Example 3·2. Consider a polynomial $f \in K[x_1, \dots, x_n]$ , where K is a valued field. We will denote by trop(f) the polynomial in ${\mathbb{T}[x_1, \dots, x_n]}$ obtained from f by valuating all coefficients of f and replacing the operations multiplication and addition with their tropical counterparts. Let $J \subseteq K[x_1, \dots, x_n]$ , then the ideal $trop(J) = \{ trop(f) \ : \ f\in J \}$ is a realisable tropical ideal.

Proof. Let $I \subseteq {\mathbb{T}[x_1^{\pm 1}, \dots, x_n^{\pm 1}]}$ be a tropical prime ideal and $J = I \cap {\mathbb{T}[x_1, \dots, x_n]}$ . Note that J is a proper ideal and can not contain a monomial. We will show that J is also prime and tropical. Assume it is not prime, that is, there are $f, g \in {\mathbb{T}[x_1, \dots, x_n]}$ such that $fg \in J$ but neither f nor g is in J. But since $J \subseteq I$ then $fg \in I$ and since I is prime then without loss of generality $f \in I$ . However, since $f \in {\mathbb{T}[x_1, \dots, x_n]}$ so $f \in J$ , contradicting the assumption that J is not prime.

Since I is a tropical ideal, for any $f, g \in J \subseteq I$ there exists some $h \in I$ which satisfy the property in Definition 3·1. Since f, g are polynomials, so is h in the definition. So J satisfies the conditions of Definition 3·1 and hence J is tropical.

Proof. Let $f, g \in \mathcal{I}_{C}$ , i.e., $bend(f), bend(g) \subseteq C$ . Consider the pair $(f+g, (f+g)_{\hat \imath})$ for each $i \in supp(f+g)$ . If i is in the support of both summands, then $(f+g, (f+g)_{\hat \imath}) = (f, f_{\hat \imath}) + (g, g_{\hat \imath}) \in C$ , since $bend(f), bend(g) \subseteq C$ . In the case when i is only in the support of one of the summands, say f, then $(f+g, (f+g)_{\hat \imath}) = (f, f_{\hat \imath}) + (g, g) \in C$ . This shows that $bend(f+g) \subseteq C$ and thus $f+g \in \mathcal{I}_{C}$ .

Let $f \in \mathcal{I}_{C}$ and $h \in {\mathbb{T}[x_1^{\pm 1}, \dots, x_n^{\pm 1}]}$ . We will show that $fh \in \mathcal{I}_{C}$ , by showing that $bend(fh) \subseteq C$ . Consider the pair $(fh, (fh)_{\hat k})$ for each $k \in supp(fh)$ . Let $k = (i, j)$ , i.e., $(fh)_k = f_ih_j$ . If $f_ih_j \neq af_sh_r$ , for all $(i,j) \neq (s,r)$ and $a\in {\mathbb{T}}$ , then by direct computation we see that $(fh, (fh)_{\hat k}) = (f, f_{\hat\imath})(h, h_{\hat\jmath})$ .

Now let $f_ig_j = af_sg_r$ for some $(i,j) \neq (s,r)$ and $a\in {\mathbb{T}}$ . Assume without loss of generality that the coefficient of $f_sg_r$ is bigger than that of $f_ig_j$ . Then we can write $(fh, (fh)_{\hat k}) = (f, f_{\hat s})(h_{\hat\jmath}, 0) + (d, d),$ where d is the polynomial

$$d = \sum_{\substack{l\in supp(f), \\l\neq i}}f_lh_j + \sum_{\substack{l\in supp(h), \\l\neq r}}f_sh_l.$$

Now $(f, f_{\hat s})(h_{\hat\jmath}, 0)$ is in C by Proposition 2·3. The pair (d,d) also in C since the diagonal is contained in every congruence and so we conclude that $(fh, (fh)_{\hat k}) \in C$ .

Example 3·5. Let I be the ideal of ${\mathbb{B}}[x^{\pm 1},y^{\pm 1}]$ generated by the set $\{x+y, x+z\}$ . Observe that $bend(y+z) \subseteq Bend(I)$ and so $y+z \in \mathcal{I}_{Bend(I)}$ but $y+z \not\in I$ .

Proof. For ${\textrm{closed}}$ ideals J, L assume that $f \in \mathcal{I}_{Bend(J \cap L)}$ . Then we have $f \in \mathcal{I}_{Bend(J)}$ and $f \in \mathcal{I}_{Bend(L)}$ . Since $\mathcal{I}_{Bend(J)} = J$ and $\mathcal{I}_{Bend(L)} = L$ we conclude that $f \in J \cap L$ .

Example 3·8.

1. (i) All tropical ideals are ${\textrm{closed}}$ by [ Reference Maclagan and Rincón16 , theorem 1·1].

2. (ii) While the intersection of tropical ideals is not necessarily tropical, it is a ${\textrm{closed}}$ ideal.

3. (iii) The unique maximal ideal of ${\mathbb{T}[x_1, \dots, x_n]}$ consisting of every non-constant polynomial is ${\textrm{closed}}$ .

Remark 3·9. To every congruence C one can associate at most one ${\textrm{closed}}$ ideal I such that $C=Bend(I)$ and this ${\textrm{closed}}$ ideal is $\mathcal{I}_{C}$ by definition.

Example 3·10. Let P be a geometric prime corresponding to the point $(b_1,\ldots,b_n)\in{\mathbb{T}}^n$ . Fix a field k with a surjective valuation $v:k\to{\mathbb{T}}$ and pick $a_1,\ldots,a_n\in k$ such that $v(a_i)=b_i$ . Let $\mathfrak{m}$ be the maximal ideal of $k[x_1\ldots,x_n]$ corresponding to the point $(a_1,\ldots,a_n)\in k^n$ . Now $\mathcal{I}_{P}$ is the tropicalisation of $\mathfrak{m}$ , and is therefore a tropical ideal. In particular, $\mathcal{I}_{P}$ consists of those polynomials which attain their maximum at least twice at P. then by Proposition 2·10 each equivalence class of the congruence P contains a single term, i.e., no polynomial will attain its maximum twice with respect to the order induced by P, hence $\mathcal{I}_{P}$ is the zero ideal.

Proof. If P is a minimal prime congruence on ${\mathbb{T}[x_1^{\pm 1}, \dots, x_n^{\pm 1}]}$ , then as noted in Example 3·10 the ideal $\mathcal{I}_{P}$ is the zero ideal so the statement is true because the zero ideal in ${\mathbb{T}[x_1^{\pm 1}, \dots, x_n^{\pm 1}]}$ is prime. Let f, g be two polynomials, $f = \sum f_l$ and $g = \sum g_q$ , such that $f, g\not\in \mathcal{I}_{P}$ but $fg \in \mathcal{I}_{P}$ and so $bend(fg) \subseteq Bend(\mathcal{I}_{P}) \subseteq P$ . If in the formal expression $fg_{\hat \jmath}+f_{\hat \imath}g$ none of the terms $f_ig_j$ and $f_sg_r$ for all pairs $(i,j) \neq (s,r)$ have the same exponent vector, then we can write

(3·1) \begin{equation} (f, f_{\hat \imath})(g, g_{\hat \jmath}) = (fg + f_{\hat \imath}g_{\hat \jmath}, fg_{\hat \jmath}+f_{\hat \imath}g) = (fg, fg_{\hat \jmath}+f_{\hat \imath}g) = (fg, (fg)_{\hat k}),\end{equation}

where i and j are in the support of f and g respectively, $(fg)_k = f_ig_j$ , and $f_{\hat \imath}$ denotes f with the term i omitted. In particular, since $(fg, (fg)_{\hat k}) \in bend(fg) \subseteq P$ and P is prime, either $(f, f_{\hat \imath}) \in P$ or $(g, g_{\hat \jmath})\in P$ .

Now assume that for two pairs $(i,j) \neq (s,r)$ the terms $f_ig_j$ and $f_sg_r$ have the same exponent vector. Without loss of generality let $f_sg_r \geq f_ig_j \;=\!:\;d$ . Then

$$(fg, (fg)_{\hat k}) + (d,d) = (f, f_{\hat \imath})(g, g_{\hat \jmath}).$$

Since both summands on the left hands side are in P then the sum is in P, and so either $(f, f_{\hat \imath}) \in P$ or $(g, g_{\hat \jmath})\in P$ . Note that when $f_sg_r = f_ig_j$ , then $(f, f_{\hat \imath})(g, g_{\hat \jmath}) = (fg, fg)$ which is also in P.

Since this holds for each pair i, j, it follows that either f or g has all their bend relations in P, or equivalently $f \in \mathcal{I}_{P}$ or $g \in \mathcal{I}_{P}$ .

Example 3·13. Consider the ideal $I \subseteq {\mathbb{T}}[x^{\pm 1},y^{\pm 1}]$ that consists of all polynomials that are not homogeneous with respect to x. This is a prime ideal since the product of homogeneous polynomials is homogeneous. Consider $f, g \in I$ , where $f = x+1$ and $g = xy+1$ and so $bend(f) = (x,1)$ and $bend(g) = (xy,1)$ are elements of Bend(I). Since Bend(I) is a congruence, in particular it is transitively closed, we get that $(x, xy) \in Bend(I)$ . However, $x + xy \not\in I$ since it is homogeneous with respect to x.

Proof. The inclusion $Bend(\mathcal{I}_{P}) \subseteq P$ is always true. For the other direction, first note that P is generated by pairs of the form $(m_1 + m_2, m_1)$ , where $m_1$ and $m_2$ are terms, since every element of its quotient can be represented by a term. Since P is not a minimal prime there must be distinct terms a, b such that $a \neq b$ and $(a,b) \in P$ . Now for any $m_1, m_2$ , such that $(m_1 + m_2, m_1) \in P$ , equivalently written $m_1+m_2 \sim_P m_1$ , we have:

(3·2) \begin{equation}am_1 + am_2 \sim_P am_1 + bm_1 \sim_P am_2 + bm_1 \Rightarrow am_1 + am_2 + bm_1 \in \mathcal{I}_{P}\end{equation}

and

(3·3) \begin{equation}am_1 \sim_P bm_1 \Rightarrow am_1 + bm_1 \in \mathcal{I}_{P}.\end{equation}

Taking the bend relations of the polynomials in Equations (3·2) and (3·3) we obtain:

(3·4) \begin{equation}am_1 + am_2 \sim_P am_1 + bm_1 \sim_P am_1 \in Bend(\mathcal{I}_{P}).\end{equation}

Since $a \in {\mathbb{T}[x_1^{\pm 1}, \dots, x_n^{\pm 1}]}$ it is invertible and we can divide the pairs in Equation (3·4) by a and we obtain that $(m_1 + m_2, m_1) \in Bend(\mathcal{I}_{P}).$

Proof. Consider the polynomial $f \in I$ , and write it as $f = a + b + c$ , where a, b are terms and $c = m_1 + \dots + m_n$ and $m_i$ are terms. If $n = 0$ then we are done. Assume $n \gt 0$ and observe that $(a+b+c)(ab + bc + ac) = (a + b)(a+c)(b+c)$ and since I is prime then one of $a + b, a+c$ and $b+c$ is in I. If $a+b \in I$ we are done. Otherwise, without loss of generality let $a+c \in I$ . We can now apply the same argument to $a + c$ , noticing it has one less term than f. Iterating the procedure we reduce to the case $n = 0$ .

Proof. To show that Bend(I) is a prime congruence it is enough to check that the quotient by it is totally ordered in view of Corollary 2·6 (ii). In other words, we need to see that for any two terms $m_1$ and $m_2$ one of the pairs $(m_1 + m_2, m_1)$ and $(m_1 + m_2, m_2)$ is in Bend(I).

Example 3·17. Consider the ideal $I \subseteq {\mathbb{T}}[x^{\pm 1}, y^{\pm 1}]$ given as the intersection of the set of all polynomials not homogeneous in x and the set of all polynomials not homogeneous in y. Observe that I is not a prime ideal because $(x+xy)(y+xy) \in I$ but neither of the factors is an element of I. On the other hand, since $f = x+y$ and $g = x+y^2$ are both elements of I, then $(x, y), (y,1) \in Bend(I)$ . Thus

$${\mathbb{T}}[x^{\pm 1}, y^{\pm 1}]/Bend(I) \cong {\mathbb{T}},$$

and since ${\mathbb{T}}$ is a totally ordered semifield by Corollary 2·6 we conclude that Bend(I) is a prime congruence.

Proof. Indeed, assume for contradiction that $\mathcal{I}_{P}$ is tropical. Let $m_3$ be a term such that in we have $1 \gt_P m_3$ . Note that such $m_3$ exists – for example $m_1^{-1}$ – and it must have a distinct exponent from $m_1$ and $m_2$ . Now the polynomials $m_1 + m_2 + m_3$ and $m_1 + m_2 + m_3^2$ attain their maximum twice at P hence they are in $\mathcal{I}_{P}$ . By the monomial elimination axiom of tropical ideals we can eliminate $m_1$ , but the resulting polynomial $c m_2 + m_3 + m_3^2$ can not attain its maximum twice at P as $m_3 \gt_P m_3^2$ and we either have $c = 0$ or $c m_2 \gt_P m_3$ .

Remark 3·19. A non-minimal prime congruence P whose matrix has more than one row is not finitely generated. Moreover, the variety of P is not determined by finitely many pairs in P as shown in [ Reference Mincheva14 , example 6·3·4]. Note that for any P, by definition $\mathbb{V}(P) = \mathbb{V}(\mathcal{I}_{P})$ . Thus the ideal $\mathcal{I}_{P}$ corresponding to a prime P with rank greater than one is not tropical because it will not have a finite tropical basis.

Proof of Theorem 3·11. For (i) if I is a non-zero closed prime then by Proposition 3·16 Bend(I) is a prime congruence, and since I is closed we have that $I = \mathcal{I}_{Bend(I)}$ . As it was explained in Example 3·10 the zero ideal can be written as $\mathcal{I}_{P}$ for any minimal prime P. This also proves one direction of (ii) and the other follows from Proposition 3·14. For (iii) again by Proposition 3·16 we have that for a non-zero closed prime I the congruence Bend(I) is prime and for a non-minimal prime congruence P the ideal $\mathcal{I}_{P}$ is prime by Proposition 3·12, non-zero by (ii) and closed by Remark 3·9. Moreover, for a closed ideal I we have that $I = \mathcal{I}_{Bend(I)}$ and for a non-minimal prime P by Proposition 3·14 we have that $P = Bend(\mathcal{I}_{P})$ , showing that the two maps are indeed inverses of each other.

For (iv) we have already seen in Example 3·10 that if P is a geometric prime then $\mathcal{I}_{P}$ is tropical. We will show that if P is not geometric, then $\mathcal{I}_{P}$ is not tropical. If the defining matrix U of P consists of multiple rows, then by Remark 3·19 the ideal $\mathcal{I}_{P}$ is not tropical. Remains to consider non-geometric primes whose matrices have a single row. Recall that a single row matrix corresponds to a geometric prime if the (1, 1) entry is non-zero and let us assume that the (1,1) entry of U is zero, or equivalently that P identifies the elements of ${\mathbb{T}}$ with 1. By Proposition 3·15 there is a binomial $m_1 + m_2 \in \mathcal{I}_{P}$ . By multiplying both with a monomial whose exponent vector is a sufficiently large multiple of the first row of U we can assume that $m_1 \sim_P m_2 \gt_P 1$ , and as we assumed that P identifies ${\mathbb{T}}$ with 1 we also have that $c m_1 \gt_P 1$ for all $c \in {\mathbb{R}}$ . Hence we can apply Lemma 3·18, and conclude that $\mathcal{I}_{P}$ is not a tropical ideal.

Proof. For the case of ${\mathbb{T}[x_1^{\pm 1}, \dots, x_n^{\pm 1}]}$ from Proposition 3·16 we know that Bend(I) is prime and so $\mathbb{V}(Bend(I))$ consists of at most a point by Remark 2·11. The theorem follows as $\mathbb{V}(I) = \mathbb{V}(Bend(I))$ which is true by definition of the sets. For ${\mathbb{T}[x_1, \dots, x_n]}$ , if I is a prime ideal containing a monomial, then it also contains at least one of the variables appearing in that monomial hence its tropical vanishing locus is in bijection with that of a tropical prime ideal in less variables. If I does not contain monomials then the statement follows from Proposition 3·3 and the ${\mathbb{T}[x_1^{\pm 1}, \dots, x_n^{\pm 1}]}$ case.

Example 3·21. We give an example of a tropical ideal whose variety is a line, explicitly showing it is not prime. Consider the ideal $J = (x-y) \subseteq k[x,y]$ , where k is a trivially valued field. Let $I = trop(J) \subseteq {\mathbb{B}}[x,y]$ , which is a realisable tropical ideal. Notice that $(x+y+1)(x+y+xy)=(x+y)(x+y+xy+1) \in I$ , but neither $(x+y+1)$ nor $(x+y+xy)$ are in I, hence I is not prime.

Remark 3·22. Not every ${\textrm{closed}}$ ideal of ${\mathbb{T}[x_1^{\pm 1}, \dots, x_n^{\pm 1}]}$ satisfies the strong Nullstellensatz, that is, it is not true that f is in every ${\textrm{closed}}$ prime containing a ${\textrm{closed}}$ ideal I if and only if some power of f is in I.

Proof of Remark 3·22. Let I be the closure of the ideal generated by the polynomials $(1 + x + x^2)y + z, (1 + x + x^2)y + zx, (1 + x + x^2)y + zx^2, (1 + x + x^2)z + y, (1 + x + x^2)z + yx, (1 + x + x^2)z + yx^2$ . From the first three polynomials we conclude that

$$(1 + x + x^2)y \sim (1 + x + x^2)y + (1 + x + x^2)z \in Bend(I).$$

Similarly, from the other three generators we obtain that

$$(1 + x + x^2)z \sim (1 + x + x^2)y + (1 + x + x^2)z \in Bend(I).$$

By transitivity we obtain that the pair $(1 + x + x^2)y \sim (1 + x + x^2)z$ is also in Bend(I) and so $(1 + x + x^2, 0) (y, z)$ in Bend(I). Every prime congruence P over Bend(I) contains $(1 + x + x^2, 0) (y, z)$ hence it contains one of the factors. Assume $(1 + x + x^2,0) \in P$ . Since P is a prime congruence on ${\mathbb{T}[x_1^{\pm 1}, \dots, x_n^{\pm 1}]}$ , in the quotient of ${\mathbb{T}[x_1^{\pm 1}, \dots, x_n^{\pm 1}]}$ by P each polynomial is congruent to its leading term; denote the leading term of $1 + x + x^2$ by m. So we have $1 + x + x^2 \sim_P m \sim_P 0$ . Multiplying by $m^{-1}$ tells us that that $(1_{\mathbb{T}}, 0_{\mathbb{T}}) \in P$ , which makes P not proper and hence not prime. Hence (y, z) is in every prime congruence containing Bend(I). Moreover, $y+z$ is in every ${\textrm{closed}}$ prime containing I. However, no power of $y + z$ is in I.

Remark 3·23. Maclagan and Rincón have announced that tropical ideals satisfy the strong Nullstellensatz; at time of writing, the result is still unpublished.

Remark 3·26. In a series of papers [ Reference Maclagan and Rincón16–Reference Maclagan and Rincón18 ] Maclagan and Rincón prove that the varieties of tropical ideals are finite balanced $\mathbb{R}$ -rational polyhedral complexes. Moreover, in [ Reference Maclagan and Rincón17 , theorem 1·4] they show that the only tropical ideal whose variety is empty is the polynomial semiring ${\mathbb{T}[x_1, \dots, x_n]}$ . Thus every non-trivial tropical ideal has a non-empty variety.

Proof. Recall that if P is a prime congruence with defining matrix U then $\dim {\mathbb{T}[x_1, \dots, x_n]}/P $ is equal to the rank of U, denoted r(U). We first observe that if C is a congruence on ${\mathbb{T}[x_1, \dots, x_n]}$ then

$$\dim {\mathbb{T}[x_1, \dots, x_n]}/C = \dim {\mathbb{T}[x_1, \dots, x_n]}/P,$$

where P is a prime congruence over C of maximal rank, in particular, P is a minimal prime over C. Thus, if P is a prime of maximal rank over Bend(I) it suffices to show that P has rank at least $d+1$ .

Proof. We will show that every prime congruence over Bend(I) has rank at most $d+1$ . We first see this in the case when every maximal rank prime P has a geometric prime over it. Let W be the defining matrix of P. Denote the rows of W by $w_1, \dots , w_{r(W)}$ . Note that they are linearly independent by definition of the defining matrix of a prime. Consider the vectors $w_1^{\prime} = w_1,\ w_2^{\prime} = w_1 + \epsilon_1^{\prime}w_2,\ \dots ,\ w_{r(W)}^{\prime} = w_1 + \dots + \epsilon_{r(W)}^{\prime}w_{r(W)}$ which are also linearly independent. We can scale each of the vectors $w_1', \dots , w_{r(W)}'$ so that the first entry is 1. Now consider the re-scaled vectors without the first entry and call them $w_1'', \dots, w_{r(W)}''$ . Note that the vectors $w_1'', \dots, w_{r(W)}''$ are affine independent and lie on the same face of X. Since we know that the dimension of X is at most d, we know that $r(W) \leq d+1$ . Remains to investigate the case when a maximal rank prime P containing Bend(I) has no geometric prime over it. Denote by U the defining matrix of P. If there is no geometric prime over P then the first entry of the first row of U is zero. If the first entry of any other row is not zero, we can add a suitable multiple of this row to the first one. This way we obtain the matrix of a different prime of the same rank, which still contains Bend(I) but has a geometric prime over it. In this case we are done by the previous discussion. Thus we can assume that the first column of U is the zero vector. Consider the prime $P^{\prime}$ with matrix $U^{\prime}$ , where

$$U' = \begin{pmatrix} 1\;\;\;\; & \boldsymbol{0} \\ \boldsymbol{0}\;\;\;\; & U\end{pmatrix}.$$

Then there are two cases. First, if $I \subseteq {\mathbb{B}}[x_1, \dots, x_n]$ , then the prime P ^′ lies above Bend(I) and clearly $r(U')\gt r(U)$ . So P is not a maximal rank prime over Bend(I). If $I \not\subseteq {\mathbb{B}}[x_1, \dots, x_n]$ , consider the prime P ^′ as before. Since the (1, 1) entry of U ^′ is 1, there exists a geometric prime over P ^′. We can infer from [ Reference Maclagan and Rincón18 , proposition∼2·12], that $\dim \mathbb{V}(I) = \dim \mathbb{V}(\phi(I)) = d+1$ where the map $\phi\;:\; {\mathbb{T}[x_1, \dots, x_n]} \rightarrow {\mathbb{B}}[x_1,\dots,x_n]$ . Notice that P ^′ is a prime over $Bend(\phi(I))$ . Then by the earlier argument $\dim {\mathbb{T}[x_1, \dots, x_n]}/P'$ is at most $d + 1$ . Now since $r(U')= r(U) + 1 \gt r(U)$ , then $r(U) \lt d+1$ and hence P is not a maximal rank prime. So we conclude that if P is a maximal rank prime with matrix U, then $r(U) \leq d+1$ .

Proof of Theorem 3·24. The theorem follows immediately from Proposition 3·27, Remark 3·26, and Proposition 3·28.

Remark 3·29. Recently, in [ Reference Maclagan and Rincón18 ] the authors show that if I is a non-necessarily realisable tropical ideal the dimension of V(I) is equal to the degree of the Hilbert polynomial of I. Their result agrees with our computation of the Krull dimension of ${\mathbb{T}}[x_1, \dots, x_n]/Bend(I)$ .