the kernel is generated by the rank 4, quadratic binomials, $$ Z_{l_0,l_1,m,n}Z_{l_0',l_1',m',n'} - Z_{l_0'',l_1'',m'',n''}Z_{l_0''',l_1''',m''',n'''}, $$ $$ \phi(l_0+l_0',l_1+l_1',m+m',n+n') = \phi(l_0''+l_0''', l_1''+l_1''', m''+m''',n''+n'''). $$

$\textbf{Edit}.$ I realize now that I am not certain whether these binomials are sufficient to generate the kernel. Certainly the kernel is generated by binomials, and the kernel includes these binomials. I believe these binomials define the closed subscheme, i.e., the saturation of the ideal generated by these binomials equals the full kernel. However, I do not immediately see how to check that there are no further binomial generators. Eisenbud and Sturmfels do give algorithms to check whether a binomial ideal is prime; this might do the trick.

$\textbf{Second Edit.}$ Actually, there is a way to check this using Macaulay. Macaulay will tell you the Hilbert polynomial of the closed subscheme cut out by given generators. So you can input the generators above and compare the Hilbert polynomial to the Hilbert polynomial of the surface $\mathbb{F}_1$, namely $(7t+2)(3t+1)/2$.